12.2 More Identities

In Section 12.1, we saw the utility of identities in finding the values of the circular functions of a given angle as well as simplifying expressions involving the circular functions. In this section, we introduce several collections of identities which have uses in this course and beyond.

Our first set of identities is the `Even / Odd' identities. We observed the even and odd properties of the circular functions graphically in Sections 11.3 and 11.5. Here, we take the time to prove these properties from first principles. We state the theorem below for reference.

Theorem

Even / Odd Identities: For all applicable angles $θ$ ,

$\cos (- θ) = \cos (θ)$
$\sec (- θ) = \sec (θ)$
$\sin (- θ) = - \sin (θ)$
$\csc (- θ) = - \csc (θ)$
$\tan (- θ) = - \tan (θ)$
$\cot (- θ) = - \cot (θ)$

We start by proving $\cos (- θ) = \cos (θ)$ and $\sin (- θ) = - \sin (θ)$ .

Consider an angle $θ$ plotted in standard position. Let $θ sub 0$ be the angle coterminal with $θ$ with $0 less than or equal to θ sub 0 less than 2 ⁢ π$ . (We can construct the angle $θ sub 0$ by rotating counter-clockwise from the positive $x$ -axis to the terminal side of $θ$ as pictured below.) Since $θ$ and $θ sub 0$ are coterminal, $cos ⁡ open parenthesis θ close parenthesis equals cos ⁡ open parenthesis θ sub 0 close parenthesis$ and $sin ⁡ open parenthesis θ close parenthesis equals sin ⁡ open parenthesis θ sub 0 close parenthesis$ .

Coordinate-plane figure. — **Figure 12.1**

We now consider the angles $- θ$ and $minus θ sub 0$ . Since $θ$ is coterminal with $θ sub 0$ , there is some integer $k$ so that $θ = θ_{0} + 2 π \cdot k$ . Hence, $- θ = - θ_{0} - 2 π \cdot k = - θ_{0} + 2 π \cdot (- k)$ . Since $k$ is an integer, so is $(- k)$ , which means $- θ$ is coterminal with $minus θ sub 0$ . Therefore, $\cos (- θ) = \cos (- θ_{0})$ and $\sin (- θ) = \sin (- θ_{0})$ .

Let $P$ and $Q$ denote the points on the terminal sides of $θ sub 0$ and $minus θ sub 0$ , respectively, which lie on the Unit Circle. By definition, the coordinates of $P$ are $open parenthesis cos ⁡ open parenthesis θ sub 0 close parenthesis, sin ⁡ open parenthesis θ sub 0 close parenthesis close parenthesis$ and the coordinates of $Q$ are $(\cos (- θ_{0}), \sin (- θ_{0}))$ .

Since $θ sub 0$ and $minus θ sub 0$ sweep out congruent central sectors of the Unit Circle, it follows that the points $P$ and $Q$ are symmetric about the $x$ -axis. Thus, $\cos (- θ_{0}) = \cos (θ_{0})$ and $\sin (- θ_{0}) = - \sin (θ_{0})$ .

Since the cosines and sines of $θ sub 0$ and $minus θ sub 0$ are the same as those for $θ$ and $- θ$ , respectively, we get $\cos (- θ) = \cos (θ)$ and $\sin (- θ) = - \sin (θ)$ , as required.

As we saw in Section 11.5, the remaining four circular functions `inherit' their even/odd nature from sine and cosine courtesy of the Reciprocal and Quotient Identities, Theorem 12.1.

Our next set of identities establish how the cosine function handles sums and differences of angles.

Theorem

Sum and Difference Identities for Cosine: For all angles $α$ and $β$ ,

$\cos (α + β) = \cos (α) \cos (β) - \sin (α) \sin (β)$
$\cos (α - β) = \cos (α) \cos (β) + \sin (α) \sin (β)$

We first prove the result for differences. As in the proof of the Even / Odd Identities, we can reduce the proof for general angles $α$ and $β$ to angles $α sub 0$ and $β sub 0$ , coterminal with $α$ and $β$ , respectively, each of which measure between $0$ and $2 ⁢ π$ radians. Since $α$ and $α sub 0$ are coterminal, as are $β$ and $β sub 0$ , it follows that $(α - β)$ is coterminal with $open parenthesis α sub 0 minus β sub 0 close parenthesis$ . Consider the case below where $α sub 0 greater than or equal to β sub 0$ .

Since the angles $P O Q$ and $A O B$ are congruent, the distance between $P$ and $Q$ is equal to the distance between $A$ and $B$ .¹ The distance formula, Equation A.1, yields

$\begin{matrix} \sqrt{{(\cos (α_{0}) - \cos (β_{0}))}^{2} + {(\sin (α_{0}) - \sin (β_{0}))}^{2}} & = & \sqrt{{(\cos (α_{0} - β_{0}) - 1)}^{2} + {(\sin (α_{0} - β_{0}) - 0)}^{2}} \end{matrix}$

Squaring both sides, we expand the left hand side of this equation as

$\begin{matrix} {(\cos (α_{0}) - \cos (β_{0}))}^{2} + {(\sin (α_{0}) - \sin (β_{0}))}^{2} & = & \cos^{2} (α_{0}) - 2 \cos (α_{0}) \cos (β_{0}) + \cos^{2} (β_{0}) \\ + \sin^{2} (α_{0}) - 2 \sin (α_{0}) \sin (β_{0}) + \sin^{2} (β_{0}) \\ = & \cos^{2} (α_{0}) + \sin^{2} (α_{0}) + \cos^{2} (β_{0}) + \sin^{2} (β_{0}) \\ - 2 \cos (α_{0}) \cos (β_{0}) - 2 \sin (α_{0}) \sin (β_{0}) \end{matrix}$

From the Pythagorean Identities, $\cos^{2} (α_{0}) + \sin^{2} (α_{0}) = 1$ and $\cos^{2} (β_{0}) + \sin^{2} (β_{0}) = 1$ , so

$\begin{matrix} {(\cos (α_{0}) - \cos (β_{0}))}^{2} + {(\sin (α_{0}) - \sin (β_{0}))}^{2} & = & 2 - 2 \cos (α_{0}) \cos (β_{0}) - 2 \sin (α_{0}) \sin (β_{0}) \end{matrix}$

Turning our attention to the right hand side of our equation, we find

$\begin{matrix} {(\cos (α_{0} - β_{0}) - 1)}^{2} + {(\sin (α_{0} - β_{0}) - 0)}^{2} & = & \cos^{2} (α_{0} - β_{0}) - 2 \cos (α_{0} - β_{0}) + 1 + \sin^{2} (α_{0} - β_{0}) \\ = & 1 + \cos^{2} (α_{0} - β_{0}) + \sin^{2} (α_{0} - β_{0}) - 2 \cos (α_{0} - β_{0}) \end{matrix}$

Once again, we simplify $\cos^{2} (α_{0} - β_{0}) + \sin^{2} (α_{0} - β_{0}) = 1$ , so that

$\begin{matrix} {(\cos (α_{0} - β_{0}) - 1)}^{2} + {(\sin (α_{0} - β_{0}) - 0)}^{2} & = & 2 - 2 \cos (α_{0} - β_{0}) \end{matrix}$

Putting it all together, we get $2 - 2 \cos (α_{0}) \cos (β_{0}) - 2 \sin (α_{0}) \sin (β_{0}) = 2 - 2 \cos (α_{0} - β_{0})$ , which simplifies to: $\cos (α_{0} - β_{0}) = \cos (α_{0}) \cos (β_{0}) + \sin (α_{0}) \sin (β_{0})$ .

Since $α$ and $α sub 0$ , $β$ and $β sub 0$ , and $(α - β)$ and $open parenthesis α sub 0 minus β sub 0 close parenthesis$ are all coterminal pairs of angles, we have established the identity: $\cos (α - β) = \cos (α) \cos (β) + \sin (α) \sin (β)$ .

For the case where $α sub 0 less than or equal to β sub 0$ , we can apply the above argument to the angle $β sub 0 minus α sub 0$ to obtain the identity $\cos (β_{0} - α_{0}) = \cos (β_{0}) \cos (α_{0}) + \sin (β_{0}) \sin (α_{0})$ . Using this formula in conjunction with the Even Identity of cosine gives us the result in this case, too:

$\begin{matrix} \cos (α_{0} - β_{0}) = \cos (- (α_{0} - β_{0})) = \cos (β_{0} - α_{0}) & = & \cos (β_{0}) \cos (α_{0}) + \sin (β_{0}) \sin (α_{0}) \\ = & \cos (α_{0}) \cos (β_{0}) + \sin (α_{0}) \sin (β_{0}) . \end{matrix}$

To get the sum identity for cosine, we use the difference formula along with the Even/Odd Identities

$\cos (α + β) = \cos (α - (- β)) = \cos (α) \cos (- β) + \sin (α) \sin (- β) = \cos (α) \cos (β) - \sin (α) \sin (β) .$

We put these newfound identities to good use in the following example.

Example 1

Find the exact value of $cos ⁡ open parenthesis 15 degrees close parenthesis$ .
Verify the identity: $\cos (\frac{π}{2} - θ) = \sin (θ)$ .
Suppose $α$ is a Quadrant I angle with $\sin (α) = \frac{3}{5}$ and $β$ is a Quadrant IV angle with $sec ⁡ open parenthesis β close parenthesis equals 4$ . Find the exact value of $\cos (α + β)$ .

In order to use Theorem 12.5 to find $cos ⁡ open parenthesis 15 degrees close parenthesis$ , we need to write $15 degrees$ as a sum or difference of angles whose cosines and sines we know. One way to do so is to write $15^{\circ} = 45^{\circ} - 30^{\circ}$ . We find:
$\begin{matrix} \cos (15^{\circ}) & = & \cos (45^{\circ} - 30^{\circ}) \\ = & \cos (45^{\circ}) \cos (30^{\circ}) + \sin (45^{\circ}) \sin (30^{\circ}) \\ = & (\frac{\sqrt{2}}{2}) (\frac{\sqrt{3}}{2}) + (\frac{\sqrt{2}}{2}) (\frac{1}{2}) \\ = & \frac{\sqrt{6} + \sqrt{2}}{4} . \end{matrix}$
Using Theorem 12.5 gives:
$\begin{matrix} \cos (\frac{π}{2} - θ) & = & \cos (\frac{π}{2}) \cos (θ) + \sin (\frac{π}{2}) \sin (θ) \\ = & (0) (\cos (θ)) + (1) (\sin (θ)) \\ = & \sin (θ) . \end{matrix}$
Per Theorem 12.5, we know $\cos (α + β) = \cos (α) \cos (β) - \sin (α) \sin (β)$ . Hence, we need to find the sines and cosines of $α$ and $β$ to complete the problem.
We are given $\sin (α) = \frac{3}{5}$ , so our first task is to find $\cos (α)$ . We can quickly get $\cos (α)$ using the Pythagorean Identity $\cos^{2} (α) = 1 - \sin^{2} (α) = 1 - {(\frac{3}{5})}^{2} = \frac{16}{25}$ . We get $\cos (α) = \frac{4}{5}$ , choosing the positive root since $α$ is a Quadrant I angle.
Next, we need the $\sin (β)$ and $\cos (β)$ . Since $sec ⁡ open parenthesis β close parenthesis equals 4$ , we immediately get $\cos (β) = \frac{1}{4}$ courtesy of the Reciprocal and Quotient Identities.
To get $\sin (β)$ , we employ the Pythagorean Identity: $\sin^{2} (β) = 1 - \cos^{2} (β) = 1 - {(\frac{1}{4})}^{2} = \frac{15}{16}$ . Here, since $β$ is a Quadrant IV angle, we get $\sin (β) = - \frac{\sqrt{15}}{4}$ .
Finally, we get: $\cos (α + β) = \cos (α) \cos (β) - \sin (α) \sin (β) = (\frac{4}{5}) (\frac{1}{4}) - (\frac{3}{5}) (- \frac{\sqrt{15}}{4}) = \frac{4 + 3 \sqrt{15}}{20}$ .

The identity verified in Example 12.2.1, namely, $\cos (\frac{π}{2} - θ) = \sin (θ)$ , is the first of the celebrated `cofunction' identities. These identities were first hinted at in Exercise in Section B.2.

From $\sin (θ) = \cos (\frac{π}{2} - θ)$ , we get: $\sin (\frac{π}{2} - θ) = \cos (\frac{π}{2} - [\frac{π}{2} - θ]) = \cos (θ)$ , which says, in words, that the `co'sine of an angle is the sine of its `co'mplement. Now that these identities have been established for cosine and sine, the remaining circular functions follow suit. The remaining proofs are left as exercises.

Theorem

Cofunction Identities: For all applicable angles $θ$ ,

$\cos (\frac{π}{2} - θ) = \sin (θ)$
$\sin (\frac{π}{2} - θ) = \cos (θ)$
$\sec (\frac{π}{2} - θ) = \csc (θ)$
$\csc (\frac{π}{2} - θ) = \sec (θ)$
$\tan (\frac{π}{2} - θ) = \cot (θ)$
$\cot (\frac{π}{2} - θ) = \tan (θ)$

The Cofunction Identities enable us to derive the sum and difference formulas for sine. We first convert to sine to cosine and expand:

$\begin{matrix} \sin (α + β) & = & \cos (\frac{π}{2} - (α + β)) \\ = & \cos ([\frac{π}{2} - α] - β) \\ = & \cos (\frac{π}{2} - α) \cos (β) + \sin (\frac{π}{2} - α) \sin (β) \\ = & \sin (α) \cos (β) + \cos (α) \sin (β) \end{matrix}$

We can derive the difference formula for sine by rewriting $\sin (α - β)$ as $\sin (α + (- β))$ and using the sum formula and the Even / Odd Identities. Again, we leave the details to the reader.

Theorem

Sum and Difference Identities for Sine: For all angles $α$ and $β$ ,

$\sin (α + β) = \sin (α) \cos (β) + \cos (α) \sin (β)$
$\sin (α - β) = \sin (α) \cos (β) - \cos (α) \sin (β)$

We try out these new identities in the next example.

Example 2

Find the exact value of $\sin (\frac{19 π}{12})$
Suppose $α$ is a Quadrant II angle with $\sin (α) = \frac{5}{13}$ , and $β$ is a Quadrant III angle with $tan ⁡ open parenthesis β close parenthesis equals 2$ . Find the exact value of $\sin (α - β)$ .
Derive a formula for $\tan (α + β)$ in terms of $\tan (α)$ and $\tan (β)$ .

As in Example 12.2.1, we need to write the angle $\frac{19 π}{12}$ as a sum or difference of common angles. The denominator of $12$ suggests a combination of angles with denominators $3$ and $4$ . One such combination² is $\frac{19 π}{12} = \frac{4 π}{3} + \frac{π}{4}$ . Applying Theorem 12.7, we get
$\begin{matrix} \sin (\frac{19 π}{12}) & = & \sin (\frac{4 π}{3} + \frac{π}{4}) \\ = & \sin (\frac{4 π}{3}) \cos (\frac{π}{4}) + \cos (\frac{4 π}{3}) \sin (\frac{π}{4}) \\ = & (- \frac{\sqrt{3}}{2}) (\frac{\sqrt{2}}{2}) + (- \frac{1}{2}) (\frac{\sqrt{2}}{2}) \\ = & \frac{- \sqrt{6} - \sqrt{2}}{4} \end{matrix}$
In order to find $\sin (α - β)$ using Theorem 12.7, we need to find $\cos (α)$ and both $\cos (β)$ and $\sin (β)$ .
To find $\cos (α)$ , we use the Pythagorean Identity $\cos^{2} (α) = 1 - \sin^{2} (α) = 1 - {(\frac{5}{13})}^{2} = \frac{144}{169}$ . We get $\cos (α) = - \frac{12}{13}$ , the negative, here, owing to the fact that $α$ is a Quadrant II angle.
We now set about finding $\sin (β)$ and $\cos (β)$ . We have several ways to proceed at this point, but since there isn't a direct way to get from $tan ⁡ open parenthesis β close parenthesis equals 2$ to either $\sin (β)$ or $\cos (β)$ , we opt for a more geometric approach as presented in Section 11.4. Since $β$ is a Quadrant III angle with $\tan (β) = 2 = \frac{- 2}{- 1}$ , we know the point $Q (x, y) = (- 1, - 2)$ is on the terminal side of $β$ as illustrated below.³
Figure 12.5 the terminal side of $β$ contains $Q (- 1, - 2)$
We find $r = \sqrt{x^{2} + y^{2}} = \sqrt{{(- 1)}^{2} + {(- 2)}^{2}} = \sqrt{5}$ , so per Theorem 11.9, $\sin (β) = \frac{- 2}{\sqrt{5}} = - \frac{2 \sqrt{5}}{5}$ and $\cos (β) = \frac{- 1}{\sqrt{5}} = - \frac{\sqrt{5}}{5}$ .
At last, we have $\sin (α - β) = \sin (α) \cos (β) - \cos (α) \sin (β) = (\frac{5}{13}) (- \frac{\sqrt{5}}{5}) - (- \frac{12}{13}) (- \frac{2 \sqrt{5}}{5}) = - \frac{29 \sqrt{5}}{65}$ .
We can start expanding $\tan (α + β)$ using a quotient identity and our sum formulas
$\begin{matrix} \tan (α + β) & = & \frac{\sin (α + β)}{\cos (α + β)} \\ = & \frac{\sin (α) \cos (β) + \cos (α) \sin (β)}{\cos (α) \cos (β) - \sin (α) \sin (β)} \end{matrix}$
Since $\tan (α) = \frac{\sin (α)}{\cos (α)}$ and $\tan (β) = \frac{\sin (β)}{\cos (β)}$ , it looks as though if we divide both numerator and denominator by $\cos (α) \cos (β)$ we will have what we want
$\begin{matrix} \tan (α + β) & = & \frac{\sin (α) \cos (β) + \cos (α) \sin (β)}{\cos (α) \cos (β) - \sin (α) \sin (β)} \cdot \frac{\frac{1}{\cos (α) \cos (β)}}{\frac{1}{\cos (α) \cos (β)}} \\ = & \frac{\frac{\sin (α) \cos (β)}{\cos (α) \cos (β)} + \frac{\cos (α) \sin (β)}{\cos (α) \cos (β)}}{\frac{\cos (α) \cos (β)}{\cos (α) \cos (β)} - \frac{\sin (α) \sin (β)}{\cos (α) \cos (β)}} \\ = & \frac{\frac{\sin (α) \cos (β)}{\cos (α) \cos (β)} + \frac{\cos (α) \sin (β)}{\cos (α) \cos (β)}}{\frac{\cos (α) \cos (β)}{\cos (α) \cos (β)} - \frac{\sin (α) \sin (β)}{\cos (α) \cos (β)}} \\ = & \frac{\tan (α) + \tan (β)}{1 - \tan (α) \tan (β)} \end{matrix}$
Naturally, this formula is limited to those cases where all of the tangents are defined.

The formula developed in Exercise 12.2.2 for $\tan (α + β)$ can be used to find a formula for $\tan (α - β)$ by rewriting the difference as a sum, $\tan (α + (- β))$ and using the odd property of tangent. (The reader is encouraged to fill in the details.) Below we summarize all of the sum and difference formulas.

Theorem

Sum and Difference Identities: For all applicable angles $α$ and $β$ ,

$\cos (α \pm β) = \cos (α) \cos (β) \mp \sin (α) \sin (β)$
$\sin (α \pm β) = \sin (α) \cos (β) \pm \cos (α) \sin (β)$
$\tan (α \pm β) = \frac{\tan (α) \pm \tan (β)}{1 \mp \tan (α) \tan (β)}$

In the statement of Theorem 12.8, we have combined the cases for the sum ` $+$ ' and difference `' of angles into one formula. The convention here is that if you want the formula for the sum ` $+$ ' of two angles, you use the top sign in the formula; for the difference, `', use the bottom sign. For example,

$\tan (α - β) = \frac{\tan (α) - \tan (β)}{1 + \tan (α) \tan (β)}$

If we set $α = β$ in the sum formulas in Theorem 12.8, we obtain the following `Double Angle' Identities:

Theorem

Double Angle Identities: For all applicable angles $θ$ ,

$\cos (2 θ) = {\begin{matrix} \cos^{2} (θ) - \sin^{2} (θ) \\ 2 \cos^{2} (θ) - 1 \\ 1 - 2 \sin^{2} (θ) \end{matrix}$
$sin ⁡ open parenthesis 2 ⁢ θ close parenthesis equals 2 ⁢ sin ⁡ open parenthesis θ close parenthesis ⁢ cos ⁡ open parenthesis θ close parenthesis$
$\tan (2 θ) = \frac{2 \tan (θ)}{1 - \tan^{2} (θ)}$

The three different forms for $cos ⁡ open parenthesis 2 ⁢ θ close parenthesis$ can be explained by our ability to `exchange' squares of cosine and sine via the Pythagorean Identity. For instance, if we substitute $\sin^{2} (θ) = 1 - \cos^{2} (θ)$ into the first formula for $cos ⁡ open parenthesis 2 ⁢ θ close parenthesis$ , we get $\cos (2 θ) = \cos^{2} (θ) - \sin^{2} (θ) = \cos^{2} (θ) - (1 - \cos^{2} (θ)) = 2 \cos^{2} (θ) - 1$ .

It is interesting to note that to determine the value of $cos ⁡ open parenthesis 2 ⁢ θ close parenthesis$ , only one piece of information is required: either $\cos (θ)$ or $\sin (θ)$ . To determine $sin ⁡ open parenthesis 2 ⁢ θ close parenthesis$ , however, it appears that we must know both $\sin (θ)$ and $\cos (θ)$ . In the next example, we show how we can find $sin ⁡ open parenthesis 2 ⁢ θ close parenthesis$ knowing just one piece of information, namely $\tan (θ)$ .

Example 3

Suppose $P ⁢ open parenthesis minus 3, 4 close parenthesis$ lies on the terminal side of $θ$ when $θ$ is plotted in standard position.
Find $cos ⁡ open parenthesis 2 ⁢ θ close parenthesis$ and $sin ⁡ open parenthesis 2 ⁢ θ close parenthesis$ and determine the quadrant in which the terminal side of the angle $2 ⁢ θ$ lies when it is plotted in standard position.
If $\sin (θ) = x$ for $- \frac{π}{2} \leq θ \leq \frac{π}{2}$ , find an expression for $sin ⁡ open parenthesis 2 ⁢ θ close parenthesis$ in terms of $x$ .
Verify the identity: $\sin (2 θ) = \frac{2 \tan (θ)}{1 + \tan^{2} (θ)}$ .
Express $cos ⁡ open parenthesis 3 ⁢ θ close parenthesis$ as a polynomial in terms of $\cos (θ)$ .

We sketch the terminal side of $θ$ below on the left. Using Theorem 11.3 from Section 11.2 with $x = - 3$ and $y equals 4$ , we find $r = \sqrt{x^{2} + y^{2}} = 5$ . Hence, $\cos (θ) = - \frac{3}{5}$ and $\sin (θ) = \frac{4}{5}$ .
Theorem 12.9 gives us three different formulas to choose from to find $cos ⁡ open parenthesis 2 ⁢ θ close parenthesis$ . Using the first formula,⁴ we get: $\cos (2 θ) = \cos^{2} (θ) - \sin^{2} (θ) = {(- \frac{3}{5})}^{2} - {(\frac{4}{5})}^{2} = - \frac{7}{25}$ . For $sin ⁡ open parenthesis 2 ⁢ θ close parenthesis$ , we get $\sin (2 θ) = 2 \sin (θ) \cos (θ) = 2 (\frac{4}{5}) (- \frac{3}{5}) = - \frac{24}{25}$ .
Since both cosine and sine of $2 ⁢ θ$ are negative, the terminal side of $2 ⁢ θ$ , when plotted in standard position, lies in Quadrant III. To see this more clearly, we plot the terminal side of $2 ⁢ θ$ , along with the terminal side of $θ$ below on the right.
Note that in order to find the point $Q (x, y)$ on the terminal side of $2 ⁢ θ$ of a circle of radius $5$ , we use Theorem 11.3 again and find $x = r \cos (2 θ) = 5 (- \frac{7}{25}) = - \frac{7}{5}$ and $y = r \sin (2 θ) = 5 (- \frac{24}{25}) = - \frac{24}{5}$ .
Figure 12.6 $P ⁢ open parenthesis minus 3, 4 close parenthesis$ is on the terminal side of $θ$ .
Figure 12.7 $Q (- \frac{7}{5}, - \frac{24}{5})$ is on the terminal side of $2 ⁢ θ$ .
If your first reaction to ` $\sin (θ) = x$ ' is `No it's not, $\cos (θ) = x$ !' then you have indeed learned something, and we take comfort in that.
While we have mostly used ` $x$ ' to represent the $x$ -coordinate of the point the terminal side of an angle $θ$ , here, ` $x$ ' represents the quantity $\sin (θ)$ and our task is to express $sin ⁡ open parenthesis 2 ⁢ θ close parenthesis$ in terms of $x$ .
Since $\sin (2 θ) = 2 \sin (θ) \cos (θ) = 2 x \cos (θ)$ , what remains is to express $\cos (θ)$ in terms of $x$ .
Substituting $\sin (θ) = x$ into the Pythagorean Identity, we get $\cos^{2} (θ) = 1 - \sin^{2} (θ) = 1 - x^{2}$ , or $\cos (θ) = \pm \sqrt{1 - x^{2}}$ . Since $- \frac{π}{2} \leq θ \leq \frac{π}{2}$ , $cos ⁡ open parenthesis θ close parenthesis greater than or equal to 0$ , and thus $\cos (θ) = \sqrt{1 - x^{2}}$ .
Our final answer is $\sin (2 θ) = 2 \sin (θ) \cos (θ) = 2 x \sqrt{1 - x^{2}}$ .
We start with the right hand side of the identity and note that $1 + \tan^{2} (θ) = \sec^{2} (θ)$ . Next, we use the Reciprocal and Quotient Identities to rewrite $\tan (θ)$ and $\sec (θ)$ in terms of $\sin (θ)$ and $\cos (θ)$ :
$\begin{matrix} \frac{2 \tan (θ)}{1 + \tan^{2} (θ)} & = & \frac{2 \tan (θ)}{\sec^{2} (θ)} = \frac{2 (\frac{\sin (θ)}{\cos (θ)})}{\frac{1}{\cos^{2} (θ)}} = 2 (\frac{\sin (θ)}{\cos (θ)}) \cos^{2} (θ) \\ = & 2 (\frac{\sin (θ)}{\cos (θ)}) \cos (θ) \cos (θ) = 2 \sin (θ) \cos (θ) = \sin (2 θ) . \end{matrix}$
In Theorem 12.9, one of the formulas for $cos ⁡ open parenthesis 2 ⁢ θ close parenthesis$ , namely $\cos (2 θ) = 2 \cos^{2} (θ) - 1$ , expresses $cos ⁡ open parenthesis 2 ⁢ θ close parenthesis$ as a polynomial in terms of $\cos (θ)$ . We are now asked to find such an identity for $cos ⁡ open parenthesis 3 ⁢ θ close parenthesis$ .
Using the sum formula for cosine, we begin with
$\begin{matrix} \cos (3 θ) & = & \cos (2 θ + θ) \\ = & \cos (2 θ) \cos (θ) - \sin (2 θ) \sin (θ) . \end{matrix}$
Our ultimate goal is to express the right hand side in terms of $\cos (θ)$ only. To that end, we substitute $\cos (2 θ) = 2 \cos^{2} (θ) - 1$ and $sin ⁡ open parenthesis 2 ⁢ θ close parenthesis equals 2 ⁢ sin ⁡ open parenthesis θ close parenthesis ⁢ cos ⁡ open parenthesis θ close parenthesis$ which yields:
$\begin{matrix} \cos (3 θ) & = & \cos (2 θ) \cos (θ) - \sin (2 θ) \sin (θ) \\ = & (2 \cos^{2} (θ) - 1) \cos (θ) - (2 \sin (θ) \cos (θ)) \sin (θ) \\ = & 2 \cos^{3} (θ) - \cos (θ) - 2 \sin^{2} (θ) \cos (θ) \end{matrix}$
Finally, we exchange $\sin^{2} (θ) = 1 - \cos^{2} (θ)$ courtesy of the Pythagorean Identity, and get
$\begin{matrix} \cos (3 θ) & = & 2 \cos^{3} (θ) - \cos (θ) - 2 \sin^{2} (θ) \cos (θ) \\ = & 2 \cos^{3} (θ) - \cos (θ) - 2 (1 - \cos^{2} (θ)) \cos (θ) \\ = & 2 \cos^{3} (θ) - \cos (θ) - 2 \cos (θ) + 2 \cos^{3} (θ) \\ = & 4 \cos^{3} (θ) - 3 \cos (θ) . \end{matrix}$
Hence, $\cos (3 θ) = 4 \cos^{3} (θ) - 3 \cos (θ)$ .

In the last problem in Example 12.2.3, we saw how we could rewrite $cos ⁡ open parenthesis 3 ⁢ θ close parenthesis$ as sums of powers of $\cos (θ)$ . In Calculus, we have occasion to do the reverse; that is, reduce the power of cosine and sine.

Solving the identity $\cos (2 θ) = 2 \cos^{2} (θ) - 1$ for $\cos^{2} (θ)$ and the identity $\cos (2 θ) = 1 - 2 \sin^{2} (θ)$ for $\sin^{2} (θ)$ results in the aptly-named `Power Reduction' formulas below.

Theorem

Power Reduction Formulas: For all angles $θ$ ,

$\cos^{2} (θ) = \frac{1 + \cos (2 θ)}{2}$
$\sin^{2} (θ) = \frac{1 - \cos (2 θ)}{2}$

Our next example is a typical application of Theorem 12.10 that you'll likely see in Calculus.

Example 4

Rewrite $\sin^{2} (θ) \cos^{2} (θ)$ as a sum and difference of cosines to the first power.

We begin with a straightforward application of Theorem 12.10

$\begin{matrix} \sin^{2} (θ) \cos^{2} (θ) & = & (\frac{1 - \cos (2 θ)}{2}) (\frac{1 + \cos (2 θ)}{2}) \\ = & \frac{1}{4} (1 - \cos^{2} (2 θ)) \\ = & \frac{1}{4} - \frac{1}{4} \cos^{2} (2 θ) \end{matrix}$

Next, we apply the power reduction formula to $cos squared ⁡ open parenthesis 2 ⁢ θ close parenthesis$ to finish the reduction

$\begin{matrix} \sin^{2} (θ) \cos^{2} (θ) & = & \frac{1}{4} - \frac{1}{4} \cos^{2} (2 θ) \\ = & \frac{1}{4} - \frac{1}{4} (\frac{1 + \cos (2 (2 θ))}{2}) \\ = & \frac{1}{4} - \frac{1}{8} - \frac{1}{8} \cos (4 θ) \\ = & \frac{1}{8} - \frac{1}{8} \cos (4 θ) \end{matrix}$

Another application of the Power Reduction Formulas is the Half Angle Formulas. To start, we apply the Power Reduction Formula to $\cos^{2} (\frac{θ}{2})$

$\cos^{2} (\frac{θ}{2}) = \frac{1 + \cos (2 (\frac{θ}{2}))}{2} = \frac{1 + \cos (θ)}{2} .$

We can obtain a formula for $\cos (\frac{θ}{2})$ by extracting square roots. In a similar fashion, we may obtain a half angle formula for sine, and by using a quotient formula, obtain a half angle formula for tangent.

We summarize these formulas below.

Theorem

Half Angle Formulas: For all applicable angles $θ$ ,

$\cos (\frac{θ}{2}) = \pm \sqrt{\frac{1 + \cos (θ)}{2}}$
$\sin (\frac{θ}{2}) = \pm \sqrt{\frac{1 - \cos (θ)}{2}}$
$\tan (\frac{θ}{2}) = \pm \sqrt{\frac{1 - \cos (θ)}{1 + \cos (θ)}}$

where the choice of $\pm$ depends on the quadrant in which the terminal side of $\frac{θ}{2}$ lies.

Example 5

Use a half angle formula to find the exact value of $cos ⁡ open parenthesis 15 degrees close parenthesis$ .
Suppose $minus π less than or equal to t less than or equal to 0$ with $\cos (t) = - \frac{3}{5}$ . Find $\sin (\frac{t}{2})$ .
Use the identity given in number of Example 12.2.3 to derive the identity
$\tan (\frac{θ}{2}) = \frac{\sin (θ)}{1 + \cos (θ)}$

To use the half angle formula, we note that $15^{\circ} = \frac{30^{\circ}}{2}$ and since $15 degrees$ is a Quadrant I angle, its cosine is positive. Thus we have
$\begin{matrix} \cos (15^{\circ}) & = & + \sqrt{\frac{1 + \cos (30^{\circ})}{2}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} \\ = & \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2} \cdot \frac{2}{2}} = \sqrt{\frac{2 + \sqrt{3}}{4}} = \frac{\sqrt{2 + \sqrt{3}}}{2} \end{matrix}$
Back in Example 12.2.1, we found $\cos (15^{\circ}) = \frac{\sqrt{6} + \sqrt{2}}{4}$ by using the difference formula for cosine. The reader is encouraged to prove that these two expressions are equal algebraically.
If $minus π less than or equal to t less than or equal to 0$ , then $- \frac{π}{2} \leq \frac{t}{2} \leq 0$ , which means $\frac{t}{2}$ corresponds to a Quadrant IV angle. Hence, $\sin (\frac{t}{2}) < 0$ , so we choose the negative root formula from Theorem 12.11:
$\begin{matrix} \sin (\frac{t}{2}) & = & - \sqrt{\frac{1 - \cos (t)}{2}} = - \sqrt{\frac{1 - (- \frac{3}{5})}{2}} \\ = & - \sqrt{\frac{1 + \frac{3}{5}}{2} \cdot \frac{5}{5}} = - \sqrt{\frac{8}{10}} = - \frac{2 \sqrt{5}}{5} \end{matrix}$
Instead of our usual approach to verifying identities, namely starting with one side of the equation and trying to transform it into the other, we will start with the identity we proved in number of Example 12.2.3 and manipulate it into the identity we are asked to prove.
The identity we are asked to start with is $\sin (2 θ) = \frac{2 \tan (θ)}{1 + \tan^{2} (θ)}$ . If we are to use this to derive an identity for $\tan (\frac{θ}{2})$ , it seems reasonable to proceed by replacing each occurrence of $θ$ with $\frac{θ}{2}$
$\begin{matrix} \sin (2 (\frac{θ}{2})) & = & \frac{2 \tan (\frac{θ}{2})}{1 + \tan^{2} (\frac{θ}{2})} \\ \sin (θ) & = & \frac{2 \tan (\frac{θ}{2})}{1 + \tan^{2} (\frac{θ}{2})} \end{matrix}$
We now have the $\sin (θ)$ we need, but we somehow need to get a factor of $1 plus cos ⁡ open parenthesis θ close parenthesis$ involved. We substitute $1 + \tan^{2} (\frac{θ}{2}) = \sec^{2} (\frac{θ}{2})$ , and continue to manipulate our given identity by converting secants to cosines.
$\begin{matrix} \sin (θ) & = & \frac{2 \tan (\frac{θ}{2})}{1 + \tan^{2} (\frac{θ}{2})} \\ \sin (θ) & = & \frac{2 \tan (\frac{θ}{2})}{\sec^{2} (\frac{θ}{2})} \\ \sin (θ) & = & 2 \tan (\frac{θ}{2}) \cos^{2} (\frac{θ}{2}) \end{matrix}$
Finally, we apply a power reduction formula, and then solve for $\tan (\frac{θ}{2})$
$\begin{matrix} \sin (θ) & = & 2 \tan (\frac{θ}{2}) \cos^{2} (\frac{θ}{2}) \\ \sin (θ) & = & 2 \tan (\frac{θ}{2}) (\frac{1 + \cos (2 (\frac{θ}{2}))}{2}) \\ \sin (θ) & = & \tan (\frac{θ}{2}) (1 + \cos (θ)) \\ \tan (\frac{θ}{2}) & = & \frac{\sin (θ)}{1 + \cos (θ)} \end{matrix}$

Our next batch of identities, the Product to Sum Formulas,⁵ are easily verified by expanding each of the right hand sides in accordance with Theorem 12.8 and as you should expect by now we leave the details as exercises. They are of particular use in Calculus, and we list them here for reference.

Theorem

Product to Sum Formulas: For all angles $α$ and $β$ ,

$\cos (α) \cos (β) = \frac{1}{2} [\cos (α - β) + \cos (α + β)]$
$\sin (α) \sin (β) = \frac{1}{2} [\cos (α - β) - \cos (α + β)]$
$\sin (α) \cos (β) = \frac{1}{2} [\sin (α - β) + \sin (α + β)]$

Related to the Product to Sum Formulas are the Sum to Product Formulas, which we will have need of in Section 12.4. These are essentially restatements of the Product to Sum Formulas (by re-labeling the arguments of the sine and cosine functions) and as such, their proofs are left as exercises.

Theorem

Sum to Product Formulas: For all angles $α$ and $β$ ,

$\cos (α) + \cos (β) = 2 \cos (\frac{α + β}{2}) \cos (\frac{α - β}{2})$
$\cos (α) - \cos (β) = - 2 \sin (\frac{α + β}{2}) \sin (\frac{α - β}{2})$
$\sin (α) \pm \sin (β) = 2 \sin (\frac{α \pm β}{2}) \cos (\frac{α \mp β}{2})$

Example 6

Write $cos ⁡ open parenthesis 2 ⁢ θ close parenthesis ⁢ cos ⁡ open parenthesis 6 ⁢ θ close parenthesis$ as a sum.
Write $sin ⁡ open parenthesis θ close parenthesis minus sin ⁡ open parenthesis 3 ⁢ θ close parenthesis$ as a product.

Identifying $α equals 2 ⁢ θ$ and $β equals 6 ⁢ θ$ , we find
$\begin{matrix} \cos (2 θ) \cos (6 θ) & = & \frac{1}{2} [\cos (2 θ - 6 θ) + \cos (2 θ + 6 θ)] \\ = & \frac{1}{2} \cos (- 4 θ) + \frac{1}{2} \cos (8 θ) \\ = & \frac{1}{2} \cos (4 θ) + \frac{1}{2} \cos (8 θ), \end{matrix}$
where the last equality is courtesy of the even identity for cosine, $\cos (- 4 θ) = \cos (4 θ)$ .
Identifying $α = θ$ and $β equals 3 ⁢ θ$ yields
$\begin{matrix} \sin (θ) - \sin (3 θ) & = & 2 \sin (\frac{θ - 3 θ}{2}) \cos (\frac{θ + 3 θ}{2}) \\ = & 2 \sin (- θ) \cos (2 θ) \\ = & - 2 \sin (θ) \cos (2 θ), \end{matrix}$
where the last equality is courtesy of the odd identity for sine, $\sin (- θ) = - \sin (θ)$ .

The reader is reminded that all of the identities presented in this section which regard the circular functions as functions of angles (in radian measure) apply equally well to the circular (trigonometric) functions regarded as functions of real numbers.

Sinusoids, Revisited

We first studied sinusoids in Section 11.3. Using the sum formulas for sine and cosine, we can expand the forms given to us in Theorem 11.6:

$S (t) = A \sin (ω t + ϕ) + B = A \sin (ω t) \cos (ϕ) + A \cos (ω t) \sin (ϕ) + B,$

and

$C (t) = A \cos (ω t + ϕ) + B = A \cos (ω t) \cos (ϕ) - A \sin (ω t) \sin (ϕ) + B .$

As we'll see in the next example, recognizing these `expanded' forms of sinusoids allows us to graph functions as sinusoids which, at first glance, don't appear to fit the forms of either $C (t)$ or $S (t)$ .

Example 7

Consider the function $f (t) = \cos (2 t) - \sqrt{3} \sin (2 t)$ . Find a formula for $f (t)$ :

in the form $C (t) = A \cos (ω t + ϕ) + B$ for $ω greater than 0$
in the form $S (t) = A \sin (ω t + ϕ) + B$ for $ω greater than 0$

Check your answers analytically using identities and using a graphing utility.

The key to this problem is to use the expanded forms of the sinusoid formulas and match up corresponding coefficients. We start by equating $f (t) = \cos (2 t) - \sqrt{3} \sin (2 t)$ with the expanded form of $C (t) = A \cos (ω t + ϕ) + B$ : $\cos (2 t) - \sqrt{3} \sin (2 t) = A \cos (ω t) \cos (ϕ) - A \sin (ω t) \sin (ϕ) + B$ .
If we take $ω equals 2$ and $B equals 0$ , we get: $\cos (2 t) - \sqrt{3} \sin (2 t) = A \cos (2 t) \cos (ϕ) - A \sin (2 t) \sin (ϕ)$ .
To determine $A$ and $ϕ$ , a bit more work is involved. We get started by equating the coefficients of the trigonometric functions on either side of the equation.
On the left hand side, the coefficient of $cos ⁡ open parenthesis 2 ⁢ t close parenthesis$ is $1$ , while on the right hand side, it is $A \cos (ϕ)$ . Since this equation is to hold for all real numbers, we must have⁶ that $A ⁢ cos ⁡ open parenthesis ϕ close parenthesis equals 1$ .
Similarly, we find by equating the coefficients of $sin ⁡ open parenthesis 2 ⁢ t close parenthesis$ that $A \sin (ϕ) = \sqrt{3}$ . In conjunction with $A ⁢ cos ⁡ open parenthesis ϕ close parenthesis equals 1$ , we have a system of two (nonlinear) equations and two unknowns.
As usual, our first task is to reduce this system of two equations and two unknowns to one equation and one unknown. We can temporarily eliminate the dependence on $ϕ$ by using a Pythagorean Identity. From $\cos^{2} (ϕ) + \sin^{2} (ϕ) = 1$ , we multiply through by $A^{2}$ to get $A^{2} \cos^{2} (ϕ) + A^{2} \sin^{2} (ϕ) = A^{2}$ .
In our case, $A ⁢ cos ⁡ open parenthesis ϕ close parenthesis equals 1$ and $A \sin (ϕ) = \sqrt{3}$ , hence $A^{2} = A^{2} \cos^{2} (ϕ) + A^{2} \sin^{2} (ϕ) = 1^{2} + {(\sqrt{3})}^{2} = 4$ so $A = \pm 2$ . In much the same way we fit a sinusoid to a graph in Example 11.3.3, we choose $A equals 2$ , and then find the phase angle $ϕ$ associated with this choice.
Substituting $A equals 2$ into our two equations, $A ⁢ cos ⁡ open parenthesis ϕ close parenthesis equals 1$ and $A \sin (ϕ) = \sqrt{3}$ , we get $2 ⁢ cos ⁡ open parenthesis ϕ close parenthesis equals 1$ and $2 \sin (ϕ) = \sqrt{3}$ . After some rearrangement, $\cos (ϕ) = \frac{1}{2}$ and $\sin (ϕ) = \frac{\sqrt{3}}{2}$ . One such angle $ϕ$ which satisfies this criteria is $ϕ = \frac{π}{3}$ .
Hence, one way to write $f (t)$ as a sinusoid is $f (t) = 2 \cos (2 t + \frac{π}{3})$ . We can check our answer using the sum formula for cosine:
$\begin{matrix} f (t) & = & 2 \cos (2 t + \frac{π}{3}) \\ = & 2 [\cos (2 t) \cos (\frac{π}{3}) - \sin (2 t) \sin (\frac{π}{3})] \\ = & 2 [\cos (2 t) (\frac{1}{2}) - \sin (2 t) (\frac{\sqrt{3}}{2})] \\ = & \cos (2 t) - \sqrt{3} \sin (2 t) . \end{matrix}$
Proceeding as before, we equate $f (t) = \cos (2 t) - \sqrt{3} \sin (2 t)$ with the expanded form of of the sinusoid $S (t) = A \sin (ω t + ϕ) + B$ to get: $\cos (2 t) - \sqrt{3} \sin (2 t) = A \sin (ω t) \cos (ϕ) + A \cos (ω t) \sin (ϕ) + B$ .
Taking $ω equals 2$ and $B equals 0$ , we get $\cos (2 t) - \sqrt{3} \sin (2 t) = A \sin (2 t) \cos (ϕ) + A \cos (2 t) \sin (ϕ)$ . We equate⁷ the coefficients of $cos ⁡ open parenthesis 2 ⁢ t close parenthesis$ on either side and get $A ⁢ sin ⁡ open parenthesis ϕ close parenthesis equals 1$ and $A \cos (ϕ) = - \sqrt{3}$ .
Using $A^{2} \cos^{2} (ϕ) + A^{2} \sin^{2} (ϕ) = A^{2}$ as before, we get $A = \pm 2$ , and again we choose $A equals 2$ .
This means $2 ⁢ sin ⁡ open parenthesis ϕ close parenthesis equals 1$ , or $\sin (ϕ) = \frac{1}{2}$ , and $2 \cos (ϕ) = - \sqrt{3}$ , so $\cos (ϕ) = - \frac{\sqrt{3}}{2}$ . One such angle which meets these criteria is $ϕ = \frac{5 π}{6}$ .
Hence, we have $f (t) = 2 \sin (2 t + \frac{5 π}{6})$ . Checking our work analytically, we have
$\begin{matrix} f (t) & = & 2 \sin (2 t + \frac{5 π}{6}) \\ = & 2 [\sin (2 t) \cos (\frac{5 π}{6}) + \cos (2 t) \sin (\frac{5 π}{6})] \\ = & 2 [\sin (2 t) (- \frac{\sqrt{3}}{2}) + \cos (2 t) (\frac{1}{2})] \\ = & \cos (2 t) - \sqrt{3} \sin (2 t) \end{matrix}$
Graphing the three formulas for $f (t)$ result in the identical curve, verifying the work done analytically.
Figure 12.8

A couple of remarks about Example 12.2.7 are in order. First, had we chosen $A = - 2$ instead of $A equals 2$ as we worked through Example 12.2.7, our final answers would have looked different. The reader is encouraged to rework Example 12.2.7 using $A = - 2$ to see what these differences are, and then for a challenging exercise, use identities to show that the formulas are all equivalent.⁸

It is important to note that in order for the technique presented in Example 12.2.7 to fit a function into one of the forms in Theorem 11.6, the frequencies of the sine and cosine terms much match. For example, in the Exercises, you'll be asked to write $f (t) = 3 \sqrt{3} \sin (3 t) - 3 \cos (3 t)$ in the form of $S (t)$ and $C (t)$ above, and since both the sine and cosine terms have frequency $3$ , this is possible.

However, a function such as $f (t) = \sin (t) - \sin (3 t)$ cannot be written in the form of $S (t)$ or $C (t)$ . The quickest way to see this is to examine its graph below which is decidedly not a sinusoid. That being said, we can still analyze this curve using identities.

Using our result from number Example 12.2.6, we may rewrite $f (t) = \sin (t) - \sin (3 t) = - 2 \sin (t) \cos (2 t)$ . Grouping factors, we can view $f (t) = [- 2 \sin (t)] \cos (2 t) = A (t) \cos (2 t)$ as the curve $y equals cos ⁡ open parenthesis 2 ⁢ t close parenthesis$ with a variable amplitude, $A (t) = - 2 \sin (t)$ .

Overlaying the graphs of $f (t)$ with the (dashed) graphs of $A sub 1 ⁢ open parenthesis t close parenthesis equals 2 ⁢ sin ⁡ open parenthesis t close parenthesis$ and $A_{2} (t) = - 2 \sin (t)$ , we can see the role these two curves play in the graph of $y = f (t)$ . They create a kind of `wave envelope' for the graph of $y = f (t)$ . This is an example of the beats phenomenon. Note that when written as a product of sinusoids, it is always the lower frequency factor which creates the `wave-envelope' of the curve.

Note that in order to rewrite a sum or difference of sine and cosine functions with different frequencies into a product using the sum to product identities, Theorem 12.13, we need the amplitudes of each term to be the same. We explore more examples of these functions and this behavior in the Exercises.

Exercises

In Exercises -, use the Even / Odd Identities to verify the identity. Assume all quantities are defined.

$\sin (3 π - 2 θ) = - \sin (2 θ - 3 π)$
$\cos (- \frac{π}{4} - 5 t) = \cos (5 t + \frac{π}{4})$
$\tan (- x^{2} + 1) = - \tan (x^{2} - 1)$
$\csc (- θ - 5) = - \csc (θ + 5)$
$\sec (- 6 x) = \sec (6 x)$
$\cot (9 - 7 θ) = - \cot (7 θ - 9)$

In Exercises -, use the Sum and Difference Identities to find the exact value. You may have need of the Quotient, Reciprocal or Even / Odd Identities as well.

$cos ⁡ open parenthesis 75 degrees close parenthesis$
$sec ⁡ open parenthesis 165 degrees close parenthesis$
$sin ⁡ open parenthesis 105 degrees close parenthesis$
$csc ⁡ open parenthesis 195 degrees close parenthesis$
$cot ⁡ open parenthesis 255 degrees close parenthesis$
$tan ⁡ open parenthesis 375 degrees close parenthesis$
$\cos (\frac{13 π}{12})$
$\sin (\frac{11 π}{12})$
$\tan (\frac{13 π}{12})$
$\cos (\frac{7 π}{12})$
$\tan (\frac{17 π}{12})$
$\sin (\frac{π}{12})$
$\cot (\frac{11 π}{12})$
$\csc (\frac{5 π}{12})$
$\sec (- \frac{π}{12})$
If $α$ is a Quadrant IV angle with $\cos (α) = \frac{\sqrt{5}}{5}$ , and $\sin (β) = \frac{\sqrt{10}}{10}$ , where $\frac{π}{2} < β < π$ , find
1. $\cos (α + β)$
2. $\sin (α + β)$
3. $\tan (α + β)$
1. $\cos (α - β)$
2. $\sin (α - β)$
3. $\tan (α - β)$
If $csc ⁡ open parenthesis α close parenthesis equals 3$ , where $0 < α < \frac{π}{2}$ , and $β$ is a Quadrant II angle with $\tan (β) = - 7$ , find
1. $\cos (α + β)$
2. $\sin (α + β)$
3. $\tan (α + β)$
1. $\cos (α - β)$
2. $\sin (α - β)$
3. $\tan (α - β)$
If $\sin (α) = \frac{3}{5}$ , where $0 < α < \frac{π}{2}$ , and $\cos (β) = \frac{12}{13}$ where $\frac{3 π}{2} < β < 2 π$ , find
1. $\sin (α + β)$
2. $\cos (α - β)$
3. $\tan (α - β)$
If $\sec (α) = - \frac{5}{3}$ , where $\frac{π}{2} < α < π$ , and $\tan (β) = \frac{24}{7}$ , where $π < β < \frac{3 π}{2}$ , find
1. $\csc (α - β)$
2. $\sec (α + β)$
3. $\cot (α + β)$

In Exercises -, use Example 12.2.7 as a guide to show that the function is a sinusoid by rewriting it in the forms $C (t) = A \cos (ω t + ϕ) + B$ and $S (t) = A \sin (ω t + ϕ) + B$ for $ω greater than 0$ and $0 less than or equal to ϕ less than 2 ⁢ π$ .

$f (t) = \sqrt{2} \sin (t) + \sqrt{2} \cos (t) + 1$
$f (t) = 3 \sqrt{3} \sin (3 t) - 3 \cos (3 t)$
$f (t) = - \sin (t) + \cos (t) - 2$
$f (t) = - \frac{1}{2} \sin (2 t) - \frac{\sqrt{3}}{2} \cos (2 t)$
$f (t) = 2 \sqrt{3} \cos (t) - 2 \sin (t)$
$f (t) = \frac{3}{2} \cos (2 t) - \frac{3 \sqrt{3}}{2} \sin (2 t) + 6$
$f (t) = - \frac{1}{2} \cos (5 t) - \frac{\sqrt{3}}{2} \sin (5 t)$
$f (t) = - 6 \sqrt{3} \cos (3 t) - 6 \sin (3 t) - 3$
$f (t) = \frac{5 \sqrt{2}}{2} \sin (t) - \frac{5 \sqrt{2}}{2} \cos (t)$
$f (t) = 3 \sin (\frac{t}{6}) - 3 \sqrt{3} \cos (\frac{t}{6})$
In Exercises -, you should have noticed a relationship between the phases $ϕ$ for the $S (t)$ and $C (t)$ . Show that if $f (t) = A \sin (ω t + α) + B$ , then $f (t) = A \cos (ω t + β) + B$ where $β = α - \frac{π}{2}$ .
Let $ϕ$ be an angle measured in radians and let $P (a, b)$ be a point on the terminal side of $ϕ$ when it is drawn in standard position. Use Theorem 11.3 and the sum identity for sine in Theorem 12.7 to show that $f (t) = a \sin (ω t) + b \cos (ω t) + B$ (with $ω greater than 0$ ) can be rewritten as $f (t) = \sqrt{a^{2} + b^{2}} \sin (ω t + ϕ) + B$ .
In Example 11.3.5 in Section 11.3, we developed two (seemingly) different formulas to model the hours of daylight, $H (t)$ : $H_{1} (t) = 9.25 \sin (\frac{π}{6} t - \frac{π}{2}) + 12.55$ and $H_{2} (t) = - 8.13 \sin (\frac{π}{6} t - 4.70) + 12.5$ . Use the difference identities for sine to expand $H sub 1 ⁢ open parenthesis t close parenthesis$ and $H sub 2 ⁢ open parenthesis t close parenthesis$ . How different are they?

In Exercises -, verify the identity.⁹

$\sin (θ + \frac{π}{2}) = \cos (t)$
$\cos (θ - \frac{π}{2}) = \sin (t)$
$\cos (θ - π) = - \cos (θ)$
$\sin (π - θ) = \sin (θ)$
$\tan (θ + \frac{π}{2}) = - \cot (θ)$
$\sin (α + β) + \sin (α - β) = 2 \sin (α) \cos (β)$
$\sin (α + β) - \sin (α - β) = 2 \cos (α) \sin (β)$
$\cos (α + β) + \cos (α - β) = 2 \cos (α) \cos (β)$
$\cos (α + β) - \cos (α - β) = - 2 \sin (α) \sin (β)$
$\frac{\sin (α + β)}{\sin (α - β)} = \frac{1 + \cot (α) \tan (β)}{1 - \cot (α) \tan (β)}$
$\frac{\cos (α + β)}{\cos (α - β)} = \frac{1 - \tan (α) \tan (β)}{1 + \tan (α) \tan (β)}$
$\frac{\tan (α + β)}{\tan (α - β)} = \frac{\sin (α) \cos (α) + \sin (β) \cos (β)}{\sin (α) \cos (α) - \sin (β) \cos (β)}$

In Exercise -, use the results from Exercise in Section 11.4: $lim θ → 0$ $\frac{\sin (θ)}{θ} = 1$ and Exercise in Section 12.1: $lim θ → 0$ $\frac{1 - \cos (θ)}{θ} = 0$ to help you find the derivatives of the sine and cosine functions.

1. Verify for $f (t) = \sin (t)$ , $\frac{f (t + h) - f (t)}{h} = \frac{\sin (t + h) - \sin (t)}{h} = \cos (t) (\frac{\sin (h)}{h}) + \sin (t) (\frac{\cos (h) - 1}{h})$
2. Show $f^{'} (t) = lim_{h \to 0} \frac{f (t + h) - f (t)}{h} = \cos (t)$ .
  HINT: $lim θ → 0$ $\frac{\cos (θ) - 1}{θ} =$ $lim θ → 0$ $(- \frac{1 - \cos (θ)}{θ}) = - (0) = 0$
3. Find the equation of the tangent line to the graph of $y = \sin (t)$ at $open parenthesis 0, 0 close parenthesis$ , $(\frac{π}{2}, 1)$ and $open parenthesis π, 0 close parenthesis$ .
  Check your answers graphically.
1. Verify for $g (t) = \cos (t)$ , $\frac{g (t + h) - g (t)}{h} = \frac{\cos (t + h) - \cos (t)}{h} = \cos (t) (\frac{\cos (h) - 1}{h}) - \sin (t) (\frac{\sin (h)}{h})$
2. Show $g^{'} (t) = lim_{h \to 0} \frac{g (t + h) - g (t)}{h} = - \sin (t)$ .
3. Find the equation of the tangent line to the graph of $y = \cos (t)$ at $open parenthesis 0, 1 close parenthesis$ , $(\frac{π}{2}, 0)$ and $(- \frac{π}{2}, 0)$ .
  Check your answers graphically.
1. Use the fact¹⁰ that $lim θ → 0$ $\frac{\sin (θ)}{θ} = 1$ to prove $lim θ → 0$ $\frac{\tan (θ)}{θ} = 1$ .
  HINT: $\frac{\tan (θ)}{θ} = \frac{\sin (θ)}{θ} \frac{1}{\cos (θ)}$ …
2. For $f (t) = \tan (t)$ , show $\frac{f (t + h) - f (t)}{h} = \frac{\tan (t + h) - \tan (t)}{h} = (\frac{\tan (h)}{h}) (\frac{\sec^{2} (t)}{1 - \tan (t) \tan (h)})$ .
3. Use part to show $f^{'} (t) = lim_{h \to 0} \frac{f (t + h) - f (t)}{h} = \sec^{2} (t)$
4. Find the equation of the tangent line to the graph of $y = \tan (t)$ at $open parenthesis 0, 0 close parenthesis$ , $(\frac{π}{4}, 1)$ and $(- \frac{π}{4}, - 1)$ .
  Check your answers graphically.

In Exercises -, use the Half Angle Formulas to find the exact value. You may have need of the Quotient, Reciprocal or Even / Odd Identities as well.

$cos ⁡ open parenthesis 75 degrees close parenthesis$ (compare with Exercise )
$sin ⁡ open parenthesis 105 degrees close parenthesis$ (compare with Exercise )
$cos ⁡ open parenthesis 67.5 degrees close parenthesis$
$sin ⁡ open parenthesis 157.5 degrees close parenthesis$
$tan ⁡ open parenthesis 112.5 degrees close parenthesis$
$\cos (\frac{7 π}{12})$ (compare with Exercise )
$\sin (\frac{π}{12})$ (compare with Exercise )
$\cos (\frac{π}{8})$
$\sin (\frac{5 π}{8})$
$\tan (\frac{7 π}{8})$

In Exercises -, use the given information about $θ$ to find the exact values of

$sin ⁡ open parenthesis 2 ⁢ θ close parenthesis$
$\sin (\frac{θ}{2})$
$cos ⁡ open parenthesis 2 ⁢ θ close parenthesis$
$\cos (\frac{θ}{2})$
$tan ⁡ open parenthesis 2 ⁢ θ close parenthesis$
$\tan (\frac{θ}{2})$
$\sin (θ) = - \frac{7}{25}$ where $\frac{3 π}{2} < θ < 2 π$
$\cos (θ) = \frac{28}{53}$ where $0 < θ < \frac{π}{2}$
$\tan (θ) = \frac{12}{5}$ where $π < θ < \frac{3 π}{2}$
$csc ⁡ open parenthesis θ close parenthesis equals 4$ where $\frac{π}{2} < θ < π$
$\cos (θ) = \frac{3}{5}$ where $0 < θ < \frac{π}{2}$
$\sin (θ) = - \frac{4}{5}$ where $π < θ < \frac{3 π}{2}$
$\cos (θ) = \frac{12}{13}$ where $\frac{3 π}{2} < θ < 2 π$
$\sin (θ) = \frac{5}{13}$ where $\frac{π}{2} < θ < π$
$\sec (θ) = \sqrt{5}$ where $\frac{3 π}{2} < θ < 2 π$
$\tan (θ) = - 2$ where $\frac{π}{2} < θ < π$

In Exercises -, verify the identity. Assume all quantities are defined.

${(\cos (θ) + \sin (θ))}^{2} = 1 + \sin (2 θ)$
${(\cos (θ) - \sin (θ))}^{2} = 1 - \sin (2 θ)$
$\tan (2 t) = \frac{1}{1 - \tan (t)} - \frac{1}{1 + \tan (t)}$
$\csc (2 θ) = \frac{\cot (θ) + \tan (θ)}{2}$
$8 \sin^{4} (x) = \cos (4 x) - 4 \cos (2 x) + 3$
$8 \cos^{4} (x) = \cos (4 x) + 4 \cos (2 x) + 3$
$\sin (3 θ) = 3 \sin (θ) - 4 \sin^{3} (θ)$
$\sin (4 θ) = 4 \sin (θ) \cos^{3} (θ) - 4 \sin^{3} (θ) \cos (θ)$
$32 \sin^{2} (t) \cos^{4} (t) = 2 + \cos (2 t) - 2 \cos (4 t) - \cos (6 t)$
$32 \sin^{4} (t) \cos^{2} (t) = 2 - \cos (2 t) - 2 \cos (4 t) + \cos (6 t)$
$\cos (4 θ) = 8 \cos^{4} (θ) - 8 \cos^{2} (θ) + 1$
$\cos (8 θ) = 128 \cos^{8} (θ) - 256 \cos^{6} (θ) + 160 \cos^{4} (θ) - 32 \cos^{2} (θ) + 1$ (HINT: Use the result to.)
$\sec (2 x) = \frac{\cos (x)}{\cos (x) + \sin (x)} + \frac{\sin (x)}{\cos (x) - \sin (x)}$
$\frac{1}{\cos (θ) - \sin (θ)} + \frac{1}{\cos (θ) + \sin (θ)} = \frac{2 \cos (θ)}{\cos (2 θ)}$
$\frac{1}{\cos (θ) - \sin (θ)} - \frac{1}{\cos (θ) + \sin (θ)} = \frac{2 \sin (θ)}{\cos (2 θ)}$
Suppose $θ$ is a Quadrant I angle with $\sin (θ) = x$ . Verify the following formulas
1. $\cos (θ) = \sqrt{1 - x^{2}}$
2. $\sin (2 θ) = 2 x \sqrt{1 - x^{2}}$
3. $\cos (2 θ) = 1 - 2 x^{2}$
Discuss with your classmates how each of the formulas, if any, in Exercise change if we change assume $θ$ is a Quadrant II, III, or IV angle.
Suppose $θ$ is a Quadrant I angle with $\tan (θ) = x$ . Verify the following formulas
1. $\cos (θ) = \frac{1}{\sqrt{x^{2} + 1}}$
2. $\sin (θ) = \frac{x}{\sqrt{x^{2} + 1}}$
1. $\sin (2 θ) = \frac{2 x}{x^{2} + 1}$
2. $\cos (2 θ) = \frac{1 - x^{2}}{x^{2} + 1}$
Discuss with your classmates how each of the formulas, if any, in Exercise change if we change assume $θ$ is a Quadrant II, III, or IV angle.
If $\sin (t) = x$ for $- \frac{π}{2} < t < \frac{π}{2}$ , find an expression for $\tan (t)$ in terms of $x$ .
If $\tan (θ) = x$ for $- \frac{π}{2} < θ < \frac{π}{2}$ , find an expression for $\sec (θ)$ in terms of $x$ .
If $\sec (θ) = x$ where $θ$ is a Quadrant II angle, find an expression for $\tan (θ)$ in terns of $x$ .
If $\sin (t) = \frac{x}{2}$ for $- \frac{π}{2} < t < \frac{π}{2}$ , find an expression for $cos ⁡ open parenthesis 2 ⁢ t close parenthesis$ in terms of $x$ .
If $\tan (θ) = \frac{x}{7}$ for $- \frac{π}{2} < θ < \frac{π}{2}$ , find an expression for $sin ⁡ open parenthesis 2 ⁢ θ close parenthesis$ in terms of $x$ .
If $\sec (t) = \frac{x}{4}$ for $0 < t < \frac{π}{2}$ , find an expression for $\ln | \sec (t) + \tan (t) |$ in terms of $x$ .
Show that $\cos^{2} (θ) - \sin^{2} (θ) = 2 \cos^{2} (θ) - 1 = 1 - 2 \sin^{2} (θ)$ for all $θ$ .
Let $θ$ be a Quadrant III angle with $\cos (θ) = - \frac{1}{5}$ . Show that this is not enough information to determine the sign of $\sin (\frac{θ}{2})$ by first assuming $3 π < θ < \frac{7 π}{2}$ and then assuming $π < θ < \frac{3 π}{2}$ and computing $\sin (\frac{θ}{2})$ in both cases.
Without using your calculator, show that $\frac{\sqrt{2 + \sqrt{3}}}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$
In part of Example 12.2.3, we wrote $cos ⁡ open parenthesis 3 ⁢ θ close parenthesis$ as a polynomial in terms of $\cos (θ)$ . In Exercise, we had you verify an identity which expresses $cos ⁡ open parenthesis 4 ⁢ θ close parenthesis$ as a polynomial in terms of $\cos (θ)$ . Can you find a polynomial in terms of $\cos (θ)$ for $cos ⁡ open parenthesis 5 ⁢ θ close parenthesis$ ? $cos ⁡ open parenthesis 6 ⁢ θ close parenthesis$ ? Can you find a pattern so that $\cos (n θ)$ could be written as a polynomial in cosine for any natural number $n$ ?
In Exercise, we has you verify an identity which expresses $sin ⁡ open parenthesis 3 ⁢ θ close parenthesis$ as a polynomial in terms of $\sin (θ)$ . Can you do the same for $sin ⁡ open parenthesis 5 ⁢ θ close parenthesis$ ? What about for $sin ⁡ open parenthesis 4 ⁢ θ close parenthesis$ ? If not, what goes wrong?

In Exercises -, verify the identity by graphing the right and left hand using a graphing utility.

$\sin^{2} (t) + \cos^{2} (t) = 1$
$\sec^{2} (x) - \tan^{2} (x) = 1$
$\cos (t) = \sin (\frac{π}{2} - t)$
$\tan (x + π) = \tan (x)$
$sin ⁡ open parenthesis 2 ⁢ t close parenthesis equals 2 ⁢ sin ⁡ open parenthesis t close parenthesis ⁢ cos ⁡ open parenthesis t close parenthesis$
$\tan (\frac{x}{2}) = \frac{\sin (x)}{1 + \cos (x)}$

In Exercises -, write the given product as a sum. Note: you may need to use an Even/Odd Identity to match the answer provided.

$cos ⁡ open parenthesis 3 ⁢ θ close parenthesis ⁢ cos ⁡ open parenthesis 5 ⁢ θ close parenthesis$
$sin ⁡ open parenthesis 2 ⁢ t close parenthesis ⁢ sin ⁡ open parenthesis 7 ⁢ t close parenthesis$
$sin ⁡ open parenthesis 9 ⁢ x close parenthesis ⁢ cos ⁡ open parenthesis x close parenthesis$
$cos ⁡ open parenthesis 2 ⁢ θ close parenthesis ⁢ cos ⁡ open parenthesis 6 ⁢ θ close parenthesis$
$sin ⁡ open parenthesis 3 ⁢ t close parenthesis ⁢ sin ⁡ open parenthesis 2 ⁢ t close parenthesis$
$cos ⁡ open parenthesis x close parenthesis ⁢ sin ⁡ open parenthesis 3 ⁢ x close parenthesis$

In Exercises -, write the given sum as a product. Note: you may need to use an Even/Odd or Cofunction Identity to match the answer provided.

$cos ⁡ open parenthesis 3 ⁢ θ close parenthesis plus cos ⁡ open parenthesis 5 ⁢ θ close parenthesis$
$sin ⁡ open parenthesis 2 ⁢ t close parenthesis minus sin ⁡ open parenthesis 7 ⁢ t close parenthesis$
$cos ⁡ open parenthesis 5 ⁢ x close parenthesis minus cos ⁡ open parenthesis 6 ⁢ x close parenthesis$
$\sin (9 θ) - \sin (- θ)$
$\sin (t) + \cos (t)$
$\cos (x) - \sin (x)$

In Exercises -, using the remarks following Example 12.2.7 on page as a guide, rewrite the given function $f (t)$ as a product of sinusoids. Identify the functions which create the `wave envelope.' Check your answer by graphing the function along with the `wave-envelope' using a graphing utility.

$f (t) = \cos (3 t) + \cos (5 t)$
$f (t) = 3 \cos (5 t) - 3 \cos (6 t)$
$f (t) = \frac{1}{2} \sin (9 t) + \frac{1}{2} \sin (t)$
$f (t) = \frac{2}{3} \sin (2 t) - \frac{2}{3} \sin (7 t)$
Verify the Even / Odd Identities for tangent, secant, cosecant and cotangent.
Verify the Cofunction Identities for tangent, secant, cosecant and cotangent.
Verify the Difference Identities for sine and tangent.
Verify the Product to Sum Identities.
Verify the Sum to Product Identities.

Answers

$\cos (75^{\circ}) = \frac{\sqrt{6} - \sqrt{2}}{4}$
$\sec (165^{\circ}) = - \frac{4}{\sqrt{2} + \sqrt{6}} = \sqrt{2} - \sqrt{6}$
$\sin (105^{\circ}) = \frac{\sqrt{6} + \sqrt{2}}{4}$
$\csc (195^{\circ}) = \frac{4}{\sqrt{2} - \sqrt{6}} = - (\sqrt{2} + \sqrt{6})$
$\cot (255^{\circ}) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3}$
$\tan (375^{\circ}) = \frac{3 - \sqrt{3}}{3 + \sqrt{3}} = 2 - \sqrt{3}$
$\cos (\frac{13 π}{12}) = - \frac{\sqrt{6} + \sqrt{2}}{4}$
$\sin (\frac{11 π}{12}) = \frac{\sqrt{6} - \sqrt{2}}{4}$
$\tan (\frac{13 π}{12}) = \frac{3 - \sqrt{3}}{3 + \sqrt{3}} = 2 - \sqrt{3}$
$\cos (\frac{7 π}{12}) = \frac{\sqrt{2} - \sqrt{6}}{4}$
$\tan (\frac{17 π}{12}) = 2 + \sqrt{3}$
$\sin (\frac{π}{12}) = \frac{\sqrt{6} - \sqrt{2}}{4}$
$\cot (\frac{11 π}{12}) = - (2 + \sqrt{3})$
$\csc (\frac{5 π}{12}) = \sqrt{6} - \sqrt{2}$
$\sec (- \frac{π}{12}) = \sqrt{6} - \sqrt{2}$
1. $\cos (α + β) = - \frac{\sqrt{2}}{10}$
2. $\sin (α + β) = \frac{7 \sqrt{2}}{10}$
1. $\tan (α + β) = - 7$
2. $\cos (α - β) = - \frac{\sqrt{2}}{2}$
1. $\sin (α - β) = \frac{\sqrt{2}}{2}$
2. $\tan (α - β) = - 1$
1. $\cos (α + β) = - \frac{4 + 7 \sqrt{2}}{30}$
2. $\sin (α + β) = \frac{28 - \sqrt{2}}{30}$
1. $\tan (α + β) = \frac{- 28 + \sqrt{2}}{4 + 7 \sqrt{2}} = \frac{63 - 100 \sqrt{2}}{41}$
2. $\cos (α - β) = \frac{- 4 + 7 \sqrt{2}}{30}$
1. $\sin (α - β) = - \frac{28 + \sqrt{2}}{30}$
2. $\tan (α - β) = \frac{28 + \sqrt{2}}{4 - 7 \sqrt{2}} = - \frac{63 + 100 \sqrt{2}}{41}$
1. $\sin (α + β) = \frac{16}{65}$
2. $\cos (α - β) = \frac{33}{65}$
3. $\tan (α - β) = \frac{56}{33}$
1. $\csc (α - β) = - \frac{5}{4}$
2. $\sec (α + β) = \frac{125}{117}$
3. $\cot (α + β) = \frac{117}{44}$
$f (t) = \sqrt{2} \sin (t) + \sqrt{2} \cos (t) + 1 = 2 \sin (t + \frac{π}{4}) + 1 = 2 \cos (t + \frac{7 π}{4}) + 1$
$f (t) = 3 \sqrt{3} \sin (3 t) - 3 \cos (3 t) = 6 \sin (3 t + \frac{11 π}{6}) = 6 \cos (3 t + \frac{4 π}{3})$
$f (t) = - \sin (t) + \cos (t) - 2 = \sqrt{2} \sin (t + \frac{3 π}{4}) - 2 = \sqrt{2} \cos (t + \frac{π}{4}) - 2$
$f (t) = - \frac{1}{2} \sin (2 t) - \frac{\sqrt{3}}{2} \cos (2 t) = \sin (2 t + \frac{4 π}{3}) = \cos (2 t + \frac{5 π}{6})$
$f (t) = 2 \sqrt{3} \cos (t) - 2 \sin (t) = 4 \sin (t + \frac{2 π}{3}) = 4 \cos (t + \frac{π}{6})$
$f (t) = \frac{3}{2} \cos (2 t) - \frac{3 \sqrt{3}}{2} \sin (2 t) + 6 = 3 \sin (2 t + \frac{5 π}{6}) + 6 = 3 \cos (2 t + \frac{π}{3}) + 6$
$f (t) = - \frac{1}{2} \cos (5 t) - \frac{\sqrt{3}}{2} \sin (5 t) = \sin (5 t + \frac{7 π}{6}) = \cos (5 t + \frac{2 π}{3})$
$f (t) = - 6 \sqrt{3} \cos (3 t) - 6 \sin (3 t) - 3 = 12 \sin (3 t + \frac{4 π}{3}) - 3 = 12 \cos (3 t + \frac{5 π}{6}) - 3$
$f (t) = \frac{5 \sqrt{2}}{2} \sin (t) - \frac{5 \sqrt{2}}{2} \cos (t) = 5 \sin (t + \frac{7 π}{4}) = 5 \cos (t + \frac{5 π}{4})$
$f (t) = 3 \sin (\frac{t}{6}) - 3 \sqrt{3} \cos (\frac{t}{6}) = 6 \sin (\frac{t}{6} + \frac{5 π}{3}) = 6 \cos (\frac{t}{6} + \frac{7 π}{6})$
1. at $open parenthesis 0, 0 close parenthesis$ : $y = x$ ; at $(\frac{π}{2}, 1)$ : $y equals 1$ at $open parenthesis π, 0 close parenthesis$ : $y = - x + π$
1. at $open parenthesis 0, 1 close parenthesis$ : $y equals 1$ ; at $(\frac{π}{2}, 0)$ : $y = - x + \frac{π}{2}$ at $(- \frac{π}{2}, 0)$ : $y = x + \frac{π}{2}$
1. at $open parenthesis 0, 0 close parenthesis$ : $y = x$ ; at $(\frac{π}{4}, 1)$ : $y = 2 x - \frac{π}{2} + 1$ at $(- \frac{π}{4}, - 1)$ : $y = 2 x + \frac{π}{2} - 1$
$\cos (75^{\circ}) = \frac{\sqrt{2 - \sqrt{3}}}{2}$
$\sin (105^{\circ}) = \frac{\sqrt{2 + \sqrt{3}}}{2}$
$\cos ({67.5}^{\circ}) = \frac{\sqrt{2 - \sqrt{2}}}{2}$
$\sin ({157.5}^{\circ}) = \frac{\sqrt{2 - \sqrt{2}}}{2}$
$\tan ({112.5}^{\circ}) = - \sqrt{\frac{2 + \sqrt{2}}{2 - \sqrt{2}}} = - 1 - \sqrt{2}$
$\cos (\frac{7 π}{12}) = - \frac{\sqrt{2 - \sqrt{3}}}{2}$
$\sin (\frac{π}{12}) = \frac{\sqrt{2 - \sqrt{3}}}{2}$
$\cos (\frac{π}{8}) = \frac{\sqrt{2 + \sqrt{2}}}{2}$
$\sin (\frac{5 π}{8}) = \frac{\sqrt{2 + \sqrt{2}}}{2}$
$\tan (\frac{7 π}{8}) = - \sqrt{\frac{2 - \sqrt{2}}{2 + \sqrt{2}}} = 1 - \sqrt{2}$
- $\sin (2 θ) = - \frac{336}{625}$
- $\sin (\frac{θ}{2}) = \frac{\sqrt{2}}{10}$
- $\cos (2 θ) = \frac{527}{625}$
- $\cos (\frac{θ}{2}) = - \frac{7 \sqrt{2}}{10}$
- $\tan (2 θ) = - \frac{336}{527}$
- $\tan (\frac{θ}{2}) = - \frac{1}{7}$
- $\sin (2 θ) = \frac{2520}{2809}$
- $\sin (\frac{θ}{2}) = \frac{5 \sqrt{106}}{106}$
- $\cos (2 θ) = - \frac{1241}{2809}$
- $\cos (\frac{θ}{2}) = \frac{9 \sqrt{106}}{106}$
- $\tan (2 θ) = - \frac{2520}{1241}$
- $\tan (\frac{θ}{2}) = \frac{5}{9}$
- $\sin (2 θ) = \frac{120}{169}$
- $\sin (\frac{θ}{2}) = \frac{3 \sqrt{13}}{13}$
- $\cos (2 θ) = - \frac{119}{169}$
- $\cos (\frac{θ}{2}) = - \frac{2 \sqrt{13}}{13}$
- $\tan (2 θ) = - \frac{120}{119}$
- $\tan (\frac{θ}{2}) = - \frac{3}{2}$
- $\sin (2 θ) = - \frac{\sqrt{15}}{8}$
- $\sin (\frac{θ}{2}) = \frac{\sqrt{8 + 2 \sqrt{15}}}{4}$
- $\cos (2 θ) = \frac{7}{8}$
- $\cos (\frac{θ}{2}) = \frac{\sqrt{8 - 2 \sqrt{15}}}{4}$
- $\tan (2 θ) = - \frac{\sqrt{15}}{7}$
- $\tan (\frac{θ}{2}) = \sqrt{\frac{8 + 2 \sqrt{15}}{8 - 2 \sqrt{15}}} \tan (\frac{θ}{2}) = 4 + \sqrt{15}$
- $\sin (2 θ) = \frac{24}{25}$
- $\sin (\frac{θ}{2}) = \frac{\sqrt{5}}{5}$
- $\cos (2 θ) = - \frac{7}{25}$
- $\cos (\frac{θ}{2}) = \frac{2 \sqrt{5}}{5}$
- $\tan (2 θ) = - \frac{24}{7}$
- $\tan (\frac{θ}{2}) = \frac{1}{2}$
- $\sin (2 θ) = \frac{24}{25}$
- $\sin (\frac{θ}{2}) = \frac{2 \sqrt{5}}{5}$
- $\cos (2 θ) = - \frac{7}{25}$
- $\cos (\frac{θ}{2}) = - \frac{\sqrt{5}}{5}$
- $\tan (2 θ) = - \frac{24}{7}$
- $\tan (\frac{θ}{2}) = - 2$
- $\sin (2 θ) = - \frac{120}{169}$
- $\sin (\frac{θ}{2}) = \frac{\sqrt{26}}{26}$
- $\cos (2 θ) = \frac{119}{169}$
- $\cos (\frac{θ}{2}) = - \frac{5 \sqrt{26}}{26}$
- $\tan (2 θ) = - \frac{120}{119}$
- $\tan (\frac{θ}{2}) = - \frac{1}{5}$
- $\sin (2 θ) = - \frac{120}{169}$
- $\sin (\frac{θ}{2}) = \frac{5 \sqrt{26}}{26}$
- $\cos (2 θ) = \frac{119}{169}$
- $\cos (\frac{θ}{2}) = \frac{\sqrt{26}}{26}$
- $\tan (2 θ) = - \frac{120}{119}$
- $\tan (\frac{θ}{2}) = 5$
- $\sin (2 θ) = - \frac{4}{5}$
- $\sin (\frac{θ}{2}) = \frac{\sqrt{50 - 10 \sqrt{5}}}{10}$
- $\cos (2 θ) = - \frac{3}{5}$
- $\cos (\frac{θ}{2}) = - \frac{\sqrt{50 + 10 \sqrt{5}}}{10}$
- $\tan (2 θ) = \frac{4}{3}$
- $\tan (\frac{θ}{2}) = - \sqrt{\frac{5 - \sqrt{5}}{5 + \sqrt{5}}} \tan (\frac{θ}{2}) = \frac{5 - 5 \sqrt{5}}{10}$
- $\sin (2 θ) = - \frac{4}{5}$
- $\sin (\frac{θ}{2}) = \frac{\sqrt{50 + 10 \sqrt{5}}}{10}$
- $\cos (2 θ) = - \frac{3}{5}$
- $\cos (\frac{θ}{2}) = \frac{\sqrt{50 - 10 \sqrt{5}}}{10}$
- $\tan (2 θ) = \frac{4}{3}$
- $\tan (\frac{θ}{2}) = \sqrt{\frac{5 + \sqrt{5}}{5 - \sqrt{5}}} \tan (\frac{θ}{2}) = \frac{5 + 5 \sqrt{5}}{10}$
$\tan (t) = \frac{x}{\sqrt{1 - x^{2}}}$
$\sec (θ) = \sqrt{1 + x^{2}}$
$\tan (θ) = - \sqrt{x^{2} - 1}$
$\cos (2 t) = 1 - \frac{x^{2}}{2}$
$\sin (2 θ) = \frac{14 x}{x^{2} + 49}$
$\ln | \sec (t) + \tan (t) | = \ln | x + \sqrt{x^{2} - 16} | - \ln (4)$
$\frac{\cos (2 θ) + \cos (8 θ)}{2}$
$\frac{\cos (5 t) - \cos (9 t)}{2}$
$\frac{\sin (8 x) + \sin (10 x)}{2}$
$\frac{\cos (4 θ) + \cos (8 θ)}{2}$
$\frac{\cos (t) - \cos (5 t)}{2}$
$\frac{\sin (2 x) + \sin (4 x)}{2}$
$2 ⁢ cos ⁡ open parenthesis 4 ⁢ θ close parenthesis ⁢ cos ⁡ open parenthesis θ close parenthesis$
$- 2 \cos (\frac{9}{2} t) \sin (\frac{5}{2} t)$
$2 \sin (\frac{11}{2} x) \sin (\frac{1}{2} x)$
$2 ⁢ cos ⁡ open parenthesis 4 ⁢ θ close parenthesis ⁢ sin ⁡ open parenthesis 5 ⁢ θ close parenthesis$
$\sqrt{2} \cos (t - \frac{π}{4})$
$- \sqrt{2} \sin (x - \frac{π}{4})$
$f ⁢ open parenthesis t close parenthesis equals open bracket 2 ⁢ cos ⁡ open parenthesis t close parenthesis close bracket ⁢ cos ⁡ open parenthesis 4 ⁢ t close parenthesis$ , $A ⁢ open parenthesis t close parenthesis equals 2 ⁢ cos ⁡ open parenthesis t close parenthesis$ , wave-envelope: $y = \pm 2 \cos (t)$ .
$f (t) = [6 \sin (\frac{1}{2} t)] \sin (\frac{11}{2} t)$ , $A (t) = 6 \sin (\frac{1}{2} t)$ , wave-envelope: $y = \pm 6 \sin (\frac{1}{2} t)$ .
$f ⁢ open parenthesis t close parenthesis equals open bracket cos ⁡ open parenthesis 4 ⁢ t close parenthesis close bracket ⁢ sin ⁡ open parenthesis 5 ⁢ t close parenthesis$ , $A ⁢ open parenthesis t close parenthesis equals cos ⁡ open parenthesis 4 ⁢ t close parenthesis$ , wave-envelope: $y = \pm \cos (4 t)$ .
$f (t) = [- \frac{4}{3} \sin (\frac{5}{2} t)] \cos (\frac{9}{2} t)$ , $A (t) = - \frac{4}{3} \sin (\frac{5}{2} t)$ , wave-envelope: $y = \pm \frac{4}{3} \sin (\frac{5}{2} t)$ .

In the picture we've drawn, the triangles $P O Q$ and $A O B$ are congruent, which is even better. However, $α sub 0 minus β sub 0$ could be $0$ or it could be $π$ , neither of which makes a triangle. It could also be larger than $π$ , which makes a triangle, just not the one we've drawn. You should think about those three cases. ↩
It takes some trial and error to find this combination. One alternative is to convert to degrees … ↩
Note that even though $\tan (β) = \frac{\sin (β)}{\cos (β)}$ , we cannot take $\sin (β) = - 2$ and $\cos (β) = - 1$ . Recall that $\sin (β)$ and $\cos (β)$ are the $y$ and $x$ coordinates on a specific circle, the Unit Circle. As we'll see shortly, $(- 1, - 2)$ lies on a circle of $the square root of 5$ , so not the Unit Circle. ↩
We invite the reader to check this answer using the other two formulas. ↩
These are also known as the Prosthaphaeresis Formulas and have a rich history. The authors recommend that you conduct some research on them as your schedule allows. ↩
This should remind you of equation coefficients of like powers of $x$ in Section 9.6. ↩
Be careful here! ↩
The general equations to fit a function of the form $f (x) = a \cos (ω x) + b \sin (ω x) + B$ into one of the forms in Theorem 11.6 are explored in Exercise . ↩
Note: numbers and are the conversion formulas stated in Theorem 11.5 in Section 11.3. ↩
See Exercise in Section 11.4. ↩

Adapted from Precalculus, Preliminary 4th Edition (integrated calculus), by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.