Exact trigonometric values

​​In a nutshell

It is possible to compute the exact values of $$\sin$$​, $$\cos$$​ and $$\tan$$​ for the angles $$30^{\circ}$$​, $$45^{\circ}$$​ and $$60^{\circ}$$​ by considering an equilateral triangle and an isoceles right-angled triangle.

Exact values from the equilateral triangle ​​

To compute $$\sin$$​, $$\cos$$​ and $$\tan$$​ of $$30^{\circ}$$ and $$60^{\circ}$$​, consider an equilateral triangle of side length $$1$$, and draw a line perpendicular to one of the sides from its midpoint to the opposite corner.

The perpendicular bisects the triangle into two congruent right-angled triangles, with angles $$30^{\circ}$$​, $$60^{\circ}$$​ and $$90^{\circ}$$. The hypotenuse has length $$1$$​, and one of the sides has length $$\dfrac{1}{2}$$. Call the length of the third side ​$$c$$. Using Pythagoras' theorem to compute $$c$$​:

​$$\begin{aligned}1^2 &= (\dfrac{1}{2})^2 + c^2 \\1 &= \dfrac{1}{4} + c^2 \\\dfrac{3}{4} &= c^2 \\\pm\dfrac{\sqrt{3}}{2} &= c\end{aligned}$$​​

However $$c$$​ is a length, therefore positive, so deduce that $$c = \dfrac{\sqrt{3}}{2}$$​.

Trigonometric ratios (SOHCAHTOA) in the right-angled triangle with side lengths $$1$$​, $$\dfrac{1}{2}$$​ and $$\dfrac{\sqrt{3}}{2}$$ allow you to compute:

​$$\boxed{\begin{aligned}\sin(30^{\circ}) &= \dfrac{1}{2}\\\cos(30^{\circ}) &= \dfrac{\sqrt{3}}{2}\\\tan(30^{\circ}) &= \dfrac{1}{\sqrt{3}}= \dfrac{\sqrt{3}}{3}\\\\\sin(60^{\circ}) &= \dfrac{\sqrt{3}}{2}\\\cos(60^{\circ}) &= \dfrac{1}{2}\\\tan(60^{\circ}) &= \sqrt{3}\end{aligned}}$$​​

​​Example 1

What is the exact value of $$\sin(150^{\circ})$$?

Express $$\sin(150^{\circ})$$ in terms of the acute angle $$30^{\circ}$$ by substituting $$\theta = 30^{\circ}$$ into the identity $$\sin(180^{\circ} - \theta) = \sin(\theta)$$:

$$\begin{aligned}\sin(180^{\circ} - \theta) &= \sin(\theta)\\\sin(180^{\circ} - 30^{\circ}) &= \sin(30^{\circ}) \\\sin(150^{\circ}) &= \sin(30^{\circ})\end{aligned}$$​​

Now use the fact that $$\sin(30^{\circ}) = \dfrac{1}{2}$$ to deduce that $$\sin(150^{\circ}) = \dfrac12$$​

$$\underline{\sin(150^{\circ}) = \dfrac12.}$$​​

​​

Exact values from the isosceles right-angled triangle

To compute $$\sin$$​, $$\cos$$​ and $$\tan$$​ of $$45^{\circ}$$, consider an isosceles right-angled triangle with a hypotenuse of length $$1$$. 

This triangle has two $$45^{\circ}$$​ angles, and one right angle. Call the length of the two shorter sides $$a$$​, and use Pythagoras' theorem to find $$a$$​:

$$\begin{aligned}1^2 &= a^2 + a^2\\1 &= 2a^2\\\dfrac{1}{2} &= a^2\\\pm\dfrac{1}{\sqrt{2}}&=a\\\pm\dfrac{\sqrt{2}}{2}&=a\end{aligned}$$​​

However $$a$$​ is a length, so must be positive, hence deduce that $$a = \dfrac{\sqrt{2}}{2}$$​.

Trigonometric ratios (SOHCAHTOA) in the isosceles right-angled triangle with a hypotenuse of length $$1$$ allow you to compute:

$$\boxed{\begin{aligned}\sin(45^{\circ}) &= \dfrac{\sqrt{2}}{2}\\\cos(45^{\circ}) &= \dfrac{\sqrt{2}}{2}\\\tan(45^{\circ})&=1\end{aligned} }$$​

Example 2

What is the exact value of $$\tan(315^{\circ})$$?

Express $$\tan(315^{\circ})$$ in terms of the acute angle $$45^{\circ}$$ by substituting $$\theta = 45^{\circ}$$ into the identity $$\tan(360^{\circ}- \theta) = -\tan(\theta)$$:

$$\begin{aligned}\tan(360^{\circ} - \theta) &= -\tan(\theta)\\\tan(360^{\circ} - 45^{\circ}) &= -\tan(45^{\circ})\\\tan(315^{\circ}) &= -\tan(45^{\circ})\end{aligned}$$​​

Now use the fact that $$\tan(45^{\circ}) = 1$$ to deduce that $$\tan(315^{\circ}) = -1$$​

$$\underline{\tan(315^{\circ}) = -1.}$$