Trig identities
The identities are printed on your formula sheet, so these questions are never about recall — they are about choosing the substitution that makes an expression collapse. Here is the full list sorted by what it is good for, a strategy for proving, and worked proofs of the four shapes that keep coming back.
- 1. The list, sorted by use
- 2. How to prove one
- 3. Four worked proofs
- 4. Identities inside equations
- 5. Verify is not prove
The list, sorted by use
The idea: do not learn them as a list of twenty. Learn what each group is for.
| Group | Identities | Use it when |
|---|---|---|
| Reciprocal | csc θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ | the expression mixes ratios — convert everything to sine and cosine |
| Quotient | tan θ = sin θ/cos θ, cot θ = cos θ/sin θ | same reason: get down to two functions |
| Pythagorean | sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, 1 + cot²θ = csc²θ | you see a squared term, or a 1 that needs to become something |
| Sum and difference | sin(α ± β), cos(α ± β), tan(α ± β) | an angle is a sum, or you need an exact value for 15°, 75°, 105° |
| Double angle | sin 2α = 2 sin α cos α; cos 2α = cos²α − sin²α = 2cos²α − 1 = 1 − 2sin²α | an angle is doubled, or you want to halve one |
The three versions of cos 2α exist so you can choose the one that leaves you with the function you want. If the rest of the expression is all sines, use 1 − 2sin²α and everything matches.
Exact values from the sum identities
sin 75° = sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30° = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4.
Pick the two special angles that add or subtract to the one you want: 75 = 45 + 30, 15 = 45 − 30, 105 = 60 + 45.
How to prove one
The idea: work on one side only, and drive it toward the other. Never move terms across the equals sign — you are not solving an equation.
- Start with the messier side. There is more to simplify, and more room to make progress.
- Convert to sine and cosine if the two sides use different functions. It nearly always unlocks the problem.
- Look for a Pythagorean pattern. sin²θ, cos²θ, a lone 1, or tan²θ next to a 1 — these are asking to be replaced.
- Combine fractions over a common denominator, then look again.
- Multiply by a conjugate when you have 1 ± something: (1 − cos θ)/(1 − cos θ) turns a sum into a difference of squares and sets up a Pythagorean substitution.
- Factor — common factors, differences of squares, and quadratics in sin θ all show up.
- State restrictions where a denominator could be zero; markers look for them.
Watch out: writing the whole identity and cross-multiplying assumes the thing you are trying to prove. Keep a vertical line down the page: left side on the left, right side untouched on the right, until they match.
Four worked proofs
The idea: almost every proof you will be given is one of these four shapes.
1. Convert to sine and cosine: (1 − cos²θ)/(sin θ cos θ) = tan θ
Left side: 1 − cos²θ = sin²θ by the Pythagorean identity, so the fraction is sin²θ/(sin θ cos θ). Cancel one sine: sin θ/cos θ = tan θ = right side. ∎ (sin θ ≠ 0, cos θ ≠ 0.)
2. Common denominator: 1/(1 + cos θ) + 1/(1 − cos θ) = 2csc²θ
Left side over the common denominator (1 + cos θ)(1 − cos θ) = 1 − cos²θ = sin²θ:
[(1 − cos θ) + (1 + cos θ)]/sin²θ = 2/sin²θ = 2csc²θ = right side. ∎
3. Conjugate: sin θ/(1 + cos θ) = (1 − cos θ)/sin θ
Multiply the left side by (1 − cos θ)/(1 − cos θ):
sin θ(1 − cos θ)/(1 − cos²θ) = sin θ(1 − cos θ)/sin²θ = (1 − cos θ)/sin θ = right side. ∎
4. Double angle: (sin 2θ)/(1 + cos 2θ) = tan θ
Choose the version of cos 2θ that makes the denominator collapse: cos 2θ = 2cos²θ − 1, so 1 + cos 2θ = 2cos²θ.
Then (2 sin θ cos θ)/(2cos²θ) = sin θ/cos θ = tan θ. ∎ Choosing 1 − 2sin²θ instead leaves 2 − 2sin²θ and a longer road.
Identities inside equations
The idea: if an equation contains two different functions, substitute an identity so only one is left — then it is an ordinary equation to solve on the unit circle.
Worked: 2cos²θ + sin θ = 1 on 0 ≤ θ < 2π
- One function only: cos²θ = 1 − sin²θ, so 2(1 − sin²θ) + sin θ = 1.
- Tidy: 2 − 2sin²θ + sin θ = 1, so 2sin²θ − sin θ − 1 = 0.
- Factor: (2 sin θ + 1)(sin θ − 1) = 0.
- Solve each: sin θ = −1/2 gives θ = 7π/6, 11π/6. sin θ = 1 gives θ = π/2.
- Answer: θ = π/2, 7π/6, 11π/6.
- Double angles: sin 2θ = cos θ becomes 2 sin θ cos θ − cos θ = 0, then cos θ(2 sin θ − 1) = 0. Factor — never divide by cos θ, or you lose the cos θ = 0 solutions.
- Check the domain the question gives, and whether it is in degrees or radians.
- Restricted values: if the original equation had tan θ or sec θ in it, discard any solution where cos θ = 0.
Verify is not prove
The idea: two different instructions, two different answers. Diploma markers single this out as a habit worth fixing.
| Word | Asks you to | Looks like |
|---|---|---|
| Verify | show it holds for one particular value | substitute θ = π/4 into both sides and show both give the same number |
| Prove | show it holds for every permitted value | transform one side into the other, with reasons |
Substituting a value into an identity you were asked to prove earns almost nothing, no matter how neat the arithmetic — one value cannot establish a general statement.