Even when an angle is given as a rational number of radians, the values taken by the trigonometric functions of the angle cannot, in most cases, be written down exactly in decimal form. (We covered some of these ideas here)
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 $\sin0=0$sin0=0 and $\tan0=0$tan0=0. Also, $\sin\pi=1$sinπ=1 and $\cos0=1$cos0=1.
We can obtain values for the trigonometric functions of the angles $\frac{\pi}{6}$π6, $\frac{\pi}{3}$π3 and $\frac{\pi}{4}$π4 by applying Pythagoras’ theorem in some special triangles.
1. Equilateral triangle with an altitude
From the diagram we can read off the following function values:
| $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$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
From this diagram we see that:
| $\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}$sinπ4=1√2 |
| $\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}$cosπ4=1√2 |
| $\tan\frac{\pi}{4}=1$tanπ4=1 |
Example:
Write the expression $\cos\frac{2\pi}{3}+\sin\frac{\pi}{4}$cos2π3+sinπ4 more simply.
The second quadrant angle $\frac{2\pi}{3}$2π3 is related to the first quadrant angle $\frac{\pi}{3}$π3. Therefore, the expression can be written $-\cos\frac{\pi}{3}+\sin\frac{\pi}{4}$−cosπ3+sinπ4. Then, using the exact values for these ratios, we have $-\frac{1}{2}+\frac{1}{\sqrt{2}}$−12+1√2. This is $-\frac{1}{2}+\frac{\sqrt{2}}{2}$−12+√22 or, more simply, $\frac{\sqrt{2}-1}{2}$√2−12.
Example
Question 1
Find the exact value of $\sin\frac{\pi}{3}$sinπ3.