4.06 Evaluating trigonometric functions: exact values

Previously, we have seen how the unit circle can allow us to rewrite any angle in terms of a reference angle which $0^\circ\le\alpha<90^\circ$0°≤α<90° or $0\le\alpha<\frac{\pi}{2}$0≤α<π2​ by using what we know about the sign of the ratios in the given quadrant.

The periodic nature of the trigonometric functions means that the same function value is obtained for many different angles or numbers in the domain of the function.  In fact, there is always an acute angle that leads to any given function value and this can be used instead of a more general angle in order to simplify calculations. 

Exploration

First, consider the table that we completed when studying exact value trig ratios of special right triangles. We can use the unit circle to extend the table to account for values of $0^\circ$0° and $90^\circ$90° angles.

$\theta$θ	$0^\circ$0° or $0$0	$30^\circ$30° or $\frac{\pi}{6}$π6​	$45^\circ$45° or $\frac{\pi}{4}$π4​	$60^\circ$60° or $\frac{\pi}{6}$π6​	$90^\circ$90° or $\frac{\pi}{2}$π2​
$\sin(\theta)$sin(θ)	$0$0	$\frac{1}{2}$12​	$\frac{\sqrt{2}}{2}$√22​	$\frac{\sqrt{3}}{2}$√32​	$1$1
$\cos(\theta)$cos(θ)	$1$1	$\frac{\sqrt{3}}{2}$√32​	$\frac{\sqrt{2}}{2}$√22​	$\frac{1}{2}$12​	$0$0
$\tan(\theta)$tan(θ)	$0$0	$\frac{\sqrt{3}}{3}$√33​	$1$1	$\sqrt{3}$√3	undefined

Notice that the $\tan90^\circ$tan90° is undefined.  This is because to evaluate $\frac{y}{x}$yx​ at this point on the unit circle would require us to divide $\frac{1}{0}$10​, which is not defined in mathematics.

Now let's consider rewriting the values in different but equivalent forms.  What patterns do you see in the table?

$\theta$θ	$0^\circ$0° or $0$0	$30^\circ$30° or $\frac{\pi}{6}$π6​	$45^\circ$45° or $\frac{\pi}{4}$π4​	$60^\circ$60° or $\frac{\pi}{6}$π6​	$90^\circ$90° or $\frac{\pi}{2}$π2​
$\sin(\theta)$sin(θ)	$\frac{\sqrt{0}}{2}$√02​	$\frac{\sqrt{1}}{2}$√12​	$\frac{\sqrt{2}}{2}$√22​	$\frac{\sqrt{3}}{2}$√32​	$\frac{\sqrt{4}}{2}$√42​
$\cos(\theta)$cos(θ)	$\frac{\sqrt{4}}{2}$√42​	$\frac{\sqrt{3}}{2}$√32​	$\frac{\sqrt{2}}{2}$√22​	$\frac{\sqrt{1}}{2}$√12​	$\frac{\sqrt{0}}{2}$√02​
$\tan(\theta)$tan(θ)	$0$0	$\frac{\sqrt{3}}{3}$√33​	$1$1	$\sqrt{3}$√3	undefined

Some students like to memorize the values in the table above and apply their knowledge of reference angles in order to evaluate trig functions more quickly. Other students prefer to remember the entire unit circle. Choose a method that makes the most sense to you!

 

Worked examples

Question 1

Evaluate $\cos855^\circ$cos855°

Think: We first need to determine which quadrant $855^\circ$855° is in, from there we can determine the sign and the reference angle.

Do: 

First let's determine a coterminal angle, $\theta$θ, where $0^\circ\le\theta<360^\circ$0°≤θ<360°.

$855^\circ-2\times360^\circ$855°−2×360°	$=$=	$855^\circ-720^\circ$855°−720°
	$=$=	$135^\circ$135°

 

So $855^\circ$855° is in quadrant $II$II and is coterminal with $135^\circ$135°. The reference angle is therefore $180^\circ-135^\circ=45^\circ$180°−135°=45°.

In quadrant $II$II, cosine is negative, so we get:

 $\cos855^\circ=-\cos45^\circ$cos855°=−cos45°

We have our two special triangles, which we can use to get the exact value.

Using the triangle on the right, and that $\cos\theta=\frac{\text{adjacent }}{\text{hypotenuse }}$cosθ=adjacent hypotenuse ​, we finish with:

$\cos855^\circ=-\frac{1}{\sqrt{2}}$cos855°=−1√2​

Question 2

Evaluate the expression $\sin240^\circ+\cos\left(-30^\circ\right)$sin240°+cos(−30°). 

Think:

The angle $240^\circ$240° is in the third quadrant.

We find its first quadrant reference angle by subtracting $180^\circ$180° to get $60^\circ$60°.

 

Similarly, the angle $-30^\circ$−30° is equivalent to $330^\circ$330°, which is in the fourth quadrant. So, we subtract this from $360^\circ$360° to obtain the reference angle $30^\circ$30°.

Do: We now use the fact that sine is negative in the third quadrant and cosine is positive in the fourth quadrant to give us:

$\sin240^\circ+\cos\left(-30^\circ\right)$sin240°+cos(−30°)	$=$=	$-\sin60^\circ+\cos30^\circ$−sin60°+cos30°	Using the unit circle
	$=$=	$-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}$−√32​+√32​	Using the special triangles
	$=$=	$0$0	Simplifying

 

Question 3

Evaluate the expression: $\sin\frac{9\pi}{4}\left(\tan\frac{5\pi}{6}+\cos\frac{5\pi}{6}\right)$sin9π4​(tan5π6​+cos5π6​).

Think: The argument $\frac{9\pi}{4}$9π4​ shares a terminal side with $\frac{\pi}{4}$π4​.

Similarly, $\frac{5\pi}{6}$5π6​ is in the second quadrant and therefore has a reference angle of $\frac{\pi}{6}$π6​. 

We can use the triangles below to help us:

Do:

$\sin\frac{9\pi}{4}\left(\tan\frac{5\pi}{6}+\cos\frac{5\pi}{6}\right)$sin9π4​(tan5π6​+cos5π6​)	$=$=	$\sin\frac{\pi}{4}\left(-\tan\frac{\pi}{6}+-\cos\frac{\pi}{6}\right)$sinπ4​(−tanπ6​+−cosπ6​)	Using reference angles and unit circle
	$=$=	$\frac{1}{\sqrt{2}}\left(-\frac{1}{\sqrt{3}}-\frac{\sqrt{3}}{2}\right)$1√2​(−1√3​−√32​)	Using the special triangles
	$=$=	$-\frac{1}{\sqrt{6}}-\frac{\sqrt{3}}{2\sqrt{2}}$−1√6​−√32√2​	Simplifying
	$=$=	$-\frac{2}{2\sqrt{6}}-\frac{3}{2\sqrt{6}}$−22√6​−32√6​	Common denominator
	$=$=	$-\frac{5}{2\sqrt{6}}$−52√6​	Simplifying
	$=$=	$-\frac{5\sqrt{6}}{12}$−5√612​	Rationalizing the denominator

 

Practice questions

QUESTION 4

QUESTION 5

QUESTION 6