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Proving trigonometric identities

Prove trigonometric identities using compound and double angles, factorisation and common denominators. Track original domains, justify cancellation and distinguish proofs from numerical checks.

Before you startCompound and double angle formulae, algebraic fractions and trig domains

01 / A proof explains equality for every allowed input

Start with a domain statement and a useful direction.

An identity is an equality that holds throughout its stated domain. An equation may hold at only some inputs. A proof must use valid algebra and known identities; checking a few calculator values cannot establish a statement for every angle.

Original domains → valid transformations → target expression

The manual proof ledger keeps every line visible and highlights one justification at a time. The angle check helps expose domain holes, while the algebra supplies the proof.

Keep the domain beside the proofExplore

IdentityTest inputHighlight a proof step

Target: sin 2x/(1 + cos 2x) = tan x.

Common domain: cos x ≠ 0.

sin 2x/(1 + cos 2x)
2sin x cos x/(2cos² x)
sin x/cos x
tan x, still with cos x ≠ 0

Begin with the left side on its original domain.

At x = 30°: left = 0.5774; right = 0.5774.

Both sides exist and agree. A numerical match illustrates the identity; it does not prove it.

All four lines stay visible. Select a step to inspect its justification. Try 0°, 90° and 270° to compare the original domains.

02 / Choose the transformation that simplifies the structure

Do not expand everything before deciding what you need.

For mixed angles, try compound or double angle formulae.
For sec, cosec, cot or tangent fractions, rewriting in sine and cosine can reveal common factors.
For 1 ± cos 2x, use squared half-angle forms.
For 1 ± sin x in a denominator, a conjugate may create cos² x.
For even powers, use sin² x + cos² x = 1 or a double-angle reduction.

Often the more complicated side offers a useful starting point. Transform it into the simpler side, or transform both sides separately into the same expression. Do not assume the desired equality as an established fact.

03 / Choose the double-angle form that creates cancellation

The cancelled factor must be nonzero on the original domain.

Prove sin 2x/(1 + cos 2x) = tan x.Worked example

1 + cos 2x = 2cos² x

The original left side requires cos x ≠ 0, matching tangent’s domain.

sin 2x/(1 + cos 2x) = (2sin x cos x)/(2cos² x)

Use the sine and cosine double-angle identities.

= sin x/cos x

Cancel 2cos x, which is nonzero on the domain.

= tan x

The original restriction cos x ≠ 0 remains.

04 / Cancellation never fills an original hole

A simpler formula may exist at more inputs.

Prove (1 − cos 2x)/sin 2x = tan x on its common domain.Worked example

The original denominator requires sin 2x ≠ 0

Hence both sin x and cos x are nonzero.

(1 − cos 2x)/sin 2x = 2sin² x/(2sin x cos x)

Use compatible double-angle forms.

= tan x, with sin x cos x ≠ 0

Cancellation keeps the original restriction.

At x = 0°, tangent is zero but the original quotient is 0/0

The identity cannot extend the original expression to this input.

Two expressions agreeing on their common domain need not define the same function if they have different domains.

Watch cancellation leave a missing input

Pause, replay or seek freely. The notes explain the same idea and stay in view.

05 / Use a conjugate to create a Pythagorean factor

State the common domain before multiplying.

Prove cos x/(1 − sin x) = (1 + sin x)/cos x on the common domain.Worked example

The right side requires cos x ≠ 0

This also ensures 1 − sin x and 1 + sin x are nonzero.

cos x/(1 − sin x) = cos x(1 + sin x)/((1 − sin x)(1 + sin x))

Multiply by the conjugate divided by itself.

= cos x(1 + sin x)/cos² x

Use 1 − sin² x = cos² x.

= (1 + sin x)/cos x

Cancel one nonzero cosine factor.

At x = 270°, the left side is 0/2 = 0 but the right side is 0/0 and undefined. This is why the common-domain qualification is necessary.

06 / Combine fractions before applying an identity

A shared denominator can reveal a constant numerator.

Prove 1/(1 − sin x) + 1/(1 + sin x) = 2sec² x.Worked example

The original denominators exclude sin x = ±1

Equivalently cos x ≠ 0, also required by secant.

The numerator after combining is (1 + sin x) + (1 − sin x) = 2

Use the common denominator (1 − sin x)(1 + sin x).

The denominator is 1 − sin² x = cos² x

Apply Pythagoras.

The result is 2/cos² x = 2sec² x

Keep the common original domain.

07 / Exploit symmetry before expanding high powers

Paired plus and minus expressions cancel odd terms.

Prove (sin x + cos x)⁴ + (sin x − cos x)⁴ = 3 − cos 4x.Worked example

Let s = sin x and c = cos x

This keeps the algebra readable.

(s + c)⁴ + (s − c)⁴ = 2(s⁴ + 6s²c² + c⁴)

The odd cross terms cancel.

= 2((s² + c²)² + 4s²c²) = 2 + 8s²c²

Use s² + c² = 1.

= 2 + 2sin² 2x

Since sin 2x = 2sc.

= 3 − cos 4x

Use 2sin² 2x = 1 − cos 4x.

No division occurred, so this identity holds for every real x.

08 / Derive a higher-angle identity from a lower one

Replace the whole input angle consistently.

Prove cos 4x = 8cos⁴ x − 8cos² x + 1.Worked example

cos 4x = 2cos² 2x − 1

Apply the double-angle formula at angle 2x.

= 2(2cos² x − 1)² − 1

Replace the inner cosine.

= 8cos⁴ x − 8cos² x + 1

Expand the square carefully.

As a check, at x = 0 both sides equal 1. That is useful for detecting a typo, but the displayed derivation is what proves the identity.

09 / Know which transformations change a claim

Squaring and taking roots can lose sign information.

From sin² x + cos² x = 1, we get √(1 − sin² x) = |cos x|. Replacing the absolute value with cos x requires cos x ≥ 0. At x = π, the positive square root is 1 and cosine is −1.

Disprove sin 2x = 2sin x as an identityWorked example

At x = 45°, the left side is sin 90° = 1

Choose a simple allowed input.

The right side is 2sin 45° = √2

The values differ.

Why is “square both sides and get the same expression” insufficient by itself to prove A = B?

Hint

Think about A = 1 and B = −1.

Worked solution

Equal squares allow A = B or A = −B. You need sign conditions or another reversible argument to conclude equality. For example, 1² = (−1)² but 1 ≠ −1.

11 / Recap

A valid proof keeps its assumptions visible.

Choose an identity or algebraic method that simplifies the structure.
Transform one side into the other without assuming the claim.
State common domains and retain original exclusions after cancellation.
Use counterexamples to disprove claims; finite numerical checks cannot prove identities.
Track signs when squaring or taking square roots.

Section 1 of 11 · A proof explains equality for every allowed input