Guides / Trigonometry

Proving trig identities: work on one side, and the moves that actually help

An identity is an equation that's true for every angle where both sides are defined. Proving one isn't solving an equation, so the usual move of doing the same thing to both sides doesn't count. Here's what does: a handful of rewrites, applied to one side until it turns into the other.

Short answer

Pick the more complicated side and work only on it. Rewrite everything in sin x and cos x, then use one of four moves: a Pythagorean identity (sin²x + cos²x = 1, 1 + tan²x = sec²x), a common denominator, factoring, or multiplying top and bottom by a conjugate like 1 + cos x. Stop when you've reached the other side exactly, and say where the identity holds if it has a denominator.

Why it works

A proof has to start from something already known to be true and end at the claim. If you start from the claim itself and cross-multiply or square both sides, you've assumed what you were asked to prove. Arriving at 1 = 1 that way shows nothing, because false equations can get there too (mistake 1 below).

Working on one side avoids the problem. Each line equals the one before it, so the first line equals the last: left side = … = right side. Everything you use along the way is already known: the Pythagorean identity, the definitions tan x = sin x/cos x, sec x = 1/cos x, csc x = 1/sin x, cot x = cos x/sin x, and the double-angle formulas sin 2x = 2 sin x cos x and cos 2x = cos²x − sin²x.

Before you start a proof, plug in one angle to make sure the identity is even true. Use an angle like π/3 or π/6, not π/4: at π/4, sine and cosine are equal, so it can't tell them apart (mistake 3).

Example 1: rewrite in sine and cosine

Prove (1 − cos²x)(1 + tan²x) = tan²x.

(1 − cos²x)(1 + tan²x) = sin²x · sec²xWork on the left side. Each factor is one half of a Pythagorean identity: 1 − cos²x = sin²x and 1 + tan²x = sec²x.
= sin²x · 1cos²xsec x = 1/cos x.
= sin²xcos²xMultiply.
= tan²xThat's the right side. It holds wherever cos x ≠ 0, which is where tan x is defined.

Example 2: multiply by the conjugate

Prove sin x1 − cos x = 1 + cos xsin x.

sin x1 − cos x · 1 + cos x1 + cos xWork on the left side. Multiplying by (1 + cos x)/(1 + cos x) is multiplying by 1, and it turns the bottom into a difference of squares.
= sin x (1 + cos x)1 − cos²x(1 − cos x)(1 + cos x) = 1 − cos²x. Leave the top factored.
= sin x (1 + cos x)sin²xPythagorean identity.
= 1 + cos xsin xCancel one factor of sin x. Valid wherever sin x ≠ 0; that also rules out cos x = 1, so the original denominator is never 0 either.

Example 3: factor

Prove cos⁴x − sin⁴x = cos 2x.

cos⁴x − sin⁴x = (cos²x − sin²x)(cos²x + sin²x)A difference of squares, with cos²x and sin²x playing a and b.
= (cos²x − sin²x) · 1Pythagorean identity on the second factor.
= cos 2xThe double-angle formula. No denominators, so this holds for every x.

Example 4: common denominator

Prove tan x + cot x = 2sin 2x.

tan x + cot x = sin xcos x + cos xsin xWork on the left side, in sine and cosine.
= sin²x + cos²xsin x cos xCommon denominator sin x cos x.
= 1sin x cos xPythagorean identity.
= 22 sin x cos x = 2sin 2xMultiply top and bottom by 2 to make the double-angle formula appear. Valid wherever sin x ≠ 0 and cos x ≠ 0.

Common mistakes, and the exact line they happen on

1. Working on both sides

"Prove" sin x = √1 − cos²x.

sin²x = 1 − cos²x, which is true, so the identity is trueThis is the mistake. Both sides were squared, and squaring can turn a false equation into a true one, just as (−1)² = 1² doesn't make −1 equal 1. The claim is false: at x = 3π/2, sin x = −1 but √(1 − 0) = 1. Ending at something true only proves anything if every step can be reversed. Working on one side never has this problem.

2. Splitting a fraction over a sum in the denominator

Simplify 1sin x + cos x.

1sin x + cos x = 1sin x + 1cos xThis is the mistake. A fraction splits over a sum on top, (a + b)/c = a/c + b/c, never over a sum on the bottom. At x = π/4 the left side is about 0.71 and the right side is about 2.83. The distribution mistakes guide has the same trap with numbers.

3. A misremembered Pythagorean identity that passes the π/4 test

Rewrite 1 + tan²x in Example 1.

1 + tan²x = csc²xThis is the mistake. It's sec²x, not csc²x. Checking at π/4 doesn't catch it: 1 + 1 = 2, and csc²(π/4) = 2 as well, because sine and cosine are equal there. At π/3, 1 + tan² = 1 + 3 = 4 = sec², while csc² = 4/3. Test with an angle where sine and cosine differ. (The cosecant one is 1 + cot²x = csc²x.)

4. Canceling a term instead of a factor

Simplify sin x + cos xsin x.

sin x + cos xsin x = cos xThis is the mistake. sin x was canceled out of one term on top, but it isn't a factor of the whole top. Split the fraction instead: sin x/sin x + cos x/sin x = 1 + cot x. At π/4 the original is 2, while cos(π/4) ≈ 0.71. It's the same rule as in simplifying rational expressions: cancel factors, not terms.

Practice

Prove sec x − cos x = sin x tan x.
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Left side: 1/cos x − cos x = (1 − cos²x)/cos x = sin²x/cos x = sin x · (sin x/cos x) = sin x tan x, wherever cos x ≠ 0.
Prove (sin x + cos x)² = 1 + sin 2x.
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Expand the left side: sin²x + 2 sin x cos x + cos²x = 1 + 2 sin x cos x = 1 + sin 2x. Don't forget the middle term.
Prove cos x/(1 − sin x) = (1 + sin x)/cos x.
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Multiply the left side by (1 + sin x)/(1 + sin x): cos x (1 + sin x)/(1 − sin²x) = cos x (1 + sin x)/cos²x = (1 + sin x)/cos x, wherever cos x ≠ 0.
Prove (1 − cos 2x)/sin 2x = tan x.
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Use cos 2x = 1 − 2 sin²x on top and sin 2x = 2 sin x cos x on the bottom: 2 sin²x/(2 sin x cos x) = sin x/cos x = tan x, wherever sin 2x ≠ 0.
Is tan²x − sin²x = tan²x sin²x an identity?
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Yes. Left side: sin²x/cos²x − sin²x = sin²x (1 − cos²x)/cos²x = sin²x · sin²x/cos²x = tan²x sin²x, wherever cos x ≠ 0. A quick check at π/3: 3 − 3/4 = 9/4 and 3 · 3/4 = 9/4.