5.3 Trigonometric Identities

What is an Identity?

Recall that an equation may be true or false, depending on the values of any variables involved. For example, the equation

$x^{2} + 3 x = 10$

is true only if $x equals 2$ or $x equals 5$ . An equation that is true only for certain values of the variable, and false for others, is called a conditional equation. When you solve a conditional equation, you are finding the values of the variable that make the equation true.

Some equations are true for all legitimate values of the variables. Such equations are called identities. Here are some examples of identities.

$\begin{aligned} 3 (x + y) & = 3 x + 3 y \\ (x + 1)^{2} & = x^{2} + 2 x + 1 \end{aligned}$

In an identity, the expressions on either side of the equal sign are equivalent expressions, because they have the same value for all values of the variable. When we simplify an algebraic expression, we are in fact creating an identity.

Which of the following equations are identities?

$(c - s) (c + s) = c^{2} - s^{2}$
$3 t squared equals 1$
$(2 c + 1) + (s - 3) = 2 c s - 6 c + s - 3$

(a) and (c) are identities.

Of course, you wouldn't be asked to solve an identity, because all values of the variable are solutions. Instead, we use identities to replace one form of an expression by a more useful form. You do this when you solve a quadratic equation by factoring. For example,

$\begin{aligned} 2 x^{2} - x - 1 & = 0 & Factor the left side. \\ (2 x + 1) (x - 1) & = 0 \end{aligned}$

Because $(2 x + 1) (x - 1)$ is equivalent to $2 x^{2} - x - 1$ , we have not changed the equation or its solutions. But now we can apply the Zero Factor principle and solve the equation.

Using identities is especially useful when we are working with trigonometric ratios.

Using Trigonometric Ratios in Identities

Because the identity

$2 x^{2} - x - 1 = (2 x + 1) (x - 1)$

is true for any value of $x$ , it is true when $x$ is replaced, for instance, by $\cos (θ)$ . This gives us a new identity

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

Expressions involving $\sin (θ), \cos (θ), or \tan (θ)$ can be manipulated by the same rules (such as the distributive law or the laws of exponents) that we use with simple variables.

Example 2

Which of the following equations are identities?

$3 \sin (θ) + 7 \sin (θ) = 10 \sin (θ)$
$5 \cos (θ) (\cos (θ) - 2 \sin (θ)) = 5 (\cos (θ))^{2} - 10 \cos (θ) \sin (θ)$
$2 \tan (θ) - 1 = 3$

These equations are obtained from the equations in the previous Example by replacing the variable $s$ with $\sin (θ)$ , the variable $c$ with $\cos (θ)$ , and the variable $t$ with $\tan (θ)$ . Equations (a) and (b) are identities, for the same reasons as before, and equation (c) is not. It is false, for example, when $θ equals 0$ .

Which of the following equations are identities?

$(\cos (θ) - \sin (θ)) (\cos (θ) + \sin (θ)) = \cos^{2} (θ) - \sin^{2} (θ)$
$3 tan squared ⁡ open parenthesis θ close parenthesis equals 1$
$(2 \cos (θ) + 1) + (\sin (θ) - 3) = 2 \cos (θ) \sin (θ) - 6 \cos (θ) + \sin (θ) - 3$

(a) and (c) are identities.

Checking Identities Graphically

Is the equation $\sqrt{x^{2}} = x$ an identity? The answer is no! Even though the equation is true for all positive values of $x$ , it is false for negative values of $x$ . For example, if $x = - 3$ , then

$\sqrt{x^{2}} = \sqrt{(- 3)^{2}} = \sqrt{9} = 3$

so $\sqrt{x^{2}} \neq x$ . Because the radical symbol $\sqrt{}$ stands for the nonnegative square root, the left side of the equation, $\sqrt{x^{2}}$ , is never negative. Therefore, $\sqrt{x^{2}}$ cannot equal $x$ when $x$ is a negative number. The equation is false for $x less than 0$ .

One way to see that $\sqrt{x^{2}}$ and $x$ are not equivalent is to compare the graphs of $Y_{1} = \sqrt{x^{2}}$ and $Y sub 2 equals x$ , shown below. You can see that $\sqrt{x^{2}}$ and $x$ do not have the same value for $x less than 0$ .

Thus, to check to whether an equation is an identity, we can compare graphs of $Y sub 1 equals$ (left side of the equation) and $Y sub 2 equals$ (right side of the equation). If the two graphs are identical, the equation is an identity. If the two graphs are not the same, the equation is not an identity.

Example 3

Which of the following equations are identities?

$sin ⁡ open parenthesis 2 α close parenthesis equals 2 sin ⁡ open parenthesis α close parenthesis$
$\cos (x + 1) = \cos (x)$

Compare the graphs of $Y sub 1 equals sin ⁡ open parenthesis 2 x close parenthesis$ and $Y sub 2 equals 2 sin ⁡ open parenthesis x close parenthesis$ . Enter the two equations in the ZTrig window and press ZOOM $7$ to see the graphs shown below.

To help distinguish which graph is which, we can use the calculator's trace feature, illustrated in the figure at right. Because there are two distinct graphs, the expressions $sin ⁡ open parenthesis 2 x close parenthesis$ and $2 sin ⁡ open parenthesis x close parenthesis$ are not equivalent, and consequently $sin ⁡ open parenthesis 2 α close parenthesis equals 2 sin ⁡ open parenthesis α close parenthesis$ is not an identity.
This time we graph $Y_{1} = \cos (x + 1)$ and $Y sub 2 equals cos ⁡ open parenthesis x close parenthesis$ . Although the graphs appear to be identical, when we use the trace feature we see that their $y$ -values differ at $x equals 0$ (and at many other values of $x$ .) See the figures below.

The graphs are so close together that calculator's resolution does not distinguish them, but tracing the graphs reveals that they are not identical. Because the two graphs differ, the equation $\cos (x + 1) = \cos (x)$ is not an identity.

Use graphs to decide which of the following equations are identities.

$cos ⁡ open parenthesis 2 θ close parenthesis equals 2 cos ⁡ open parenthesis θ close parenthesis$
$\cos (2 θ) = \cos^{2} (θ) - \sin^{2} (θ)$
$\cos (θ^{2}) = \cos^{2} (θ)$

(b) is an identity; (a) and (c) are not identities.

Pythagorean Identity

All of the trigonometric functions are related. This is plausible when you think about it, because the three trig ratios all involve the three sides of a right triangle. Taking advantage of these relationships will simplify many calculations involving the trig functions.

We'll begin by considering the relationship between $\sin (θ)$ and $\cos (θ)$ . Complete the following table with exact values.

$θ$	$\cos (θ)$	$\sin (θ)$	$\cos^{2} (θ)$	$\sin^{2} (θ)$	$\cos^{2} (θ) + \sin^{2} (θ)$
$0 degrees$
$30 degrees$
$45 degrees$
$60 degrees$
$90 degrees$

You should find that all the entries in the last column are 1.

For all of the angles in the table,

$\cos^{2} (θ) + \sin^{2} (θ) = 1$

We can verify that this equation holds for all angles by graphing the expressions on either side of the equal sign. The graphs of these two functions in the ZTrig window are shown at right.

We see that the graph of $Y_{1} = \cos^{2} (θ) + \sin^{2} (θ)$ appears identical to the horizontal line $Y sub 2 equals 1$ . In fact, the graphs are identical, and the equation $\cos^{2} (θ) + \sin^{2} (θ) = 1$ is an identity. It is important enough to earn a special name.

As you might guess from its name, the Pythagorean identity is true because it is related to the Pythagorean theorem. We have not actually proved the identity, and a skeptical student may wonder if $\cos^{2} (θ) + \sin^{2} (θ)$ is only very close to $1$ , or if it equals $1$ for only some values of $θ$ . Homework Problem 77 offers a proof of the Pythagorean identity.

Evaluate $5 open parenthesis cos squared ⁡ open parenthesis 12 degrees close parenthesis plus sin squared ⁡ open parenthesis 12 degrees close parenthesis close parenthesis$ without using a calculator.

$5$

When we solve more complicated trigonometric equations in later chapters, we will need to simplify trigonometric expressions so that they involve only one of the trig functions. The Pythagorean identity is useful when we wish to write an equivalent expression for either $\cos^{2} (θ)$ or for $\sin^{2} (θ)$ . Note that we can write the identity in two alternate forms:

Note

Alternate Forms of the Pythagorean Identity

$\begin{aligned} \cos^{2} (θ) & = 1 - \sin^{2} (θ) \\ \sin^{2} (θ) & = 1 - \cos^{2} (θ) \end{aligned}$

Example 5

Rewrite $\sin (θ) \cos^{2} (θ)$ as an expression involving $\sin (θ)$ only.

Using one of the alternate forms of the Pythagorean identity, we replace $\cos^{2} (θ)$ by $1 minus sin squared ⁡ open parenthesis θ close parenthesis$ to get

$\begin{aligned} \sin (θ) \cos^{2} (θ) & = \sin (θ) (1 - \sin^{2} (θ)) & Apply the distributive law. \\ = \sin (θ) - \sin^{3} (θ) \end{aligned}$

In the previous Example, we have shown that

$\sin (θ) \cos^{2} (θ) = \sin (θ) - \sin^{3} (θ)$

is an identity.

If we graph

$\begin{aligned} Y_{1} & = \sin (X) \cos^{2} (X) \\ Y_{2} & = \sin (X) - \sin^{3} (X) \end{aligned}$

for $minus 360 degrees less than or equal to X less than or equal to 360 degrees,$ we see that they have the same graph, as shown in the figure.

Rewrite $\sin^{2} (α) \cos^{2} (α)$ as an expression in $\cos (α)$ .
Verify your identity by graphing.

Replace $\sin^{2} (α)$ by $1 minus cos squared ⁡ open parenthesis α close parenthesis$ to get $open parenthesis 1 minus cos squared ⁡ open parenthesis α close parenthesis close parenthesis cos squared ⁡ open parenthesis α close parenthesis$ and apply the distributive law to get $\cos^{2} (α) - \cos^{4} (α)$ .

Tangent Identity

There is also a relationship between the tangent ratio and the sine and cosine. Complete the following table with exact values. You should find that the entries in the last two columns are identical.

$θ$	$\cos (θ)$	$\sin (θ)$	$\frac{\sin (θ)}{\cos (θ)}$	$\tan (θ)$
$0 degrees$
$30 degrees$
$45 degrees$
$60 degrees$

For the angles in the table,

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

Because $cos ⁡ open parenthesis 90 degrees close parenthesis equals 0$ , the quotient $\frac{\sin (θ)}{\cos (θ)}$ is undefined for $θ equals 90 degrees$ , and $tan ⁡ 90 degrees$ is undefined, too. The same is true for $θ equals 270 degrees$ . For all other angles between $0 degrees$ and $360 degrees$ , $\tan (θ) = \frac{\sin (θ)}{\cos (θ)}$ . A proof of the tangent identity is outlined in Homework Problem 78.

Now we can see how to use identities to simplify trigonometric expressions. One strategy for simplifying a trigonometric expression is to reduce the number of different trig ratios involved. We can use the tangent identity to replace the tangent ratio by sines and cosines.

Example 6

Simplify the expression $\cos (θ) \tan (θ) + \sin (θ)$ .

We apply the tangent identity to replace $\tan (θ)$ by $\frac{\sin (θ)}{\cos (θ)}$ and obtain

$\begin{aligned} \cos (θ) \tan (θ) + \sin (θ) & = \cos (θ) (\frac{\sin (θ)}{\cos (θ)}) + \sin (θ) \\ = \sin (θ) + \sin (θ) = 2 \sin (θ) \end{aligned}$

Simplify the expression $(1 + \tan^{2} (x)) (1 - \cos^{2} (x))$ .

Multiply the binomials to obtain

$1 - \cos^{2} (x) + \tan^{2} (x) - \tan^{2} (x) \cos^{2} (x)$

Because $\tan (x) = \frac{\sin (x)}{\cos (x)}$ , we can replace the second $\tan^{2} (x)$ by $\frac{\sin^{2} (x)}{\cos^{2} (x)}$ , leaving us with

$1 - \cos^{2} (x) + \tan^{2} (x) - \sin^{2} (x)$

Finally, because $\sin^{2} (x) + \cos^{2} (x) = 1$ , we are left with $\tan^{2} (x)$ .

Trig Ratios are Related

All three of the trigonometric functions of an angle are related. If we know the value of one of the three, we can calculate the other two (up to sign) by using the Pythagorean and tangent identities. We do not need to find the angle itself in order to do this. We need only know in which quadrant the angle lies to determine the correct sign for the trig ratios.

Example 7

If $\sin (θ) = \frac{7}{25}$ and $90 degrees less than θ less than 180 degrees,$ find $\cos (θ)$ and $\tan (θ)$ .

We substitute $\sin (θ) = \frac{7}{25}$ into the Pythagorean identity $\cos^{2} (θ) + \sin^{2} (θ) = 1$ to get

$\begin{aligned} \cos^{2} (θ) + {(\frac{7}{25})}^{2} & = 1 & Isolate \cos^{2} (θ) . \\ \cos^{2} (θ) & = 1 - {(\frac{7}{25})}^{2} & 1 - \frac{49}{625} = \frac{625 - 49}{625} \\ \cos^{2} (θ) & = \frac{576}{625} & Take square roots. \\ \cos θ & = \pm \sqrt{\frac{576}{625}} \\ = \pm \frac{24}{25} \end{aligned}$

Because $θ$ is between $90 degrees$ and $180 degrees$ , $\cos (θ)$ must be negative, so $\cos (θ) = \frac{- 24}{25}$ . To find $\tan (θ)$ , we use the tangent identity.

$\tan (θ) = \frac{\sin (θ)}{\cos (θ)} = \frac{7 / 25}{- 24 / 25} = \frac{- 7}{24}$

We can also work the previous Example by sketching an appropriate triangle. We begin by drawing an obtuse angle $θ$ and its reference triangle. Because $\sin (θ) = \frac{7}{25}$ , we can choose a point on the terminal side to have $y$ -coordinate 7 and $r equals 25$ , as shown below. In order to calculate $\cos (θ)$ and $\tan (θ)$ , we must find the $x$ -coordinate of the point. By the Pythagorean theorem,

$\begin{aligned} x^{2} + 7^{2} & = 25^{2} & Isolate x^{2} . \\ x^{2} & = 576 & Take square roots. \\ x & = - 24 \end{aligned}$

Note that $x$ is negative because the point $open parenthesis x, 7 close parenthesis$ is in the second quadrant. Then

$\cos (θ) = \frac{x}{r} = \frac{- 24}{25} and \tan (θ) = \frac{y}{x} = \frac{7}{- 24} = \frac{- 7}{24}$

If $\cos (θ) = \frac{5}{13}$ and $minus 90 degrees less than θ less than 0 degrees,$ find $\sin (θ)$ and $\tan (θ)$ .
Find $\sin (θ)$ and $\tan (θ)$ by using an appropriate sketch.

$y^{2} = 13^{2} - 5^{2}$ , so $y = - 12$ . Thus, $\sin (θ) = \frac{- 12}{13}$ , and $\tan (θ) = \frac{- 12}{5}$

Solving Equations

Now we'll see how identities are useful for solving trigonometric equations. So far we have only solved equations that involve a single trigonometric ratio. If the equation involves more than one trig function, we use identities to rewrite the equation in terms of a single trig function.

Example 8

Solve $\cos^{2} (A) - \sin^{2} (A) - \sin (A) = 0$ for $0 degrees less than or equal to A less than or equal to 360 degrees$ .

The equation involves both the cosine and sine functions, and we will rewrite the left side in terms of the sine only. To eliminate the cosines, we use the Pythagorean identity $\cos^{2} (A) = 1 - \sin^{2} (A)$ .

$\begin{aligned} \cos^{2} (A) - \sin^{2} (A) - \sin (A) & = 0 & \begin{matrix} Replace \cos^{2} (A) by \\ 1 - \sin^{2} (A) . \end{matrix} \\ (1 - \sin^{2} (A)) - \sin^{2} (A) - \sin (A) & = 0 & Combine like terms. \\ 1 - 2 \sin^{2} (A) - \sin (A) & = 0 & Multiply both sides by - 1 . \\ 2 \sin^{2} (A) + \sin (A) - 1 & = 0 & Factor the left side. \\ (2 \sin (A) - 1) (\sin (A) + 1) & = 0 \end{aligned}$

Set each factor equal to zero to see that either $\sin (A) = \frac{1}{2}$ or $\sin (A) = - 1$ . Now, $\sin (A) = \frac{1}{2}$ when $A equals 30 degrees$ or $A equals 150 degrees$ , and $\sin (A) = - 1$ when $A equals 270 degrees$ . The solutions are $A = 30^{\circ}, A = 150^{\circ}$ , or $A equals 270 degrees$ .

Solve $1 - \sin^{2} (θ) + \cos (θ) = 0$ for $0 degrees less than or equal to θ less than 360 degrees$ .

Replace $1 minus sin squared ⁡ open parenthesis θ close parenthesis$ by $\cos^{2} (θ)$ to get the equation

$\cos^{2} (θ) + \cos (θ) = 0$

Solve this quadratic equation by factoring to find $cos ⁡ open parenthesis θ close parenthesis equals 0$ and $\cos (θ) = - 1$ . The solutions are $90 degrees$ , $180 degrees$ , and $270 degrees$

Like the Pythagorean identity, the tangent identity can be helpful in solving trigonometric equations.

Example 9

Solve $\cos (B) + \sqrt{3} \sin (B) = 0$ for $0 degrees less than or equal to B less than or equal to 360 degrees$ .

We first try to write the equation in terms of a single trig function.

$\begin{aligned} \sqrt{3} \sin (B) & = - \cos (B) & Divide both sides by \cos (B) and \sqrt{3} . \\ \frac{\sin (B)}{\cos (B)} & = \frac{- 1}{\sqrt{3}} & Apply the tangent identity. \\ \tan (B) & = \frac{- 1}{\sqrt{3}} \end{aligned}$

Because $\tan (30^{\circ}) = \frac{- 1}{\sqrt{3}}$ , we see that $B$ has a reference angle of $30 degrees$ . And because the tangent is negative in the second and fourth quadrants, we have $B equals 150 degrees$ or $B equals 330 degrees$ , as shown in the figure.

Solve $2 \tan (β) \cos (β) - \sqrt{3} = 0$ for $0 degrees less than or equal to β less than 360 degrees$ .

Because $\tan (β) = \frac{\sin (β)}{\cos (β)}$ , the equation reduces to $\sin (β) = \sqrt{3}$ . The solutions are $60 degrees$ and $120 degrees$ .

Proving Identities

When we show that one trigonometric expression is equivalent to another expression, we have proved a trigonometric identity. In a previous example we proved that the equation

$\cos (θ) \tan (θ) + \sin (θ) = 2 \sin (θ)$

is an identity; it is true for all values of $θ$ . A common strategy for proving an identity is to transform one side of the equation using equivalent expressions until it is identical to the other side. To help us choose the transformations at each step of the proof, we try to match the algebraic form of the final expression.

Example 10

Prove the identity $1 + \tan^{2} (t) = \frac{1}{\cos^{2} (t)}$

By manipulating the left side of the equation, we will show that the expression $1 plus tan squared ⁡ open parenthesis t close parenthesis$ is equivalent to $\frac{1}{\cos^{2} (t)}$ . First, we use the tangent identity to write the expression in terms of sines and cosines:

$1 + \tan^{2} (t) = 1 + {(\frac{\sin (t)}{\cos (t)})}^{2} = 1 + \frac{\sin^{2} (t)}{\cos^{2} (t)}$

Next, we notice that the right side of the proposed identity has only one term, so we combine the terms on the left side. So that the fractions have the same denominator, we write 1 as $\frac{\cos^{2} (t)}{\cos^{2} (t)}$ .

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

Finally, we apply the Pythagorean identity to the numerator.

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

Thus, $1 + \tan^{2} (t) = \frac{1}{\cos^{2} (t)}$ , and the identity is proved.

When you write out the proof of an identity, you should transform the expression on one side of the identity into the expression on the other side, showing one step of the calculation on each line of your proof. You can provide a justification for each step to the right of the calculation. The proof of the identity in the previous example would look like this:

$\begin{aligned} 1 + \tan^{2} (t) & = 1 + {(\frac{\sin (t)}{\cos (t)})}^{2} & Replace \tan (t) by \frac{\sin (t)}{\cos (t)} . \\ = 1 + \frac{\sin^{2} (t)}{\cos^{2} (t)} & Square the fraction. \\ = \frac{\cos^{2} (t)}{\cos^{2} (t)} + \frac{\sin^{2} (t)}{\cos^{2} (t)} & Replace 1 by \frac{\cos^{2} (t)}{\cos^{2} (t)} . \\ = \frac{\cos^{2} (t) + \sin^{2} (t)}{\cos^{2} (t)} & Add fractions. \\ = \frac{1}{\cos^{2} (t)} & Apply Pythagorean identity. \end{aligned}$

Prove the identity $2 \cos^{2} (x) - 1 = 1 - 2 \sin^{2} (x)$

$\begin{aligned} 2 \cos^{2} (x) - 1 & = 2 (1 - \sin^{2} (x)) \\ = 2 - 2 \sin^{2} (x) - 1 \\ = 1 - 2 \sin^{2} (x) \end{aligned}$

Review the following skills you will need for this section.

Algebra Refresher

Write each expression as a single fraction in simplest form.

$\frac{1}{x} + \frac{1}{y}$
$1 - \frac{1}{x}$
$\frac{1}{x - 1} + \frac{1}{x + 1}$
$x - \frac{x}{x + 1}$
$\frac{\frac{y}{x} - \frac{x}{y}}{\frac{y}{x} + 1}$
$\frac{1 - \frac{x^{2}}{y^{2}}}{1 + \frac{x^{2}}{y^{2}}}$
$\frac{\frac{2 a}{x}}{1 - \frac{a^{2}}{x^{2}}}$
$\frac{\frac{a}{x} + \frac{b}{y}}{1 - \frac{a b}{x y}}$

$\underset{―}{}$

Algebra Refresher Answers

$\frac{x + y}{x y}$
$\frac{x - 1}{x}$
$\frac{2 x}{x^{2} - 1}$
$\frac{x^{2}}{x + 1}$
$\frac{y - x}{y}$
$\frac{y^{2} - x^{2}}{y^{2} + x^{2}}$
$\frac{2 a x}{x^{2} - a^{2}}$
$\frac{a y + b x}{x y - a b}$

Section 5.3 Summary

Vocabulary

Conditional equation
Identity
Equivalent expressions

Concepts

An equation that is true only for certain values of the variable, and false for others, is called a conditional equation. An equation that is true for all legitimate values of the variables is called an identity.
The expressions on either side of the equal sign in an identity are called equivalent expressions, because they have the same value for all values of the variable.
We often use identities to replace one form of an expression by a more useful form.
To check to whether an equation is an identity we can compare graphs of $Y sub 1 equals$ (left side of the equation) and $Y sub 2 equals$ (right side of the equation). If the two graphs agree, the equation is an identity. If the two graphs are not the same, the equation is not an identity.
Note
Pythagorean Identity
For any angle $θ$ ,
$\cos^{2} (θ) + \sin^{2} (θ) = 1$
Alternate forms:
$\begin{aligned} \cos^{2} (θ) & = 1 - \sin^{2} (θ) \\ \sin^{2} (θ) & = 1 - \cos^{2} (θ) \end{aligned}$
Note
Tangent Identity
For any angle not coterminal with $90 degrees$ or $270 degrees$ ,
$\tan (θ) = \frac{\sin (θ)}{\cos (θ)}$
To solve an equation involving more than one trig function, we use identities to rewrite the equation in terms of a single trig function.
To prove an identity, we write one side of the equation in equivalent forms until it is identical to the other side of the equation.

Study Questions

What is the difference between a conditional equation and an identity? Give an example of each.
What happens when you try to "solve" an identity?
Explain how to use graphs to verify an identity.
Delbert claims that since $\cos^{2} (θ) + \sin^{2} (θ) = 1$ , we can take the square root of both sides to get the simpler form $\cos (θ) + \sin (θ) = 1$ . Is he correct? Why or why not?
If $\frac{a}{b} = \frac{2}{5}$ , is it necessarily true that $a equals 2$ and $b equals 5$ ? Explain. If $\tan (θ) = \frac{3}{11}$ , is it true that $sin ⁡ open parenthesis θ close parenthesis equals 3$ and $cos ⁡ open parenthesis θ close parenthesis equals 11$ ?

Skills

Recognize identities #1–16, 41–46
Verify identities #17–26, 73–78
Rewrite expressions using identities #27–34, 47–50
Use identities to evaluate expressions #35–40
olve trigonometric equations #51–58, 67–72
Given one trig ratio, find the others #59–72

Homework 5.3

For Problems 1–8, decide which of the following equations are identities. Explain your reasoning.

${(\sqrt{a} + \sqrt{b})}^{2} = a + b$

not an identity

$\sqrt{a^{2} - b^{2}} = a - b$

$\frac{1}{a + b} = \frac{1}{a} + \frac{1}{b}$

not an identity

$\frac{a + b}{a} = b$

$\tan (α + β) = \frac{\sin (α + β)}{\cos (α + β)}$

identity

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

$(1 + \tan (θ))^{2} = 1 + \tan^{2} (θ)$

not an identity

$\sqrt{1 - \sin^{2} (ϕ)} = 1 - \sin (ϕ)$

For Problems 9–16, use graphs to decide which of the following equations are identities.

$sin ⁡ open parenthesis 2 t close parenthesis equals 2 sin ⁡ open parenthesis t close parenthesis$

not an identity

$\cos (θ) + \sin (θ) = 1$

$\sin (30^{\circ} + β) = \frac{1}{2} + \sin (β)$

not an identity

$\cos (90^{\circ} - C) = \sin C$

$\tan (90^{\circ} - θ) = \frac{1}{\tan (θ)}$

identity

$\tan (2 θ) = \frac{2 \tan (θ)}{1 - \tan^{2} (θ)}$

$\frac{\tan^{2} (x)}{1 + \tan^{2} (x)} = \sin^{2} (x)$

identity

$\tan (x) + \frac{1}{\tan (x)} = \sin (x) \cos (x)$

For Problems 17–26, show that the equation is an identity by transforming the left side into the right side.

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

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

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

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

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

$\sin^{2} (x) - \cos^{2} (x) = 1 - 2 \cos^{2} (x)$

$\tan (θ) \cos (θ) = \sin (θ)$

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

$\frac{\sin (μ)}{\tan (μ)} = \cos (μ)$

$\cos^{4} (x) - \sin^{4} (x) = \cos^{2} (x) - \sin^{2} (x)$

$\begin{aligned} \cos^{4} (x) - \sin^{4} (x) & = (\cos^{2} (x) - \sin^{2} (x)) (\cos^{2} (x) + \sin^{2} (x)) \\ = (\cos^{2} (x) - \sin^{2} (x)) (1) = \cos^{2} (x) - \sin^{2} (x) \end{aligned}$

$1 - 2 \cos^{2} (v) + \cos^{4} (v) = \sin^{4} (v)$

$\frac{\sin (u)}{1 + \cos (u)} = \frac{1 - \cos (u)}{\sin (u)}$

Multiply numerator and denominator of the left side by $1 minus cos ⁡ open parenthesis u close parenthesis$ .

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

$\frac{\sin (v)}{1 - \sin (v)} = \frac{\tan (v) (1 + \sin (v))}{\cos (v)}$

Multiply numerator and denominator of the left side by $1 plus sin ⁡ open parenthesis v close parenthesis$ .

For Problems 27–34, simplify, using identities as necessary.

$\frac{1}{\cos^{2} (β)} - \frac{\sin^{2} (β)}{\cos^{2} (β)}$

$1$

$\frac{1}{\sin^{2} (ϕ)} - \frac{1}{\tan^{2} (ϕ)}$

$cos squared ⁡ open parenthesis α close parenthesis open parenthesis 1 plus tan squared ⁡ open parenthesis α close parenthesis close parenthesis$

$1$

$\cos^{3} (ϕ) + \sin^{2} (ϕ) \cos (ϕ)$

$\tan^{2} (A) - \tan^{2} (A) \sin^{2} (A)$

$\sin^{2} (A)$

$\cos^{2} (B) \tan^{2} (B) + \cos^{2} (B)$

$\frac{1 - \cos^{2} (z)}{\cos^{2} (z)}$

$\tan^{2} (z)$

$\frac{\sin (t)}{\cos (t) \tan (t)}$

For Problems 35–40, evaluate without using a calculator.

$3 cos squared ⁡ open parenthesis 1.7 degrees close parenthesis plus 3 sin squared ⁡ open parenthesis 1.7 degrees close parenthesis$

$3$

$4 - \cos^{2} (338^{\circ}) - \sin^{2} (338^{\circ})$

$(\cos^{2} (20^{\circ}) + \sin^{2} (20^{\circ}))^{4}$

$1$

$\frac{18}{\cos^{2} (17^{\circ}) + \sin^{2} (17^{\circ})}$

$\frac{6}{\cos^{2} (53^{\circ})} - 6 \tan^{2} (53^{\circ})$

$6$

$\frac{1}{\sin^{2} (102^{\circ})} - \frac{\cos^{2} (102^{\circ})}{\sin^{2} (102^{\circ})}$

For Problems 41–46, one side of an identity is given. Graph the expression, and make a conjecture about the other side of the identity.

$2 \cos^{2} (θ) - 1 = ?$

$cos ⁡ open parenthesis 2 θ close parenthesis$

$1 - 2 \sin^{2} (\frac{θ}{2}) = ?$

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

$\cos (θ)$

$\frac{\sin (x) \cos (x)}{1 - \sin^{2} (x)} = ?$

$2 tan ⁡ open parenthesis t close parenthesis cos squared ⁡ open parenthesis t close parenthesis equals ?$

$sin ⁡ open parenthesis 2 t close parenthesis$

$\frac{2 \tan (t)}{1 - \tan^{2} (t)} = ?$

For Problems 47–50, use identities to rewrite each expression.

$2 - \cos^{2} (θ) + 2 \sin (θ)$ as an expression in $\sin (θ)$ only

$1 + 2 \sin (θ) + \sin^{2} (θ)$

$3 \sin^{2} (B) + 2 \cos (B) - 4$ as an expression in $\cos (B)$ only

$cos squared ⁡ open parenthesis ϕ close parenthesis minus 2 sin squared ⁡ open parenthesis ϕ close parenthesis$ as an expression in $\cos (ϕ)$ only

$3 cos squared ⁡ open parenthesis ϕ close parenthesis minus 2$

$\cos^{2} (ϕ) \sin^{2} (ϕ)$ as an expression in $\sin (ϕ)$ only

For Problems 51–58, solve the equation for $0 degrees less than or equal to θ less than 360 degrees$ . Round angles to three decimal places if necessary.

$\cos (θ) - \sin^{2} (θ) + 1 = 0$

$θ = 90^{\circ}, θ = 180^{\circ}, θ = 270^{\circ}$

$4 \sin (θ) + 2 \cos^{2} (θ) - 3 = - 1$

$1 - \sin (θ) - 2 \cos^{2} (θ) = 0$

$θ = 90^{\circ}, θ = 210^{\circ}, θ = 330^{\circ}$

$3 \cos^{2} (θ) - \sin^{2} (θ) = 2$

$2 \cos (θ) \tan (θ) + 1 = 0$

$θ = 210^{\circ}, θ = 330^{\circ}$

$\cos (θ) - \sin (θ) = 0$

$\frac{1}{3} \cos (θ) = \sin (θ)$

$θ = {18.43}^{\circ}, θ = {198.43}^{\circ}$

$5 sin ⁡ open parenthesis θ close parenthesis equals 2 cos ⁡ open parenthesis θ close parenthesis$

For Problems 59–62, use identities to find exact values for the other two trig ratios.

$\cos (A) = \frac{12}{13}$ and $270 degrees less than A less than 360 degrees$

$\sin (A) = \frac{- 5}{13}, \tan (A) = \frac{- 5}{12}$

$\sin (B) = \frac{- 3}{5}$ and $180 degrees less than B less than 270 degrees$

$\sin (ϕ) = \frac{1}{7}$ and $90 degrees less than ϕ less than 180 degrees$

$\cos (ϕ) = \frac{- 4 \sqrt{3}}{7}, \tan (ϕ) = \frac{- 1}{4 \sqrt{3}}$

$\cos (t) = \frac{- 2}{3}$ and $180 degrees less than t less than 270 degrees$

For Problems 63–66, use the identity below to find the sine and cosine of the angle.

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

$\tan (θ) = - \frac{1}{2}$ and $270 degrees less than θ less than 360 degrees$

$\sin (θ) = \frac{- 1}{\sqrt{5}}$ , $\cos (θ) = \frac{2}{\sqrt{5}}$

$tan ⁡ open parenthesis θ close parenthesis equals 2$ and $180 degrees less than θ less than 270 degrees$

$\tan (θ) = \frac{3}{4}$ and $180 degrees less than θ less than 270 degrees$

$\sin (θ) = \frac{- 3}{5}$ , $\cos (θ) = \frac{- 4}{5}$

$\tan (θ) = - 3$ and $90 degrees less than θ less than 180 degrees$

For Problems 67–72, find exact values for the sine, cosine, and tangent of the angle.

$2 \cos (A) + 9 = 8$ and $90 degrees less than A less than 180 degrees$

$\sin (θ) = \frac{\sqrt{3}}{2}$ , $\cos (θ) = \frac{- 1}{2}$ , $\tan (θ) = \sqrt{3}$

$25 \sin (B) + 8 = - 12$ and $180 degrees less than B less than 270 degrees$

$8 \tan (β) + 5 = - 11$ and $90 degrees less than β less than 180 degrees$

$\sin (β) = \frac{2}{\sqrt{5}}$ , $\cos (β) = \frac{- 1}{\sqrt{5}}$ , $\tan (β) = - 2$

$6 (\tan (β) - 4) = - 24$ and $90 degrees less than β less than 270 degrees$

$\tan^{2} (C) - \frac{1}{4} = 0$ and $0 degrees less than C less than 180 degrees$

$\begin{aligned} \sin (C) = \frac{1}{\sqrt{5}}, \cos (C) = \frac{2}{\sqrt{5}}, \tan (C) = \frac{1}{2} \\ or & \sin (C) = \frac{1}{\sqrt{5}}, \cos (C) = \frac{- 2}{\sqrt{5}}, \tan (C) = \frac{- 1}{2} \end{aligned}$

$4 \cos^{2} (A) - \cos (A) = 0$ and $0 degrees less than A less than 180 degrees$

For Problems 73–76, prove the identity by rewriting tangents in terms of sines and cosines. (These problems involve simplifying complex fractions. See the Algebra Refresher to review this skill.)

$\frac{\tan (α)}{1 + \tan (α)} = \frac{\sin (α)}{\sin (α) + \cos (α)}$

$\frac{\tan (α)}{1 + \tan (α)} = \frac{\frac{\sin (α)}{\cos (α)}}{1 + \frac{\sin (α)}{\cos (α)}} \cdot \frac{\cos (α)}{\cos (α)} = \frac{\sin (α)}{\sin (α) + \cos (α)}$

$\frac{1 - \tan (u)}{1 + \tan (u)} = \frac{\cos (u) - \sin (u)}{\cos (u) + \sin (u)}$

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

$\frac{1 + \tan^{2} (β)}{1 - \tan^{2} (β)} = \frac{\frac{1}{\cos^{2} (β)}}{1 - \frac{\sin^{2} (β)}{\cos^{2} (β)}} \cdot \frac{\cos^{2} (β)}{\cos^{2} (β)} = \frac{1}{\cos^{2} (β) - \sin^{2} (β)}$

$\tan^{2} (v) - \sin^{2} (v) = \tan^{2} (v) \sin^{2} (v)$

Prove the Pythagorean identity $\cos^{2} (θ) + \sin^{2} (θ) = 1$ by carrying out the following steps. Sketch an angle $θ$ in standard position, and label a point $(x, y)$ on the terminal side, at a distance $r$ from the vertex.

Begin with the equation $\sqrt{x^{2} + y^{2}} = r$ , and square both sides.
Divide both sides of your equation from part (a) by $r^{2}$ .
Write the left side of the equation as the sum of the squares of two fractions.
Substitute the appropriate trigonometric ratio for each fraction.

By the distance formula, $\sqrt{x^{2} + y^{2}} = r$ , or $x^{2} + y^{2} = r^{2}$ .
$\frac{x^{2}}{r^{2}} + \frac{y^{2}}{r^{2}} = 1$
${(\frac{x}{r})}^{2} + {(\frac{y}{r})}^{2} = 1$
$(\cos (θ))^{2} + (\sin (θ))^{2} = 1$

Prove the tangent identity $\tan (θ) = \frac{\sin (θ)}{\cos (θ)}$ by carrying out the following steps. Sketch an angle $θ$ in standard position, and label a point $(x, y)$ on the terminal side, at a distance $r$ from the vertex.

Write $\sin (θ)$ in terms of $y$ and $r$ , and solve for $y$ .
Write $\cos (θ)$ in terms of $x$ and $r$ , and solve for $x$ .
Write $\tan (θ)$ in terms of $x$ and $y$ , then substitute your results from parts (a) and (b).
Simplify your fraction in part (c).

Trigonometry by Katherine Yoshiwara (yoshiwarabooks.org), GNU Free Documentation License 1.2 or later. Adapted for the XYZ HTML edition with the authors' permission (recorded 2026-07-04). License: GFDL-1.2-or-later.