Math 30-1 · Trig identities
Trigonometric identities
An identity is an equation that is true for every allowed value of x. There are only a handful worth knowing, and almost every question is solved by the same move: rewrite everything in sine and cosine.
Reciprocal and quotient identities
The idea: every trig function can be written using just sin x and cos x. When a question looks messy, do that first.
| Function | In sine and cosine |
|---|---|
| tan x | sin x / cos x |
| cot x | cos x / sin x |
| sec x | 1 / cos x |
| csc x | 1 / sin x |
Worked example
Simplify csc x · tan x.
(1/sin x) · (sin x/cos x) = 1/cos x = sec x.
Watch out: sec goes with cos and csc goes with sin — the opposite of what the letters suggest. Each reciprocal pairs with the function whose name starts differently.
Pythagorean identities
The idea: on the unit circle, sin θ and cos θ are the legs of a right triangle with hypotenuse 1, so sin²θ + cos²θ = 1. The other two come from dividing that by cos²θ or sin²θ.
| Identity | Rearranged forms you will use |
|---|---|
| sin²x + cos²x = 1 | 1 − sin²x = cos²x · 1 − cos²x = sin²x |
| 1 + tan²x = sec²x | sec²x − 1 = tan²x |
| 1 + cot²x = csc²x | csc²x − 1 = cot²x |
Spot the pattern and substitute: (1 − sin²x)/cos x = cos²x/cos x = cos x, and (sec²x − 1)/tan x = tan²x/tan x = tan x.
Sum, difference and double-angle identities
The idea: these let you find exact values for angles you do not know, like 75° or 105°, by building them from angles you do know.
- sin(A ± B) = sin A cos B ± cos A sin B
- cos(A ± B) = cos A cos B ∓ sin A sin B (note the sign flips)
- tan(A ± B) = (tan A ± tan B) / (1 ∓ tan A tan B)
- sin 2A = 2 sin A cos A
- cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A
Exact values
cos 105° = cos(60° + 45°) = cos 60° cos 45° − sin 60° sin 45° = (1/2)(√2/2) − (√3/2)(√2/2) = (√2 − √6)/4. It comes out negative, which makes sense for Quadrant II.
Recognise the pattern backwards
2 sin 15° cos 15° is sin 2(15°) = sin 30° = 1/2. cos²(π/8) − sin²(π/8) is cos(π/4) = √2/2. Spotting these saves all the arithmetic.
Pick the form of cos 2A that fits
If you only know cos x = 1/3, use the version with only cosine: cos 2x = 2(1/3)² − 1 = −7/9.
Watch out: sin 2x is not 2 sin x, and sin(A + B) is not sin A + sin B. Functions do not distribute.
Proving identities and solving equations
The idea: to prove, work each side separately until they match. To solve, use an identity to get one function, then factor.
Proving: the playbook
- Start with the more complicated side.
- Rewrite in sine and cosine.
- Use a Pythagorean identity, find a common denominator, or factor.
- Never move terms across the equals sign — you have not shown it is an equation yet.
Checking one value, like x = π/4, only shows the equation holds for that value. It is a useful sanity check, never a proof.
Non-permissible values
An identity only holds where both sides are defined. For sin x/(1 − cos x), the denominator is zero when cos x = 1, so x ≠ 2πn. Tangent and secant are undefined wherever cos x = 0.
Solving with identities
Solve sin 2x = sin x for 0 ≤ x < 2π:
- Double-angle identity: 2 sin x cos x = sin x.
- Move everything to one side and factor: sin x(2cos x − 1) = 0.
- sin x = 0 gives 0 and π; cos x = 1/2 gives π/3 and 5π/3.
Four solutions: 0, π/3, π, 5π/3.
Watch out: dividing both sides by sin x instead of factoring. It throws away the solutions where sin x = 0.