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Exact values of trigonometric ratios

The exact surd values of sin, cos and tan at 0, 30, 45, 60 and 90 degrees, where they come from (two special triangles plus the unit circle), and how to use them to answer "find the exact value" questions without a calculator.

Pure · Trigonometry

Exact values of trigonometric ratios

The exact surd values of sin, cos and tan at 0, 30, 45, 60 and 90 degrees, where they come from (two special triangles plus the unit circle), and how to use them to answer "find the exact value" questions without a calculator.

Why it works

A handful of angles — 0°,30°,45°,60°,90°0°, 30°, 45°, 60°, 90° — turn up so often that you're expected to know their sine, cosine and tangent exactly, as fractions and surds, not as calculator decimals. They aren't worth memorising blindly, because they all fall out of just two triangles and the unit circle.

The 45°45° triangle. Take a right-angled triangle with both short sides equal to 11. The two non-right angles are each 45°45°, and the hypotenuse is 12+12=2\sqrt{1^2 + 1^2} = \sqrt2 (Pythagoras). So sin⁡45°=cos⁡45°=12=22,tan⁡45°=11=1.\sin 45° = \cos 45° = \frac{1}{\sqrt2} = \frac{\sqrt2}{2}, \qquad \tan 45° = \frac{1}{1} = 1.

The 30°30°–60°60° triangle. Take an equilateral triangle with side 22 and cut it in half down the middle. You get a right-angled triangle with hypotenuse 22, base 11, and height 22−12=3\sqrt{2^2 - 1^2} = \sqrt3. The angle at the top is 30°30° and at the bottom 60°60°, so sin⁡30°=12,cos⁡30°=32,tan⁡30°=1/23/2=13=33,\sin 30° = \tfrac12,\quad \cos 30° = \tfrac{\sqrt3}{2},\quad \tan 30° = \frac{1/2}{\sqrt3/2} = \frac{1}{\sqrt3} = \frac{\sqrt3}{3}, sin⁡60°=32,cos⁡60°=12,tan⁡60°=3/21/2=3.\sin 60° = \tfrac{\sqrt3}{2},\quad \cos 60° = \tfrac12,\quad \tan 60° = \frac{\sqrt3/2}{1/2} = \sqrt3.

Notice the symmetry: sin⁡60°=cos⁡30°\sin 60° = \cos 30° and sin⁡30°=cos⁡60°\sin 30° = \cos 60° — the bigger angle has the bigger sine and the smaller cosine.

The ends — 0°0° and 90°90° — from the unit circle. The point at angle θ\theta on a circle of radius 11 is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta). At 0°0° it sits at (1,0)(1, 0), and at 90°90° at (0,1)(0, 1): sin⁡0°=0, cos⁡0°=1, tan⁡0°=0,sin⁡90°=1, cos⁡90°=0.\sin 0° = 0,\ \cos 0° = 1,\ \tan 0° = 0, \qquad \sin 90° = 1,\ \cos 90° = 0. tan⁡90°\tan 90° is undefined — it would be 10\tfrac{1}{0} (this is the asymptote you saw on the tangent graph).

θ0∘30∘45∘60∘90∘sin⁡θ01222321cos⁡θ13222120tan⁡θ03313undef.\begin{array}{c|ccccc} \theta & 0^\circ & 30^\circ & 45^\circ & 60^\circ & 90^\circ \\ \hline \sin\theta & 0 & \tfrac12 & \tfrac{\sqrt2}{2} & \tfrac{\sqrt3}{2} & 1 \\ \cos\theta & 1 & \tfrac{\sqrt3}{2} & \tfrac{\sqrt2}{2} & \tfrac12 & 0 \\ \tan\theta & 0 & \tfrac{\sqrt3}{3} & 1 & \sqrt3 & \text{undef.} \end{array}

Two habits keep the marks. Give exact answers — a surd or fraction, never a rounded decimal; that is the whole point of the question. And rationalise: write 13\tfrac{1}{\sqrt3} as 33\tfrac{\sqrt3}{3} and 12\tfrac{1}{\sqrt2} as 22\tfrac{\sqrt2}{2}.

Angles beyond 90°90°. For an angle outside 0°0°–90°90°, use its related acute angle and fix the sign from the quadrant (the symmetry of the graphs): e.g. sin⁡150°=sin⁡30°=12\sin 150° = \sin 30° = \tfrac12, and cos⁡120°=−cos⁡60°=−12\cos 120° = -\cos 60° = -\tfrac12 (cosine is negative in the second quadrant). The exact size always comes from the table; only the sign changes.