Trigonometry

Pythagoras' theorem

In any right-angled triangle the three side lengths are locked together by a single relationship. Know two of them and the third is fixed — no angle needed.

The theorem

Label the two shorter sides a and b and the hypotenuse c. Then the square on the hypotenuse equals the sum of the squares on the other two sides: a² + b² = c². Rearranging lets you find a shorter side by subtraction instead: a² = c² − b².

Shortest distance to a line

The shortest distance from a point to a line is measured along the perpendicular. That perpendicular creates a right-angled triangle, so Pythagoras — or a ratio from the next section — finds it.

Worked example: a 5 m ladder rests against a vertical wall with its foot 1.4 m from the base of the wall. The wall, ground and ladder form a right-angled triangle, with the ladder as the hypotenuse: height² = 5² − 1.4² = 25 − 1.96 = 23.04, so height = √23.04 = 4.8 m — a whole number here, but always keep full accuracy until the end.

Right-angled trigonometry

Add one known angle to a right-angled triangle and the three side ratios — sine, cosine and tangent — connect the sides to that angle. The mnemonic is SOHCAHTOA.

The three ratios

Name the sides relative to the angle θ: the opposite faces it, the adjacent lies alongside it, and the hypotenuse is opposite the right angle: sin θ = opp/hyp, cos θ = adj/hyp, tan θ = opp/adj. To find an unknown angle, apply the inverse function (sin⁻¹, cos⁻¹, tan⁻¹).

Worked example: from a point 40 m from the foot of a tower, the angle of elevation of the top is 32°. The height is opposite the 32° angle; the 40 m is adjacent — so use tan: tan 32° = height ÷ 40, so height = 40 × tan 32° = 40 × 0.6249 = 25.0 m (3 s.f.) — taller than a house, as the angle suggests.

Exact trigonometric values

Two special triangles — the half-square and the half-equilateral — give the exact sine, cosine and tangent of the key angles without a calculator.

Where they come from

A right-angled isosceles triangle with two sides of 1 has hypotenuse √2, giving the ratios for 45°. Cutting an equilateral triangle of side 2 in half gives sides 1, √3 and 2, which yield 30° and 60°.

Worked example: a right-angled triangle has a hypotenuse of 10 cm and an angle of 30°. Find the exact length of the side opposite the 30° angle: opposite = hyp × sin θ = 10 × sin 30°. Since sin 30° = 1⁄2 exactly, opposite = 10 × 1⁄2 = 5 cm — no rounding, because the value was exact from the start.

Graphs and equations

Extend the ratios beyond a right-angled triangle and each becomes a curve. Their shape and symmetry over 0°–360° is what lets you solve a trigonometric equation completely.

The three curves

Over 0° ≤ x ≤ 360°, y = sin x and y = cos x wave between −1 and 1, one quarter-turn apart. y = tan x rises without bound, repeating every 180° with vertical asymptotes at 90° and 270°.

Solving an equation

Take the inverse function for the first solution, then use symmetry: sin has a second solution at 180° − x, and cos at 360° − x, within the range.

Worked example: solve sin x = 0.5 for 0° ≤ x ≤ 360°. First solution: x = sin⁻¹(0.5) = 30°. sin is also positive in the second quadrant: x = 180° − 30° = 150°. Both lie in range, so both count — a lone 30° would drop half the marks.

Sine and cosine rules

Once a triangle has no right angle, SOHCAHTOA no longer applies. Two rules take over: use the cosine rule when the odd one out is an included angle or three sides, and the sine rule otherwise: a⁄sin A = b⁄sin B = c⁄sin C, and a² = b² + c² − 2bc cos A.

Worked example: a triangular plot has two sides of 8 m and 11 m meeting at an angle of 35°. The 35° is between the two given sides, so use area = ½ab sin C = ½ × 8 × 11 × sin 35° = 44 × 0.5736 = 25.2 m² (3 s.f.) — the included angle is essential; two sides alone are not enough.

Trigonometry in 3D

Three-dimensional problems need no new formulae. The whole method is to find a right-angled triangle inside the solid, draw it flat, and use the tools from the earlier sections.

A two-step routine

First use Pythagoras on the base to find a length you cannot see directly — often a diagonal. Then place that length in a vertical right-angled triangle and take a ratio for the angle or height you want.

Worked example: a pyramid has a square base of side 6 cm and its apex 8 cm above the centre. Base diagonal = √(6² + 6²) = √72, so half of it = 4.243 cm. The edge, height and half-diagonal form a right-angled triangle: tan θ = 8 ÷ 4.243, so θ = tan⁻¹(1.886) = 62.1° (1 d.p.) — measured up from the base, as required.