Ctrl+k

Determine exact trigonometric ratios for special angles

Understanding exact trigonometric ratios for the special angles 0∘,30∘,45∘,60∘,0^\circ, 30^\circ, 45^\circ, 60^\circ, and 90∘90^\circ, together with their related angles on the unit circle, connects sine, cosine, and tangent to coordinates and the side ratios of 30∘ ⁣− ⁣60∘ ⁣− ⁣90∘30^\circ\!-\!60^\circ\!-\!90^\circ and 45∘ ⁣− ⁣45∘ ⁣− ⁣90∘45^\circ\!-\!45^\circ\!-\!90^\circ triangles. Values are expressed symbolically using fractions and radicals rather than decimal approximations, with signs determined by quadrant and tangent identified as undefined where cosine is zero; more general angle formulas and advanced exact-value techniques are not included.

Detailed Explanation: Determine exact trigonometric ratios for special angles

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

  • A 30∘30^\circ–60∘60^\circ–90∘90^\circ triangle has side ratio 1:3:21:\sqrt{3}:2.
  • A 45∘45^\circ–45∘45^\circ–90∘90^\circ triangle has side ratio 1:1:21:1:\sqrt{2}.

On the unit circle, cosine is the xx-coordinate and sine is the yy-coordinate. Tangent is found using

tan⁡θ=sin⁡θcos⁡θ.\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⁡\sin, cos⁡\cos, and tan⁡\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 sin⁡150∘\sin150^\circ, cos⁡150∘\cos150^\circ, and tan⁡150∘\tan150^\circ.

Step 1: Find the reference angle.

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

180∘−150∘=30∘.180^\circ-150^\circ=30^\circ.

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

Step 2: Determine the signs.

In Quadrant II:

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

Step 3: Use the 30∘30^\circ values.

Since

sin⁡30∘=12,cos⁡30∘=32,\sin30^\circ=\frac12,\qquad \cos30^\circ=\frac{\sqrt3}{2},

we get

sin⁡150∘=12\sin150^\circ=\frac12

and

cos⁡150∘=−32.\cos150^\circ=-\frac{\sqrt3}{2}.

Step 4: Find tangent.

Use tan⁡θ=sin⁡θcos⁡θ\tan\theta=\frac{\sin\theta}{\cos\theta}:

tan⁡150∘=12−32=−13=−33.\tan150^\circ =\frac{\frac12}{-\frac{\sqrt3}{2}} =-\frac{1}{\sqrt3} =-\frac{\sqrt3}{3}.

Therefore,

sin⁡150∘=12,cos⁡150∘=−32,tan⁡150∘=−33.\boxed{\sin150^\circ=\frac12,\qquad \cos150^\circ=-\frac{\sqrt3}{2},\qquad \tan150^\circ=-\frac{\sqrt3}{3}}.

Learn by doing: Determine exact trigonometric ratios for special angles

Click a topic below to practice the foundational skills you'll need, learn the steps, or master this skill

    Practice with unlimited practice problems

    Trigonometry, Unit Circle Dimensions as Sin/Cos Ratio of Angle Radians

      ?