Trig — Need to Know

sin A = opphyp   cos A = adjhyp   tan A = oppadj

H² = O² + A²

Doubling the angle does not double the sine, cosine or tan. The formulas are in the tables:

sin 2A = 2 sin A cos A

sin 2A = 2 tan A1 + tan² A

You also need cos A. Draw the right-angled triangle from the given ratio, find the missing side by Pythagoras, then sub into 2 sin A cos A.

Use the tan-form of sin 2A — it has only one unknown. Let t = tan A, form a quadratic, factor and solve, then convert each tan A back to sin A by triangle.

There are two answers.

cos(A+B) does not split into cos A + cos B. Use the four compound-angle formulas from the tables.

For exact sin / cos / tan of 30°, 45°, 60° use two reference triangles:

45° isosceles — legs 1, hypotenuse √2.

30°/60° equilateral split — hypotenuse 2, base 1, height √3.

Write the angle as a sum or difference of known angles (e.g. 15° = 45° − 30°, 105° = 45° + 60°), apply the compound formula, then rationalise the denominator.

Sub into the tan(A+B) formula, let t = tan B, cross-multiply and solve.

There is no cos 3A formula in the tables. Write 3A = A + 2A, apply cos(A+B), expand cos 2A and sin 2A, then convert everything to cos A using sin² A = 1 − cos² A:

cos 3A = 4 cos³ A − 3 cos A

sin A + sin B = 2 sinA+B2 cosA−B2

sin A − sin B = 2 cosA+B2 sinA−B2

cos A + cos B = 2 cosA+B2 cosA−B2

cos A − cos B = −2 sinA+B2 sinA−B2

2 cos A cos B = cos(A+B) + cos(A−B)

2 sin A cos B = sin(A+B) + sin(A−B)

2 sin A sin B = cos(A−B) − cos(A+B)

2 cos A sin B = sin(A+B) − sin(A−B)

cos A = x,   sin A = y,   tan A = yx = sin Acos A

Q1 All, Q2 Sin, Q3 Tan, Q4 Cos — positive.

Ask: sin, cos or tan? And positive or negative?

Use ASTC to pick the two quadrants. The reference angle is the Q1 angle taken from the positive value.

Degrees:  Q2 = 180 − A,  Q3 = 180 + A,  Q4 = 360 − A.

Radians:  Q2 = π − A,  Q3 = π + A,  Q4 = 2π − A.

π = 180°,  one full circle = 2π.

Find the basic quadrant solutions in [0°, 360°) or [0, 2π) using ASTC, then add 360°n (degrees) or 2nπ (radians) to each.

Multiply the range by the coefficient (2 for 2A, 3 for 3A). Find every solution in that bigger range — keep adding 360° until you exceed it.

For general solutions, divide the +2nπ by the coefficient too (e.g. +2nπ3 for 3A).

f(x) = a + b sin cx  →  range = [a−b,  a+b],  period = 360c

f(x) = a + b cos cx  →  range = [a−b,  a+b],  period = 360c = 2πc

Range = [ how low , how high ]. Period = how quickly it repeats.

−1 ≤ sin θ ≤ 1,   −1 ≤ cos θ ≤ 1

asin A = bsin B

a² = b² + c² − 2bc cos A

½ ab sin C

Opposites full? Yes ⇒ Sine Rule. None ⇒ Cosine Rule.

The largest angle is opposite the longest side.

You need 3 pieces of information to find the 3rd.

A = πr²(θ360)

ℓ = 2πr(θ360)

A = ½ r²θ

ℓ = rθ

Compute the full sector area, then subtract the triangle area ½ ab sin C.

Read carefully — is the angle in degrees or radians? Pick the matching formula.

Given an arc and asked for an area, switch to radians via θ = ℓr first — it's almost always faster.

Every 3D problem becomes a 2D one. Pick the right triangle inside the shape, then use one of four tools: Pythagoras, Sine Rule, Cosine Rule, Area.

Wherever a vertical line meets horizontal ground you have a right angle.

One length or angle feeds the next triangle — e.g. Sine Rule for a base length, then Pythagoras for a vertical, then Cosine Rule for the angle between.

The angle of elevation sits in the vertical right-angled triangle: tan = heighthorizontal.

Find the half-diagonal of the base first (corner to centre). Then stand the apex up as a right-angled triangle to get the slant edge. Face area by ½ ab sin C.

Chain Pythagoras: find the base diagonal d first (across L × W), then the space diagonal D as the hypotenuse of (d, height).

Before using the radian sector formulas, convert (e.g. 60° = π3), then plug in.

If each centre lies on the other circle, the centre-to-intersection triangle has all sides = r — equilateral — so the half-angle is π3.

The overlap (a lens) = 2 × segment, and each segment = sector − triangle.

Arc → central angle via ℓ = rθ, then chord by the Cosine Rule on the isosceles radius triangle.

Equal chords subtend equal central angles; the central angles round the centre sum to 2π.

Produce each quantity separately, then take the ratio (or difference) the question asks for.

cos⁻¹ x = A — cos⁻¹ is really an angle.

Add · Subtract,  Multiply · Divide,  Indices · Logs,  Cos · cos⁻¹.

sin(cos⁻¹ x) = √(1 − x²)

tan(sin⁻¹ x) = x√(1 − x²)

Tables — rewrite tan, sec, cot, csc as sin / cos.

Common denominator — if fractions sit on top of each other.

Pythagoras — spot sin² + cos² becoming 1, or 1 − sin² becoming cos².

Difference of squares — (a+b)(a−b) = a² − b², either direction.

Factor — pull out a common factor like sin θ.

Double angles — pick the cos 2A form that matches: 1 + cos 2A → 2 cos² A, 1 − cos 2A → 2 sin² A.

Take two points on a unit circle: P = (cos A, sin A), Q = (cos B, sin B). Find |PQ|² two ways — by the Cosine Rule and by the distance formula — then equate:

cos(A − B) = cos A cos B + sin A sin B

cos(A+B): replace B with −B.

sin(A+B): replace A with 90 − A and use cos(90 − A) = sin A.

sin(A−B): replace B with −B.

Write tan as sin(A±B)cos(A±B), then divide top and bottom by cos A cos B.

Drop the perpendicular h. Write it two ways: h = b sin A = a sin B, then equate ⇒ asin A = bsin B.

Drop the perpendicular h, splitting the base into x and c − x. Find h² from both right triangles: h² = b² − x² = a² − (c − x)². Equate and sub x = b cos A ⇒

a² = b² + c² − 2bc cos A