12.1 The Pythagorean Identities

In section Section 11.4, we first encountered the concept of an identity when discussing Theorem 11.7. Recall that an identity is an equation which is true regardless of the choice of variable. Identities are important in mathematics because they facilitate changing forms.¹

We take a moment to generalize Theorem 11.7 below.

Theorem

Reciprocal and Quotient Identities: The following relationships hold for all angles $θ$ provided each side of each equation is defined.

$\sec (θ) = \frac{1}{\cos (θ)}$
$\cos (θ) = \frac{1}{\sec (θ)}$
$\csc (θ) = \frac{1}{\sin (θ)}$
$\sin (θ) = \frac{1}{\csc (θ)}$

$\tan (θ) = \frac{\sin (θ)}{\cos (θ)}$
$\cot (θ) = \frac{\cos (θ)}{\sin (θ)}$
$\cot (θ) = \frac{1}{\tan (θ)}$
$\tan (θ) = \frac{1}{\cot (θ)}$

It is important to remember that the equivalences stated in Theorem 12.1 are valid only when all quantities described therein are defined. As an example, $tan ⁡ open parenthesis 0 close parenthesis equals 0$ , but $\tan (0) \neq \frac{1}{\cot (0)}$ since $cot ⁡ open parenthesis 0 close parenthesis$ is undefined.

When it comes down to it, the Reciprocal and Quotient Identities amount to giving different ratios on the Unit Circle different names. The main focus of this section is on a more algebraic relationship between certain pairs of the circular functions: the Pythagorean Identities.

Recall in Definition 11.2, the cosine and sine of an angle is defined as the $x$ and $y$ -coordinate, respectively, of a point on the Unit Circle. Since the coordinates of all points $(x, y)$ on the Unit Circle satisfy the equation $x^{2} + y^{2} = 1$ , we get for all angles $θ$ , ${(\cos (θ))}^{2} + {(\sin (θ))}^{2} = 1$ . An unfortunate² convention, which the authors are compelled to perpetuate, is to write ${(\cos (θ))}^{2}$ as $\cos^{2} (θ)$ and ${(\sin (θ))}^{2}$ as $\sin^{2} (θ)$ . Rewriting the identity using this convention results in the following theorem, which is without a doubt one of the most important results in Trigonometry.

The moniker `Pythagorean' brings to mind the Pythagorean Theorem, from which both the Distance Formula and the equation for a circle are ultimately derived.³ The word `Identity' reminds us that, regardless of the angle $θ$ , the equation in Theorem 12.2 is always true.

If one of $\cos (θ)$ or $\sin (θ)$ is known, Theorem 12.2 can be used to determine the other, up to a ( $\pm$ ) sign. If, in addition, we know where the terminal side of $θ$ lies when in standard position, then we can remove the ambiguity of the ( $\pm$ ) and completely determine the missing value.⁴ We illustrate this approach in the following example.

Example 1

Use Theorem 12.2 and the given information to find the indicated value.

If $θ$ is a Quadrant II angle with $\sin (θ) = \frac{3}{5}$ , find $\cos (θ)$ .
If $π < t < \frac{3 π}{2}$ with $\cos (t) = - \frac{\sqrt{5}}{5}$ , find $\sin (t)$ .
If $sin ⁡ open parenthesis θ close parenthesis equals 1$ , find $\cos (θ)$ .

When we substitute $\sin (θ) = \frac{3}{5}$ into The Pythagorean Identity, $\cos^{2} (θ) + \sin^{2} (θ) = 1$ , we obtain $\cos^{2} (θ) + \frac{9}{25} = 1$ . Solving, we find $\cos (θ) = \pm \frac{4}{5}$ . Since $θ$ is a Quadrant II angle, we know $cos ⁡ open parenthesis θ close parenthesis less than 0$ . Hence, we select $\cos (θ) = - \frac{4}{5}$ .
Here we're using the variable $t$ instead $θ$ which usually corresponds to a real number variable instead of an angle. As usual, we associate real numbers $t$ with angles $θ$ measuring $t$ radians,⁵ so the the Pythagorean Identity works equally well for all real numbers $t$ as it does for all angles $θ$ .
Substituting $\cos (t) = - \frac{\sqrt{5}}{5}$ into $\cos^{2} (t) + \sin^{2} (t) = 1$ gives $\sin (t) = \pm \frac{2}{\sqrt{5}} = \pm \frac{2 \sqrt{5}}{5}$ . Since $π < t < \frac{3 π}{2}$ , we know $t$ corresponds to a Quadrant III angle, so $sin ⁡ open parenthesis t close parenthesis less than 0$ . Hence, $\sin (t) = - \frac{2 \sqrt{5}}{5}$ .
When we substitute $sin ⁡ open parenthesis θ close parenthesis equals 1$ into $\cos^{2} (θ) + \sin^{2} (θ) = 1$ , we find $cos ⁡ open parenthesis θ close parenthesis equals 0$ .

The reader is encouraged to compare and contrast the solution strategies demonstrated in Example 12.1.1 with those showcases in Examples 11.2.3 and 11.2.5 in Section 11.2.

As with many tools in mathematics, identities give us a different way to approach and solve problems.⁶ As always, the key is to determine which approach makes the most sense (is more efficient, for instance) in the given scenario.

Our next task is to use use the Reciprocal and Quotient Identities found in Theorem 12.1 coupled with the Pythagorean Identity found in Theorem 12.2 to derive new Pythagorean-like identities for the remaining four circular functions.

Assuming $cos ⁡ open parenthesis θ close parenthesis not equal to 0$ , we may start with $\cos^{2} (θ) + \sin^{2} (θ) = 1$ and divide both sides by $\cos^{2} (θ)$ to obtain $1 + \frac{\sin^{2} (θ)}{\cos^{2} (θ)} = \frac{1}{\cos^{2} (θ)}$ . Using properties of exponents along with the Reciprocal and Quotient Identities, this reduces to $1 + \tan^{2} (θ) = \sec^{2} (θ)$ .

If $sin ⁡ open parenthesis θ close parenthesis not equal to 0$ , we can divide both sides of the identity $\cos^{2} (θ) + \sin^{2} (θ) = 1$ by $\sin^{2} (θ)$ , apply Theorem 12.1 once again, and obtain $\cot^{2} (θ) + 1 = \csc^{2} (θ)$ .

These three Pythagorean Identities are worth memorizing and they, along with some of their other common forms, are summarized in the following theorem.

Theorem

The Pythagorean Identities:

$\cos^{2} (θ) + \sin^{2} (θ) = 1$ .
Common Alternate Forms:
- $1 - \sin^{2} (θ) = \cos^{2} (θ)$
- $1 - \cos^{2} (θ) = \sin^{2} (θ)$
$1 + \tan^{2} (θ) = \sec^{2} (θ)$ , provided $cos ⁡ open parenthesis θ close parenthesis not equal to 0$ .
Common Alternate Forms:
- $\sec^{2} (θ) - \tan^{2} (θ) = 1$
- $\sec^{2} (θ) - 1 = \tan^{2} (θ)$
$1 + \cot^{2} (θ) = \csc^{2} (θ)$ , provided $sin ⁡ open parenthesis θ close parenthesis not equal to 0$ .
Common Alternate Forms:
- $\csc^{2} (θ) - \cot^{2} (θ) = 1$
- $\csc^{2} (θ) - 1 = \cot^{2} (θ)$

As usual, the formulas states in Theorem 12.3 work equally well for (the applicable) angles as well as real numbers.

Example 2

Use Theorems 12.1 and 12.3 to find the indicated values.

If $θ$ is a Quadrant IV angle with $sec ⁡ open parenthesis θ close parenthesis equals 3$ , find $\tan (θ)$ .
Find $\csc (t)$ if $π < t < \frac{3 π}{2}$ and $cot ⁡ open parenthesis t close parenthesis equals 2$ .
If $θ$ is a Quadrant II angle with $\cos (θ) = - \frac{3}{5}$ , find the exact values of the remaining circular functions.

Per Theorem 12.3, $\tan^{2} (θ) = \sec^{2} (θ) - 1$ . Since $sec ⁡ open parenthesis θ close parenthesis equals 3$ , we have $\tan^{2} (θ) = {(3)}^{2} - 1 = 8$ , or $\tan (θ) = \pm \sqrt{8} = \pm 2 \sqrt{2}$ . Since $θ$ is a Quadrant IV angle, we know $tan ⁡ open parenthesis θ close parenthesis less than 0$ so $\tan (θ) = - 2 \sqrt{2}$ .
Again, using Theorem 12.3, we have $\csc^{2} (t) = 1 + \cot^{2} (t)$ , so we have $\csc^{2} (t) = 1 + {(2)}^{2} = 5$ . This gives $\csc (t) = \pm \sqrt{5}$ . Since $π < t < \frac{3 π}{2}$ , $t$ corresponds to a Quadrant III angle, so $\csc (t) = - \sqrt{5}$ .
With five function values to find, we have our work cut out for us. From Theorem 12.1, we know $\sec (θ) = \frac{1}{\cos (θ)}$ , so we (quickly) get $\sec (θ) = \frac{1}{- 3 / 5} = - \frac{5}{3}$ .
Next, we go after $\sin (θ)$ since between $\sin (θ)$ and $\cos (θ)$ , we can get all of the remaining values courtesy of Theorem 12.1.
From Theorem 12.3, we have $\sin^{2} (θ) = 1 - \cos^{2} (θ)$ , so $\sin^{2} (θ) = 1 - {(\frac{3}{5})}^{2} = 1 - \frac{9}{25} = \frac{16}{25}$ . Hence, $\sin (θ) = \pm \frac{4}{5}$ but since $θ$ is a Quadrant II angle, we select $\sin (θ) = \frac{4}{5}$ .
Back to Theorem 12.1, we get $\csc (θ) = \frac{1}{\sin (θ)} = \frac{1}{4 / 5} = \frac{5}{4}$ , $\tan (θ) = \frac{\sin (θ)}{\cos (θ)} = \frac{4 / 5}{- 3 / 5} = - \frac{4}{3}$ , and $\cot (θ) = \frac{\cos (θ)}{\sin (θ)} = \frac{- 3 / 5}{4 / 5} = - \frac{3}{4}$ .

Again, the reader is encouraged to study the solution methodology illustrated in Example 12.1.2 as compared with that employed in Example 11.4.2 in Section 11.4.

Trigonometric identities play an important role in not just Trigonometry, but in Calculus as well. We'll use them in this book to find the values of the circular functions of an angle and solve equations and inequalities. In Calculus, they are needed to simplify otherwise complicated expressions. In the next example, we make good use of the Theorems 12.1 and 12.3.

Example 3

Verify the following identities. Assume that all quantities are defined.

$\tan (θ) = \sin (θ) \sec (θ)$
$(\tan (t) - \sec (t)) (\tan (t) + \sec (t)) = - 1$
$\sin^{2} (x) \cos^{3} (x) = \sin^{2} (x) (1 - \sin^{2} (x)) \cos (x)$
$\frac{\sec (t)}{1 - \tan (t)} = \frac{1}{\cos (t) - \sin (t)}$
$6 \sec (x) \tan (x) = \frac{3}{1 - \sin (x)} - \frac{3}{1 + \sin (x)}$
$\frac{\sin (θ)}{1 - \cos (θ)} = \frac{1 + \cos (θ)}{\sin (θ)}$

In verifying identities, we typically start with the more complicated side of the equation and use known identities to transform it into the other side of the equation.

Starting with the right hand side of $\tan (θ) = \sin (θ) \sec (θ)$ , we use $\sec (θ) = \frac{1}{\cos (θ)}$ and find:
$\sin (θ) \sec (θ) = \sin (θ) \frac{1}{\cos (θ)} = \frac{\sin (θ)}{\cos (θ)} = \tan (θ),$
where the last equality is courtesy of Theorem 12.1.
Expanding the left hand side, we get: $(\tan (t) - \sec (t)) (\tan (t) + \sec (t)) = \tan^{2} (t) - \sec^{2} (t)$ . From Theorem 12.3, we know $\sec^{2} (t) - \tan^{2} (t) = 1$ , which isn't quite what we have. We are off by a negative sign (), so we factor it out:
$(\tan (t) - \sec (t)) (\tan (t) + \sec (t)) = \tan^{2} (t) - \sec^{2} (t) = (- 1) (\sec^{2} (t) - \tan^{2} (t)) = (- 1) (1) = - 1 .$
Starting with the right hand side,⁷ we notice we have a quantity we can immediately simplify per Theorem 12.3: $1 - \sin^{2} (x) = \cos^{2} (x)$ . This increases the number of factors of cosine, (which is part of our goal in looking at the left hand side), so we proceed:
$\sin^{2} (x) (1 - \sin^{2} (x)) \cos (x) = \sin^{2} (x) \cos^{2} (x) \cos (x) = \sin^{2} (x) \cos^{3} (x) .$
While both sides of our next identity contain fractions, the left side affords us more opportunities to use our identities.⁸ Substituting $\sec (t) = \frac{1}{\cos (t)}$ and $\tan (t) = \frac{\sin (t)}{\cos (t)}$ , we get:
$\begin{matrix} \frac{\sec (t)}{1 - \tan (t)} & = & \frac{\frac{1}{\cos (t)}}{1 - \frac{\sin (t)}{\cos (t)}} = \frac{\frac{1}{\cos (t)}}{1 - \frac{\sin (t)}{\cos (t)}} \cdot \frac{\cos (t)}{\cos (t)} \\ = & \frac{(\frac{1}{\cos (t)}) (\cos (t))}{(1 - \frac{\sin (t)}{\cos (t)}) (\cos (t))} = \frac{1}{(1) (\cos (t)) - (\frac{\sin (t)}{\cos (t)}) (\cos (t))} \\ = & \frac{1}{\cos (t) - \sin (t)}, \end{matrix}$
which is exactly what we had set out to show.
Starting with the right hand side, we can get started by obtaining common denominators to add:
$\begin{matrix} \frac{3}{1 - \sin (x)} - \frac{3}{1 + \sin (x)} & = & \frac{3 (1 + \sin (x))}{(1 - \sin (x)) (1 + \sin (x))} - \frac{3 (1 - \sin (x))}{(1 + \sin (x)) (1 - \sin (x))} \\ = & \frac{3 + 3 \sin (x)}{1 - \sin^{2} (x)} - \frac{3 - 3 \sin (x)}{1 - \sin^{2} (x)} \\ = & \frac{(3 + 3 \sin (x)) - (3 - 3 \sin (x))}{1 - \sin^{2} (x)} \\ = & \frac{6 \sin (x)}{1 - \sin^{2} (x)} \end{matrix}$
At this point, we have at least reduced the number of fractions from two to one, it may not be clear how to proceed. When this happens, it isn't a bad idea to start working with the other side of the identity to get some clues how to proceed.
Using a reciprocal and quotient identity, we find $6 \sec (x) \tan (x) = 6 (\frac{1}{\cos (x)}) (\frac{\sin (x)}{\cos (x)}) = \frac{6 \sin (x)}{\cos^{2} (x)}$ .
Theorem 12.3 tells us $1 - \sin^{2} (x) = \cos^{2} (x)$ , which means to our surprise and delight, we are much closer to our goal that we may have originally thought:
$\begin{matrix} \frac{3}{1 - \sin (x)} - \frac{3}{1 + \sin (x)} & = & \frac{6 \sin (x)}{1 - \sin^{2} (x)} = \frac{6 \sin (x)}{\cos^{2} (x)} \\ = & 6 (\frac{1}{\cos (x)}) (\frac{\sin (x)}{\cos (x)}) = 6 \sec (x) \tan (x) . \end{matrix}$
It is debatable which side of the identity is more complicated. One thing which stands out is that the denominator on the left hand side is $1 minus cos ⁡ open parenthesis θ close parenthesis$ , while the numerator of the right hand side is $1 plus cos ⁡ open parenthesis θ close parenthesis$ . This suggests the strategy of starting with the left hand side and multiplying the numerator and denominator by the quantity $1 plus cos ⁡ open parenthesis θ close parenthesis$ . Theorem 12.3 comes to our aid once more when we simplify $1 - \cos^{2} (θ) = \sin^{2} (θ)$ :
$\begin{matrix} \frac{\sin (θ)}{1 - \cos (θ)} & = & \frac{\sin (θ)}{(1 - \cos (θ))} \cdot \frac{(1 + \cos (θ))}{(1 + \cos (θ))} = \frac{\sin (θ) (1 + \cos (θ))}{(1 - \cos (θ)) (1 + \cos (θ))} \\ = & \frac{\sin (θ) (1 + \cos (θ))}{1 - \cos^{2} (θ)} = \frac{\sin (θ) (1 + \cos (θ))}{\sin^{2} (θ)} \\ = & \frac{\sin (θ) (1 + \cos (θ))}{\sin (θ) \sin (θ)} = \frac{1 + \cos (θ)}{\sin (θ)} \end{matrix}$

In Example 12.1.3 number above, we see that multiplying $1 minus cos ⁡ open parenthesis θ close parenthesis$ by $1 plus cos ⁡ open parenthesis θ close parenthesis$ produces a difference of squares that can be simplified to one term using Theorem 12.3.

This is exactly the same kind of phenomenon that occurs when we multiply expressions such as $1 - \sqrt{2}$ by $1 + \sqrt{2}$ or $3 minus 4 ⁢ i$ by $3 plus 4 ⁢ i$ . In algebra, these sorts of expressions were called `conjugates.'⁹

For this reason, the quantities $open parenthesis 1 minus cos ⁡ open parenthesis θ close parenthesis close parenthesis$ and $open parenthesis 1 plus cos ⁡ open parenthesis θ close parenthesis close parenthesis$ are called `Pythagorean Conjugates.' Below is a list of other common Pythagorean Conjugates.

Pythagorean Conjugates

$1 minus cos ⁡ open parenthesis θ close parenthesis$ and $1 plus cos ⁡ open parenthesis θ close parenthesis$ : $(1 - \cos (θ)) (1 + \cos (θ)) = 1 - \cos^{2} (θ) = \sin^{2} (θ)$
$1 minus sin ⁡ open parenthesis θ close parenthesis$ and $1 plus sin ⁡ open parenthesis θ close parenthesis$ : $(1 - \sin (θ)) (1 + \sin (θ)) = 1 - \sin^{2} (θ) = \cos^{2} (θ)$
$sec ⁡ open parenthesis θ close parenthesis minus 1$ and $sec ⁡ open parenthesis θ close parenthesis plus 1$ : $(\sec (θ) - 1) (\sec (θ) + 1) = \sec^{2} (θ) - 1 = \tan^{2} (θ)$
$\sec (θ) - \tan (θ)$ and $\sec (θ) + \tan (θ)$ : $(\sec (θ) - \tan (θ)) (\sec (θ) + \tan (θ)) = \sec^{2} (θ) - \tan^{2} (θ) = 1$
$csc ⁡ open parenthesis θ close parenthesis minus 1$ and $csc ⁡ open parenthesis θ close parenthesis plus 1$ : $(\csc (θ) - 1) (\csc (θ) + 1) = \csc^{2} (θ) - 1 = \cot^{2} (θ)$
$\csc (θ) - \cot (θ)$ and $\csc (θ) + \cot (θ)$ : $(\csc (θ) - \cot (θ)) (\csc (θ) + \cot (θ)) = \csc^{2} (θ) - \cot^{2} (θ) = 1$

Verifying trigonometric identities requires a healthy mix of tenacity and inspiration. You will need to spend many hours struggling with them just to become proficient in the basics.

Like many things in life, there is no short-cut here – there is no complete algorithm for verifying identities. Nevertheless, a summary of some strategies which may be helpful (depending on the situation) is provided below and ample practice is provided for you in the Exercises.

Strategies for Verifying Identities

Try working on the more complicated side of the identity.
Use the Reciprocal and Quotient Identities in Theorem 12.1 to write functions on one side of the identity in terms of the functions on the other side of the identity.
Simplify the resulting complex fractions.
Add rational expressions with unlike denominators by obtaining common denominators.
Use the Pythagorean Identities in Theorem 12.3 to `exchange' sines and cosines, secants and tangents, cosecants and cotangents, and simplify sums or differences of squares to one term.
Multiply numerator and denominator by Pythagorean Conjugates in order to take advantage of the Pythagorean Identities in Theorem 12.3.
If you find yourself stuck working with one side of the identity, try starting with the other side of the identity and see if you can find a way to bridge the two parts of your work.
Try something. The more you work with identities, the better you'll get with identities.

Exercises

In Exercises -, use the Reciprocal and Quotient Identities (Theorem 12.1) along with the Pythagorean Identities (Theorem 12.3), to find the value of the circular function requested below. (Find the exact value unless otherwise indicated.)

If $\sin (θ) = \frac{\sqrt{5}}{5}$ , find $\csc (θ)$ .
If $\sec (θ) = - 4$ , find $\cos (θ)$ .
If $tan ⁡ open parenthesis t close parenthesis equals 3$ , find $\cot (t)$ .
If $θ$ is a Quadrant IV angle with $\cos (θ) = \frac{5}{13}$ , find $\sin (θ)$ .
If $θ$ is a Quadrant III angle with $tan ⁡ open parenthesis θ close parenthesis equals 2$ , find $\sec (θ)$ .
If $\frac{π}{2} < t < π$ with $\cot (t) = - 2$ , find $\csc (t)$ .
If $sec ⁡ open parenthesis θ close parenthesis equals 3$ and $sin ⁡ open parenthesis θ close parenthesis less than 0$ , find $\tan (θ)$ .
If $\sin (θ) = - \frac{2}{3}$ but $tan ⁡ open parenthesis θ close parenthesis greater than 0$ , find $\cos (θ)$ .
If $0 < t < \frac{π}{2}$ and $sin ⁡ open parenthesis t close parenthesis equals 0.42$ , find $\cos (t)$ , rounded to four decimal places.
If $θ$ is Quadrant IV angle with $sec ⁡ open parenthesis θ close parenthesis equals 1.17$ , find $\tan (θ)$ , rounded to four decimal places.
If $π < t < \frac{3 π}{2}$ with $cot ⁡ open parenthesis t close parenthesis equals 4.2$ , find $\csc (t)$ , rounded to four decimal places.

In Exercises -, use the Reciprocal and Quotient Identities (Theorem 12.1) along with the Pythagorean Identities (Theorem 12.3), to find the exact values of the remaining circular functions. (Compare your methods with how you solved Exercises - in Section 11.4.)

$\sin (θ) = \frac{3}{5}$ with $θ$ in Quadrant II
$\tan (θ) = \frac{12}{5}$ with $θ$ in Quadrant III
$\csc (θ) = \frac{25}{24}$ with $θ$ in Quadrant I
$sec ⁡ open parenthesis θ close parenthesis equals 7$ with $θ$ in Quadrant IV
$\csc (θ) = - \frac{10 \sqrt{91}}{91}$ with $θ$ in Quadrant III
$\cot (θ) = - 23$ with $θ$ in Quadrant II
$\tan (θ) = - 2$ with $θ$ in Quadrant IV.
$\sec (θ) = - 4$ with $θ$ in Quadrant II.
$\cot (θ) = \sqrt{5}$ with $θ$ in Quadrant III.
$\cos (θ) = \frac{1}{3}$ with $θ$ in Quadrant I.
$cot ⁡ open parenthesis t close parenthesis equals 2$ with $0 < t < \frac{π}{2}$ .
$csc ⁡ open parenthesis t close parenthesis equals 5$ with $\frac{π}{2} < t < π$ .
$\tan (t) = \sqrt{10}$ with $π < t < \frac{3 π}{2}$ .
$\sec (t) = 2 \sqrt{5}$ with $\frac{3 π}{2} < t < 2 π$ .
Skippy claims $\cos (θ) + \sin (θ) = 1$ is an identity because when $θ equals 0$ , the equation is true. Is Skippy correct? Explain.

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

$cos ⁡ open parenthesis θ close parenthesis ⁢ sec ⁡ open parenthesis θ close parenthesis equals 1$
$\tan (t) \cos (t) = \sin (t)$
$sin ⁡ open parenthesis θ close parenthesis ⁢ csc ⁡ open parenthesis θ close parenthesis equals 1$
$tan ⁡ open parenthesis t close parenthesis ⁢ cot ⁡ open parenthesis t close parenthesis equals 1$
$\csc (x) \cos (x) = \cot (x)$
$\frac{\sin (t)}{\cos^{2} (t)} = \sec (t) \tan (t)$
$\frac{\cos (θ)}{\sin^{2} (θ)} = \csc (θ) \cot (θ)$
$\frac{1 + \sin (x)}{\cos (x)} = \sec (x) + \tan (x)$
$\frac{1 - \cos (θ)}{\sin (θ)} = \csc (θ) - \cot (θ)$
$\frac{\cos (t)}{1 - \sin^{2} (t)} = \sec (t)$
$\frac{\sin (x)}{1 - \cos^{2} (x)} = \csc (x)$
$\frac{\sec (t)}{1 + \tan^{2} (t)} = \cos (t)$
$\frac{\csc (θ)}{1 + \cot^{2} (θ)} = \sin (θ)$
$\frac{\tan (x)}{\sec^{2} (x) - 1} = \cot (x)$
$\frac{\cot (t)}{\csc^{2} (t) - 1} = \tan (t)$
$4 \cos^{2} (θ) + 4 \sin^{2} (θ) = 4$
$9 - \cos^{2} (t) - \sin^{2} (t) = 8$
$\tan^{3} (t) = \tan (t) \sec^{2} (t) - \tan (t)$
$\sin^{5} (x) = {(1 - \cos^{2} (x))}^{2} \sin (x)$
$\sec^{10} (t) = {(1 + \tan^{2} (t))}^{4} \sec^{2} (t)$
$\cos^{2} (x) \tan^{3} (x) = \tan (x) - \sin (x) \cos (x)$
$\sec^{4} (t) - \sec^{2} (t) = \tan^{2} (t) + \tan^{4} (t)$
$\frac{\cos (θ) + 1}{\cos (θ) - 1} = \frac{1 + \sec (θ)}{1 - \sec (θ)}$
$\frac{\sin (t) + 1}{\sin (t) - 1} = \frac{1 + \csc (t)}{1 - \csc (t)}$
$\frac{1 - \cot (x)}{1 + \cot (x)} = \frac{\tan (x) - 1}{\tan (x) + 1}$
$\frac{1 - \tan (t)}{1 + \tan (t)} = \frac{\cos (t) - \sin (t)}{\cos (t) + \sin (t)}$
$\tan (θ) + \cot (θ) = \sec (θ) \csc (θ)$
$\csc (t) - \sin (t) = \cot (t) \cos (t)$
$\cos (x) - \sec (x) = - \tan (x) \sin (x)$
$\cos (x) (\tan (x) + \cot (x)) = \csc (x)$
$\sin (t) (\tan (t) + \cot (t)) = \sec (t)$
$\frac{1}{1 - \cos (θ)} + \frac{1}{1 + \cos (θ)} = 2 \csc^{2} (θ)$
$\frac{1}{\sec (t) + 1} + \frac{1}{\sec (t) - 1} = 2 \csc (t) \cot (t)$
$\frac{1}{\csc (x) + 1} + \frac{1}{\csc (x) - 1} = 2 \sec (x) \tan (x)$
$\frac{1}{\csc (t) - \cot (t)} - \frac{1}{\csc (t) + \cot (t)} = 2 \cot (t)$
$\frac{\cos (θ)}{1 - \tan (θ)} + \frac{\sin (θ)}{1 - \cot (θ)} = \sin (θ) + \cos (θ)$
$\frac{1}{\sec (t) + \tan (t)} = \sec (t) - \tan (t)$
$\frac{1}{\sec (x) - \tan (x)} = \sec (x) + \tan (x)$
$\frac{1}{\csc (t) - \cot (t)} = \csc (t) + \cot (t)$
$\frac{1}{\csc (θ) + \cot (θ)} = \csc (θ) - \cot (θ)$
$\frac{1}{1 - \sin (x)} = \sec^{2} (x) + \sec (x) \tan (x)$
$\frac{1}{1 + \sin (t)} = \sec^{2} (t) - \sec (t) \tan (t)$
$\frac{1}{1 - \cos (θ)} = \csc^{2} (θ) + \csc (θ) \cot (θ)$
$\frac{1}{1 + \cos (x)} = \csc^{2} (x) - \csc (x) \cot (x)$
$\frac{\cos (t)}{1 + \sin (t)} = \frac{1 - \sin (t)}{\cos (t)}$
$\csc (θ) - \cot (θ) = \frac{\sin (θ)}{1 + \cos (θ)}$
$\frac{1 - \sin (x)}{1 + \sin (x)} = {(\sec (x) - \tan (x))}^{2}$

In Exercises -, verify the identity. You may need to consult Sections 1.3 and 7.3 for a review of the properties of absolute value and logarithms before proceeding.

$\ln | \sec (x) | = - \ln | \cos (x) |$
$- \ln | \csc (x) | = \ln | \sin (x) |$
$- \ln | \sec (x) - \tan (x) | = \ln | \sec (x) + \tan (x) |$
$- \ln | \csc (x) + \cot (x) | = \ln | \csc (x) - \cot (x) |$
1. What indeterminate form is present in the limit $lim θ → 0$ $\frac{1 - \cos (θ)}{θ}$ ?
2. Graph $f (θ) = \frac{1 - \cos (θ)}{θ}$ near $θ equals 0$ . What appears to be $lim θ → 0$ $\frac{1 - \cos (θ)}{θ}$ ?
3. Verify the identity: $\frac{1 - \cos (θ)}{θ} = \frac{\sin (θ)}{1 + \cos (θ)} \frac{\sin (θ)}{θ}$ .
4. Use the fact¹⁰ that $lim θ → 0$ $\frac{\sin (θ)}{θ} = 1$ along with part to help you find $lim θ → 0$ $\frac{1 - \cos (θ)}{θ}$ .

Answers

$\csc (θ) = \sqrt{5}$ .
$\cos (θ) = - \frac{1}{4}$ .
$\cot (t) = \frac{1}{3}$ .
$\sin (θ) = - \frac{12}{13}$ .
$\sec (θ) = - \sqrt{5}$ .
$\csc (t) = \sqrt{5}$ .
$\tan (θ) = - 2 \sqrt{2}$ .
$\cos (θ) = - \frac{\sqrt{5}}{3}$ .
$cos ⁡ open parenthesis t close parenthesis approximately equals 0.9075$ .
$\tan (θ) \approx - 0.6074$ .
$\csc (t) \approx - 4.3174$ .
$\sin (θ) = \frac{3}{5}, \cos (θ) = - \frac{4}{5}, \tan (θ) = - \frac{3}{4}, \csc (θ) = \frac{5}{3}, \sec (θ) = - \frac{5}{4}, \cot (θ) = - \frac{4}{3}$
$\sin (θ) = - \frac{12}{13}, \cos (θ) = - \frac{5}{13}, \tan (θ) = \frac{12}{5}, \csc (θ) = - \frac{13}{12}, \sec (θ) = - \frac{13}{5}, \cot (θ) = \frac{5}{12}$
$\sin (θ) = \frac{24}{25}, \cos (θ) = \frac{7}{25}, \tan (θ) = \frac{24}{7}, \csc (θ) = \frac{25}{24}, \sec (θ) = \frac{25}{7}, \cot (θ) = \frac{7}{24}$
$\sin (θ) = \frac{- 4 \sqrt{3}}{7}, \cos (θ) = \frac{1}{7}, \tan (θ) = - 4 \sqrt{3}, \csc (θ) = - \frac{7 \sqrt{3}}{12}, \sec (θ) = 7, \cot (θ) = - \frac{\sqrt{3}}{12}$
$\sin (θ) = - \frac{\sqrt{91}}{10}, \cos (θ) = - \frac{3}{10}, \tan (θ) = \frac{\sqrt{91}}{3}, \csc (θ) = - \frac{10 \sqrt{91}}{91}, \sec (θ) = - \frac{10}{3}, \cot (θ) = \frac{3 \sqrt{91}}{91}$
$\sin (θ) = \frac{\sqrt{530}}{530}, \cos (θ) = - \frac{23 \sqrt{530}}{530}, \tan (θ) = - \frac{1}{23}, \csc (θ) = \sqrt{530}, \sec (θ) = - \frac{\sqrt{530}}{23}, \cot (θ) = - 23$
$\sin (θ) = - \frac{2 \sqrt{5}}{5}, \cos (θ) = \frac{\sqrt{5}}{5}, \tan (θ) = - 2, \csc (θ) = - \frac{\sqrt{5}}{2}, \sec (θ) = \sqrt{5}, \cot (θ) = - \frac{1}{2}$
$\sin (θ) = \frac{\sqrt{15}}{4}, \cos (θ) = - \frac{1}{4}, \tan (θ) = - \sqrt{15}, \csc (θ) = \frac{4 \sqrt{15}}{15}, \sec (θ) = - 4, \cot (θ) = - \frac{\sqrt{15}}{15}$
$\sin (θ) = - \frac{\sqrt{6}}{6}, \cos (θ) = - \frac{\sqrt{30}}{6}, \tan (θ) = \frac{\sqrt{5}}{5}, \csc (θ) = - \sqrt{6}, \sec (θ) = - \frac{\sqrt{30}}{5}, \cot (θ) = \sqrt{5}$
$\sin (θ) = \frac{2 \sqrt{2}}{3}, \cos (θ) = \frac{1}{3}, \tan (θ) = 2 \sqrt{2}, \csc (θ) = \frac{3 \sqrt{2}}{4}, \sec (θ) = 3, \cot (θ) = \frac{\sqrt{2}}{4}$
$\sin (t) = \frac{\sqrt{5}}{5}, \cos (t) = \frac{2 \sqrt{5}}{5}, \tan (t) = \frac{1}{2}, \csc (t) = \sqrt{5}, \sec (t) = \frac{\sqrt{5}}{2}, \cot (t) = 2$
$\sin (t) = \frac{1}{5}, \cos (t) = - \frac{2 \sqrt{6}}{5}, \tan (t) = - \frac{\sqrt{6}}{12}, \csc (t) = 5, \sec (t) = - \frac{5 \sqrt{6}}{12}, \cot (t) = - 2 \sqrt{6}$
$\sin (t) = - \frac{\sqrt{110}}{11}, \cos (t) = - \frac{\sqrt{11}}{11}, \tan (t) = \sqrt{10}, \csc (t) = - \frac{\sqrt{110}}{10}, \sec (t) = - \sqrt{11}, \cot (t) = \frac{\sqrt{10}}{10}$
$\sin (t) = - \frac{\sqrt{95}}{10}, \cos (t) = \frac{\sqrt{5}}{10}, \tan (t) = - \sqrt{19}, \csc (t) = - \frac{2 \sqrt{95}}{19}, \sec (t) = 2 \sqrt{5}, \cot (t) = - \frac{\sqrt{19}}{19}$
No, Skippy is not correct. In order to be an identity, an equation must hold for all applicable angles. For example, $\cos (θ) + \sin (θ) = 1$ does not hold when $θ = π$ .
1. As $θ → 0$ , $\frac{1 - \cos (θ)}{θ} \to \frac{0}{0}$ .
2. The graph of $f (θ) = \frac{1 - \cos (θ)}{θ}$ approaches $open parenthesis 0, 0 close parenthesis$ , so $lim θ → 0$ $\frac{1 - \cos (θ)}{θ}$ appears to be $0$ .
3. $lim θ → 0$ $\frac{1 - \cos (θ)}{θ}$ $equals lim θ → 0$ $\frac{\sin (θ)}{1 + \cos (θ)} \frac{\sin (θ)}{θ} = (\frac{0}{1 + \cos (0)}) (1) = (\frac{0}{2}) (1) = 0$ .

We've seen the utility of changing form throughout the text, most recently when we completed the square in Chapter 8 to put general quadratic equations into standard form in order to graph them. ↩
This is unfortunate from a `function notation' perspective. See Section 12.3. ↩
See Sections A.3 and 8.3 for details. ↩
See the illustration following Example 11.4.1 to refresh yourself which circular functions are positive in which quadrants. ↩
See page the related section if you need a review of how we associate real numbers with angles in radian measure. ↩
For example, factoring, completing the square, and the quadratic formula are three different (yet equivalent) ways to solve a quadratic equation. See Section A.10 for a refresher. ↩
We hope by this point a shift of variable to ` $x$ ' instead of ` $θ$ ' or ` $t$ ' is a non-issue. ↩
Or, to put to another way, earn more partial credit if this were an exam question! ↩
See Sections A.11 and A.13. ↩
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.