Exact Sine, Cosine and Tangent at 30, 45 and 60°

Values that stay exact

A calculator gives sin 60° = 0.8660254…, a decimal that never ends, so any value it shows is rounded. For three angles, 30°, 45° and 60°, the sine, cosine and tangent can be written exactly, with square roots, and two simple shapes give all nine of them.

Half a square

Take a square with sides of 1 and cut it along a diagonal, from one corner to the opposite corner. The diagonal is a line of symmetry of the square, so it cuts each right angle it passes through into two equal halves: 90 ÷ 2 = 45°.

Each half is a right triangle with angles of 45°, 45° and 90°. Its two shorter sides, the legs, are sides of the square, so both are 1. Its hypotenuse is the diagonal. By Pythagoras’ theorem, the hypotenuse squared is $1^2+1^2=2$1² + 1² = 2, so the hypotenuse is $\sqrt{2}$√2.

A square with sides of 1, cut along its diagonal into two right triangles.

One half of the square: legs of 1, a hypotenuse of $\sqrt{2}$√2, and two angles of 45°.

The ratios at 45°

Stand at either 45° corner. The opposite side is 1, the adjacent side is 1 and the hypotenuse is $\sqrt{2}$√2. So $\sin 45^{\circ }=\frac{1}{\sqrt{2}}$sin 45° = 1/√2, $\cos 45^{\circ }=\frac{1}{\sqrt{2}}$cos 45° = 1/√2 and $\tan 45^{\circ }=\frac{1}{1}=1$tan 45° = 1/1 = 1.

The sine and cosine are equal because the two legs are equal. $\frac{1}{\sqrt{2}}$1/√2 can also be written $\frac{\sqrt{2}}{2}$√2/2: multiply the numerator and the denominator by $\sqrt{2}$√2 to rationalize the denominator. Both forms are exact, and both are about 0.7071.

Half an equilateral triangle

Now take an equilateral triangle with sides of 2. All three of its angles are 60°. Fold it in half along the line from the top corner to the middle of the base. That line is a line of symmetry, so the two halves are identical.

The fold halves the base, so each half has a base of 2 ÷ 2 = 1. It halves the top angle, so each half has an angle of 60 ÷ 2 = 30° at the top. And the two angles where the fold meets the base are equal and together make a straight line, so each is 90°. Each half is a right triangle with angles of 30°, 60° and 90°.

The side lengths of each half are: a hypotenuse of 2, which is a side of the original triangle, a short side of 1, and the fold line, whose length h is unknown. By Pythagoras’ theorem, $1^2+h^2=2^2$1² + h² = 2², so $h^2=4-1=3$h² = 4 − 1 = 3 and $h=\sqrt{3}$h = √3.

The side of 2 was chosen so that the half base comes out as a whole number, 1.

An equilateral triangle with sides of 2, and its fold line from the top corner to the middle of the base.

One half, turned to lie on its long leg. The side of 1 is across from the 30° angle, and the side of $\sqrt{3}$√3 is across from the 60° angle.

The ratios at 30° and 60°

At the 30° corner, the opposite side is 1, the adjacent side is $\sqrt{3}$√3 and the hypotenuse is 2. So $\sin 30^{\circ }=\frac{1}{2}$sin 30° = 1/2, $\cos 30^{\circ }=\frac{\sqrt{3}}{2}$cos 30° = √3/2 and $\tan 30^{\circ }=\frac{1}{\sqrt{3}}$tan 30° = 1/√3, which is $\frac{\sqrt{3}}{3}$√3/3.

At the 60° corner, the two legs swap roles: the opposite side is $\sqrt{3}$√3 and the adjacent side is 1. So $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$sin 60° = √3/2, $\cos 60^{\circ }=\frac{1}{2}$cos 60° = 1/2 and $\tan 60^{\circ }=\frac{\sqrt{3}}{1}=\sqrt{3}$tan 60° = √3/1 = √3.

An equilateral triangle with sides of 1 is this one at half the size, so its height is $\frac{\sqrt{3}}{2}$√3/2, the sine of 60°. Found directly by Pythagoras, that height is $\sqrt{1^2-0.5^2}=\sqrt{0.75}$√(1² − 0.5²) = √0.75, which is 0.866 to three decimal places. The two agree: $\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{\sqrt{4}}=\sqrt{\frac{3}{4}}=\sqrt{0.75}$√3/2 = √3/√4 = √(3/4) = √0.75.

The table

Here are all nine values together. Down the sine column the values grow, $\frac{1}{2}$1/2, then $\frac{1}{\sqrt{2}}$1/√2, then $\frac{\sqrt{3}}{2}$√3/2, and down the cosine column the same values come in the reverse order. Each tangent is the sine divided by the cosine.

If you forget one, draw the shape again: half a square for 45°, half an equilateral triangle of side 2 for 30° and 60°.

The sine, cosine and tangent of 30°, 45° and 60°, all exact.

A check with Pythagoras

In a right triangle with a hypotenuse of 1, the two legs are $\cos \theta$cos θ and $\sin \theta$sin θ. By Pythagoras’ theorem, the squares of the legs add up to the square of the hypotenuse, so ${\left(\cos \theta \right)}^2+{\left(\sin \theta \right)}^2=1$(cos θ)² + (sin θ)² = 1. That gives a check on each row of the table.

At 60°: ${\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{1}{4}+\frac{3}{4}=1$(1/2)² + (√3/2)² = 1/4 + 3/4 = 1. At 45°: ${\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{1}{2}+\frac{1}{2}=1$(1/√2)² + (1/√2)² = 1/2 + 1/2 = 1. At 30° the two values are the same as at 60°, swapped, so the check works again.

The radius of 1 at 30° has legs of cos 30° and sin 30°, with a square drawn on each. The bar on the right stacks the two areas: 0.75 + 0.25 = 1, which is ${\left(\frac{\sqrt{3}}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2$(√3/2)² + (1/2)². Turn the radius until the two squares are equal: that happens at 45°, where each is ${\left(\frac{1}{\sqrt{2}}\right)}^2=0.5$(1/√2)² = 0.5.

The usual mistakes

Taking the wrong leg. At 30°, the side of 1 is opposite and the side of $\sqrt{3}$√3 is adjacent, so $\sin 30^{\circ }=\frac{1}{2}$sin 30° = 1/2 and $\cos 30^{\circ }=\frac{\sqrt{3}}{2}$cos 30° = √3/2, not the other way round.

Taking the wrong corner. The 30° and 60° angles are in the same triangle, so their values are easy to swap by mistake. The larger angle faces the longer leg, so sin 60° is the larger of the two sines, $\frac{\sqrt{3}}{2}$√3/2.

Taking the wrong shape. $\frac{1}{\sqrt{2}}$1/√2 belongs to 45°, the half square. The half equilateral triangle, for 30° and 60°, has no $\sqrt{2}$√2 in it.

Giving a decimal. 0.866 is sin 60° rounded; $\frac{\sqrt{3}}{2}$√3/2 is its exact value.

Two steps for the application

An angle of elevation is the angle between the horizontal and the line of sight up to something, such as the top of a cliff seen from the ground.

A ratio can also give a side. If the height of the cliff is h and the distance to its foot is x, then $\tan 60^{\circ }=\frac{h}{x}$tan 60° = h/x. Multiply both sides by x: $h=x\tan 60^{\circ }=\sqrt{3}x$h = x tan 60° = √3 x.