8.04 Angles with exact values

Even when an angle is given as a rational number of degrees, the values taken by the trigonometric functions of the angle cannot, in most cases, be written down exactly in decimal form. 

However, there are a few special angles whose sine, cosine or tangent can be expressed as rational numbers, and there are some whose sine, cosine or tangent can be written exactly using surds.

It is easy to check that $\sin0^\circ=0$sin0°=0 and $\tan0^\circ=0$tan0°=0. Also, $\sin90^\circ=1$sin90°=1  and $\cos0^\circ=1$cos0°=1. 

It's possible to obtain values for the trigonometric functions of the angles $30^\circ$30°, $60^\circ$60° and $45^\circ$45° by applying Pythagoras’ theorem in some special triangles.

1.  Equilateral triangle with side length 2 units. Draw in the altitude to split the triangle in two, and use Pythagoras' theorem to find the length of the altitude.

From the diagram, the following function values can be read:

$\sin60^\circ=\frac{\sqrt{3}}{2}$sin60°=√32​	$\sin30^\circ=\frac{1}{2}$sin30°=12​
$\cos60^\circ=\frac{1}{2}$cos60°=12​	$\cos30^\circ=\frac{\sqrt{3}}{2}$cos30°=√32​
$\tan60^\circ=\sqrt{3}$tan60°=√3	$\tan30^\circ=\frac{1}{\sqrt{3}}$tan30°=1√3​

OR

$\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$sinπ3​=√32​	$\sin\frac{\pi}{6}=\frac{1}{2}$sinπ6​=12​
$\cos\frac{\pi}{3}=\frac{1}{2}$cosπ3​=12​	$\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$cosπ6​=√32​
$\tan\frac{\pi}{3}=\sqrt{3}$tanπ3​=√3	$\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}$tanπ6​=1√3​

2.  Isosceles right-angled triangle, with the equal sides of length $1$1 unit. The hypotenuse can be found using Pythagoras' theorem.

From this diagram:

$\sin45^\circ=\frac{1}{\sqrt{2}}$sin45°=1√2​

$\cos45^\circ=\frac{1}{\sqrt{2}}$cos45°=1√2​

$\tan45^\circ=1$tan45°=1

OR

$\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}$sinπ4​=1√2​

$\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}$cosπ4​=1√2​

$\tan\frac{\pi}{4}=1$tanπ4​=1

These can be summarised by the following table, (the denominator for each has been rationalised):

Angle in Degrees	$0^\circ$0°	$30^\circ$30°	$45^\circ$45°	$60^\circ$60°	$90^\circ$90°
Angle in Radians	$0$0	$\frac{\pi}{6}$π6​	$\frac{\pi}{4}$π4​	$\frac{\pi}{3}$π3​	$\frac{\pi}{2}$π2​
sin	$0$0	$\frac{1}{2}$12​	$\frac{\sqrt{2}}{2}$√22​	$\frac{\sqrt{3}}{2}$√32​	$1$1
cos	$1$1	$\frac{\sqrt{3}}{2}$√32​	$\frac{\sqrt{2}}{2}$√22​	$\frac{1}{2}$12​	$0$0
tan	$0$0	$\frac{\sqrt{3}}{3}$√33​	$1$1	$\sqrt{3}$√3	$undefined$undefined

Beyond the first quadrant

If evaluating a trigonometric function for any multiple of $30^\circ$30° or $45^\circ$45°, the answer can be written in an exact form using rational numbers or surds.  Previously symmetry of the trigonometric functions was used to write equivalent statements using angles in the first quadrant. The same principles can be applied here to find an equivalent statement then use the special triangles, the summary table or diagrams such as:

Multiples of $45^\circ$45° $\left(\frac{\pi}{4}\right)$(π4​)	Multiples of $30^\circ$30° $\left(\frac{\pi}{6}\right)$(π6​)

Recall the coordinates are in the form $\left(\sin\theta,\cos\theta\right)$(sinθ,cosθ).

Worked example

Example 1

Evaluate $\cos\left(\frac{19\pi}{6}\right)$cos(19π6​), giving your answer in exact form.

$\frac{19\pi}{6}$19π6​ is more than a full circle. Subtracting $2\pi$2π an equivalent rotation would be $\frac{7\pi}{6}$7π6​. We could look this value up in the unit circle diagram above, looking for the $x$x-coordinate.

Hence:

$\cos\left(\frac{19\pi}{6}\right)$cos(19π6​)	$=$=	$\cos\left(\frac{7\pi}{6}\right)$cos(7π6​)
	$=$=	$-\frac{\sqrt{3}}{2}$−√32​

Alternatively, we can write the angle in terms of a first quadrant angle and then evaluate. 

$\frac{7\pi}{6}$7π6​ would have the same magnitude $x$x-coordinate as $\frac{\pi}{6}$π6​ but the opposite sign. 

Hence:

$\cos\left(\frac{19\pi}{6}\right)$cos(19π6​)	$=$=	$\cos\left(\frac{7\pi}{6}\right)$cos(7π6​)
	$=$=	$-\cos\left(\frac{\pi}{6}\right)$−cos(π6​)
	$=$=	$-\frac{\sqrt{3}}{2}$−√32​

 

Example 2

Evaluate the expression: $\sin\left(405^\circ\right)\left(\tan\left(150^\circ\right)+\cos\left(150^\circ\right)\right)$sin(405°)(tan(150°)+cos(150°)).

Think: The angle $405^\circ$405° is equivalent to $45^\circ$45°, one full circle of $360^\circ$360° then an extra $45^\circ$45°

So, $\sin\left(405^\circ\right)$sin(405°) is the same as $\sin\left(45^\circ\right)=\frac{\sqrt{2}}{2}$sin(45°)=√22​. 

Similarly, $150^\circ$150° is $30^\circ$30° before $180^\circ$180°. It therefore has related acute angle of $30^\circ$30° but the cosine and tangent will be negative because it is in the second quadrant.

Hence, $\tan\left(150^\circ\right)=-\frac{\sqrt{3}}{3}$tan(150°)=−√33​ and $\cos\left(150^\circ\right)=-\frac{\sqrt{3}}{2}$cos(150°)=−√32​. 

Do: Putting the pieces together, we have:

$\sin\left(405^\circ\right)\left(\tan\left(150^\circ\right)+\cos\left(150^\circ\right)\right)$sin(405°)(tan(150°)+cos(150°))	$=$=	$\frac{\sqrt{2}}{2}\left(-\frac{\sqrt{3}}{3}-\frac{\sqrt{3}}{2}\right)$√22​(−√33​−√32​)
	$=$=	$-\frac{\sqrt{6}}{6}-\frac{\sqrt{6}}{4}$−√66​−√64​
	$=$=	$-\frac{2\sqrt{6}}{12}-\frac{3\sqrt{6}}{12}$−2√612​−3√612​
	$=$=	$-\frac{5\sqrt{6}}{12}$−5√612​

 

Reflect: Try and express the angle as a multiple of $30$30 or $45$45 degrees and use a unit circle diagram to ensure you have the correct sign for the sine, cosine or tangent. 

 

Practice questions

Question 1

Question 2

Question 3