5.04 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. 

We can 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 we can read off the following function values:

$\sin60^\circ=\frac{\sqrt{3}}{2}$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. We can find the hypotenuse using Pythagoras' theorem.

From this diagram we see that:

$\sin45^\circ=\frac{1}{\sqrt{2}}$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 we evaluate 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.  In our sets on the unit circle and degrees and radians we used the symmetry of the trigonometric functions to write equivalent statements using angles in the first quadrant. We can apply the same principles 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 examples

Example 1

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

Method 1:

Think: $2\pi=\frac{12\pi}{6}$2π=12π6​ therefore $\frac{19\pi}{6}$19π6​ is more than a full circle. Subtracting $\frac{12\pi}{6}$12π6​ 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.

Do:

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

Method 2:

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

$\frac{7\pi}{6}$7π6​ is $\frac{\pi}{6}$π6​ past $\pi$π. By symmetry, it will have the same magnitude $x$x-coordinate as $\frac{7\pi}{6}$7π6​ but the opposite sign. We call $\frac{\pi}{6}$π6​ the reference angle or related acute angle. 

Do:

$\cos\left(\frac{19\pi}{6}\right)$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

Question 4