Skill: Determine exact trigonometric ratios for special angles

Special angles have exact values that come from the side ratios of two triangles:

• A $30^\circ$–$60^\circ$–$90^\circ$ triangle has side ratio $1:\sqrt{3}:2$.
• A $45^\circ$–$45^\circ$–$90^\circ$ triangle has side ratio $1:1:\sqrt{2}$.

On the unit circle, cosine is the $x$-coordinate and sine is the $y$-coordinate. Tangent is found using

$$\tan\theta=\frac{\sin\theta}{\cos\theta}.$$

Special-angle values

For angles outside the first quadrant, use the related acute angle, called the reference angle, and then apply the signs from the quadrant:

• Quadrant I: $\sin$, $\cos$, and $\tan$ are positive.
• Quadrant II: sine is positive; cosine and tangent are negative.
• Quadrant III: sine and cosine are negative; tangent is positive.
• Quadrant IV: sine is negative; cosine and tangent are positive.

Worked example

Find the exact values of $\sin150^\circ$, $\cos150^\circ$, and $\tan150^\circ$.

Step 1: Find the reference angle.

The angle $150^\circ$ is in Quadrant II. Its reference angle is

$$180^\circ-150^\circ=30^\circ.$$

So we use the special-angle values for $30^\circ$.

Step 2: Determine the signs.

In Quadrant II:

• sine is positive,
• cosine is negative,
• tangent is negative.

Step 3: Use the $30^\circ$ values.

Since

$$\sin30^\circ=\frac12,\qquad \cos30^\circ=\frac{\sqrt3}{2},$$

we get

$$\sin150^\circ=\frac12$$

and

$$\cos150^\circ=-\frac{\sqrt3}{2}.$$

Step 4: Find tangent.

Use $\tan\theta=\frac{\sin\theta}{\cos\theta}$:

$$\tan150^\circ =\frac{\frac12}{-\frac{\sqrt3}{2}} =-\frac{1}{\sqrt3} =-\frac{\sqrt3}{3}.$$

Therefore,

$$\boxed{\sin150^\circ=\frac12,\qquad \cos150^\circ=-\frac{\sqrt3}{2},\qquad \tan150^\circ=-\frac{\sqrt3}{3}}.$$