IGCSE Maths Trigonometry notes

In any right-angled triangle with hypotenuse c (the side opposite the right angle — always the longest), c² = a² + b². Finding the hypotenuse means adding the squares; finding a shorter side means subtracting: a² = c² − b². Keeping the operation straight is half the battle — a hypotenuse answer smaller than a leg is an instant red flag.

The theorem also earns marks inside other topics: the diagonal of a rectangle, the slant height of a cone, and the face diagonals of cuboids are all Pythagoras in disguise. Learn the small triples — 3-4-5, 5-12-13, 8-15-17 and their multiples — so exact answers jump out on non-calculator papers.

Label the sides from an acute angle θ: the hypotenuse H (opposite the right angle), the opposite O (across from θ) and the adjacent A (the remaining leg). Then sin θ = O⁄H, cos θ = A⁄H, tan θ = O⁄A. Exactly two of the three labels appear in each question — the given pair picks the ratio. To find a side, substitute and rearrange; to find an angle, use the inverse buttons sin⁻¹, cos⁻¹, tan⁻¹.

Two habits protect the marks: check the calculator is in degree mode, and give angles to 1 decimal place (sides 3 s.f.) — the syllabus convention. In word problems, sketch the triangle first; “the angle the ramp makes with the ground” is the angle between ramp and horizontal.

Extended. Six exact values come from two triangles: the 30-60-90 (sides 1, √3, 2) and the 45-45-90 (sides 1, 1, √2). They give sin 30 = ½, sin 45 = √2⁄2, sin 60 = √3⁄2; cos runs the same list backwards (cos 60 = ½, cos 45 = √2⁄2, cos 30 = √3⁄2); and tan 30 = 1⁄√3, tan 45 = 1, tan 60 = √3. The graphs explain why equations have two answers: y = sin x rises to 1 at 90° and falls back through zero at 180°, so any height below 1 is hit twice; y = cos x starts at 1 and mirrors about 180°.

To solve sin x = k for 0° ⩽ x ⩽ 360°, find the calculator answer θ = sin⁻¹ k, then the partner 180° − θ. For cos x = k, the partner is 360° − θ. For tan x = k, the partner is θ + 180°. Always finish by checking every solution sits inside the requested interval — and discard none without checking.

Extended. When no right angle is available: the sine rule a⁄sin A = b⁄sin B pairs each side with its opposite angle (use it with an angle–side opposite pair plus one more); the cosine rule a² = b² + c² − 2bc cos A handles two sides and the included angle, or any three sides when hunting an angle. Area = ½ab sin C, where C sits between sides a and b. Rearranging the cosine rule for an angle: cos A = (b² + c² − a²)⁄(2bc).

Two traps: sin of an obtuse angle is negative (cos 130° = −cos 50°), so a negative cosine just means the angle is obtuse; and the sine rule can have two valid answers (θ and 180° − θ) when the unknown angle could be obtuse — test the third angle to see which triangle survives. In 3-D, never use the 3-D angle directly: reduce the solid to a chain of flat right-angled triangles, often via a face diagonal, and for the “angle between a line and a plane”, drop a perpendicular and use sin = height⁄line-length.