Use relative acute angles to find exact values (deg)

Ideas in this chapter extend material that has been presented on the unit-circle definitions of the trigonometric functions, on exact values and on relative acute angles.

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. 

For example, we may wish to evaluate an expression like $\sin240^\circ+\cos\left(-30^\circ\right)$sin240°+cos(−30°). An ordinary scientific calculator will do this without any preliminary work, but you can check that the following manipulation gives the same result. These manipulations are justified by reference to the unit circle definitions of the trigonometric functions.

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°.

 

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°. We now use the fact that sine is negative in the third quadrant and cosine is positive in the fourth quadrant, to write the expression $-\sin60^\circ+\cos30^\circ$−sin60°+cos30°, which is identical in value to the original expression.

Since $60^\circ$60° and $30^\circ$30° have exact values for our three main trigonometric functions, it turns out that our original expression $\sin240^\circ+\cos\left(-30^\circ\right)$sin240°+cos(−30°) can be evaluated exactly!

Recalling that $\sin60^\circ=\frac{\sqrt{3}}{2}$sin60°=√32​ and $\cos30^\circ=\frac{\sqrt{3}}{2}$cos30°=√32​ also, we have $-\sin60^\circ+\cos30^\circ$−sin60°+cos30°$=$=$-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=0$−√32​+√32​=0.

Example:

Evaluate the expression: $\sin405^\circ\left(\tan150^\circ+\cos150^\circ\right)$sin405°(tan150°+cos150°).

The argument $405^\circ$405° is equivalent to $45^\circ$45°.

So, $\sin405^\circ$sin405° is the same as $\sin45^\circ=\frac{1}{\sqrt{2}}$sin45°=1√2​. 

Similarly, $150^\circ$150° is in the second quadrant and is therefore related to $30^\circ$30°. Hence, $\tan150^\circ=-\frac{1}{\sqrt{3}}$tan150°=−1√3​ and $\cos150^\circ=-\frac{\sqrt{3}}{2}$cos150°=−√32​. 

Putting the pieces together, we have $\sin405^\circ\left(\tan150^\circ+\cos150^\circ\right)=\frac{1}{\sqrt{2}}\left(-\frac{1}{\sqrt{3}}-\frac{\sqrt{3}}{2}\right)=$sin405°(tan150°+cos150°)=1√2​(−1√3​−√32​)=$-\frac{5}{2\sqrt{6}}$−52√6​.

This can be written with rationalised denominator as $-\frac{5\sqrt{6}}{12}$−5√612​.

Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3