Exact Trigonometric Values in Radians

The special angles in radians

Half a turn is 180°, which is $\pi$π radians. 30° is a sixth of that half turn, so $30^{\circ }=\frac{\pi }{6}$30° = π/6. In the same way 45° is a quarter of it, $\frac{\pi }{4}$π/4, 60° is a third, $\frac{\pi }{3}$π/3, and 90° is a half, $\frac{\pi }{2}$π/2.

Going on round in steps of $\frac{\pi }{6}$π/6, the angles 120° and 150° are $\frac{2\pi }{3}$2π/3 and $\frac{5\pi }{6}$5π/6, and 135° is $\frac{3\pi }{4}$3π/4.

Half a turn cut into six equal angles. Each is 180° ÷ 6 = 30°, which is $\frac{\pi }{6}$π/6, and the radii are at $\frac{\pi }{6}$π/6, $\frac{\pi }{3}$π/3, $\frac{\pi }{2}$π/2, $\frac{2\pi }{3}$2π/3, $\frac{5\pi }{6}$5π/6 and $\pi$π.

The same triangles, renamed

The exact values come from two triangles, and changing the unit of the angle does not change either triangle. Half of a square with sides of 1 has legs of 1 and 1, a hypotenuse of $\sqrt{2}$√2, and two angles of 45°, which are $\frac{\pi }{4}$π/4. So $\sin \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$sin(π/4) = 1/√2, $\cos \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$cos(π/4) = 1/√2 and $\tan \left(\frac{\pi }{4}\right)=1$tan(π/4) = 1, the same as sin 45°, cos 45° and tan 45°.

Half of an equilateral triangle with sides of 2 has a short leg of 1, a long leg of $\sqrt{3}$√3 and a hypotenuse of 2. Its angles are 30° and 60°, which are $\frac{\pi }{6}$π/6 and $\frac{\pi }{3}$π/3. The short leg is opposite $\frac{\pi }{6}$π/6, and the long leg is opposite $\frac{\pi }{3}$π/3.

Half of a square: legs of 1, a hypotenuse of $\sqrt{2}$√2, and two angles of $\frac{\pi }{4}$π/4.

Half of an equilateral triangle with sides of 2. The side of 1 is across from the angle $\frac{\pi }{6}$π/6, and the side of $\sqrt{3}$√3 is across from the angle $\frac{\pi }{3}$π/3.

The nine values

At $\frac{\pi }{6}$π/6 the opposite side is 1, the adjacent side is $\sqrt{3}$√3 and the hypotenuse is 2. So $\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}$sin(π/6) = 1/2, $\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}$cos(π/6) = √3/2 and $\tan \left(\frac{\pi }{6}\right)=\frac{1}{\sqrt{3}}$tan(π/6) = 1/√3.

At $\frac{\pi }{4}$π/4, $\sin \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$sin(π/4) = 1/√2, $\cos \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$cos(π/4) = 1/√2 and $\tan \left(\frac{\pi }{4}\right)=1$tan(π/4) = 1.

At $\frac{\pi }{3}$π/3 the two legs swap roles. So $\sin \left(\frac{\pi }{3}\right)=\frac{\sqrt{3}}{2}$sin(π/3) = √3/2, $\cos \left(\frac{\pi }{3}\right)=\frac{1}{2}$cos(π/3) = 1/2 and $\tan \left(\frac{\pi }{3}\right)=\sqrt{3}$tan(π/3) = √3.

Check one on a calculator set to radians: $\frac{\pi }{6}$π/6 is about 0.5236, and sin 0.5236 is 0.5000. Check a row with Pythagoras: at $\frac{\pi }{6}$π/6, ${\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{1}{4}+\frac{3}{4}=1$(1/2)² + (√3/2)² = 1/4 + 3/4 = 1.

Exam papers give these angles in radians without saying so. An expression such as $2\sin \left(\frac{\pi }{6}\right)+\tan \left(\frac{\pi }{4}\right)$2 sin(π/6) + tan(π/4) is then $2\times \frac{1}{2}+1=2$2 × 1/2 + 1 = 2, worked exactly with no calculator.

The quarter turn and zero

On a circle of radius 1, an angle $\theta$θ from the positive x-axis puts the point at $\left(\cos \theta ,\sin \theta \right)$(cos θ, sin θ). At the angle 0 the point is at (1, 0), so cos 0 = 1, sin 0 = 0 and tan 0 = 0.

A quarter turn, $\frac{\pi }{2}$π/2, puts the point at the top of the circle, (0, 1). So $\sin \left(\frac{\pi }{2}\right)=1$sin(π/2) = 1 and $\cos \left(\frac{\pi }{2}\right)=0$cos(π/2) = 0. The tangent is $\sin /\cos$sin / cos, and dividing by 0 has no value, so $\tan \left(\frac{\pi }{2}\right)$tan(π/2) is undefined.

A quarter turn on a circle of radius 1. The point is at the top, (0, 1): its height, $\sin \left(\frac{\pi }{2}\right)$sin(π/2), is 1, and its distance across, $\cos \left(\frac{\pi }{2}\right)$cos(π/2), is 0.

Past the quarter turn

The angle $\frac{2\pi }{3}$2π/3 is $\pi -\frac{\pi }{3}$π − π/3: it falls short of a half turn by $\frac{\pi }{3}$π/3. Its point on the circle is the mirror image, in the y-axis, of the point at $\frac{\pi }{3}$π/3. The height is the same and the distance across changes sign, so $\sin \left(\frac{2\pi }{3}\right)=\frac{\sqrt{3}}{2}$sin(2π/3) = √3/2, $\cos \left(\frac{2\pi }{3}\right)=-\frac{1}{2}$cos(2π/3) = −1/2 and $\tan \left(\frac{2\pi }{3}\right)=-\sqrt{3}$tan(2π/3) = −√3.

In the same way $\frac{5\pi }{6}$5π/6 is $\pi -\frac{\pi }{6}$π − π/6, so $\sin \left(\frac{5\pi }{6}\right)=\frac{1}{2}$sin(5π/6) = 1/2, $\cos \left(\frac{5\pi }{6}\right)=-\frac{\sqrt{3}}{2}$cos(5π/6) = −√3/2 and $\tan \left(\frac{5\pi }{6}\right)=-\frac{1}{\sqrt{3}}$tan(5π/6) = −1/√3. And $\frac{3\pi }{4}$3π/4 is $\pi -\frac{\pi }{4}$π − π/4, so $\sin \left(\frac{3\pi }{4}\right)=\frac{1}{\sqrt{2}}$sin(3π/4) = 1/√2, $\cos \left(\frac{3\pi }{4}\right)=-\frac{1}{\sqrt{2}}$cos(3π/4) = −1/√2 and $\tan \left(\frac{3\pi }{4}\right)=-1$tan(3π/4) = −1.

At a half turn, $\pi$π, the point is at (−1, 0), so $\sin \pi =0$sin π = 0 and $\cos \pi =-1$cos π = −1. Between $\frac{\pi }{2}$π/2 and $\pi$π the sizes are the familiar ones; the sine stays positive, and the cosine and the tangent are negative.

The point at 60°, which is $\frac{\pi }{3}$π/3: its height is $\sin \left(\frac{\pi }{3}\right)=0.87$sin(π/3) = 0.87 and its distance across is $\cos \left(\frac{\pi }{3}\right)=0.5$cos(π/3) = 0.5, to 2 decimal places. Drag it to 90°, which is $\frac{\pi }{2}$π/2, then on to 120°, which is $\frac{2\pi }{3}$2π/3: the height comes back to 0.87 and the distance across becomes −0.5.

The usual mistakes

Swapping sine and cosine. At $\frac{\pi }{6}$π/6 the side of 1 is opposite and the side of $\sqrt{3}$√3 is adjacent, so $\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}$sin(π/6) = 1/2 and $\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}$cos(π/6) = √3/2.

Taking a value from the wrong triangle. $\frac{1}{\sqrt{2}}$1/√2 belongs to $\frac{\pi }{4}$π/4, from the half-square; $\frac{\pi }{6}$π/6 and $\frac{\pi }{3}$π/3 come from the halved equilateral triangle, and their values contain $\frac{1}{2}$1/2 or $\sqrt{3}$√3.

Dividing by the hypotenuse for a tangent. $\tan \left(\frac{\pi }{4}\right)$tan(π/4) is 1 ÷ 1 = 1, not $\frac{1}{\sqrt{2}}$1/√2.

Working out $\sin \left(\frac{\pi }{6}\right)$sin(π/6) on a calculator set to degrees. That gives the sine of 0.5236°, about 0.0091, instead of 0.5.

Forgetting the sign past $\frac{\pi }{2}$π/2. $\cos \left(\frac{2\pi }{3}\right)$cos(2π/3) is $-\frac{1}{2}$−1/2, not $\frac{1}{2}$1/2, because the point is left of the y-axis.

A clock pendulum

In the application below, a pendulum swings to $\frac{\pi }{6}$π/6 on each side of the vertical. The arc of a swing uses $s=r\theta$s = rθ, and the height the bob rises uses the exact value $\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}$cos(π/6) = √3/2.