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Trig Identities Solver

Prove and simplify trigonometric identities with AI-powered step-by-step working
Prove sin(x)/(1 + cos(x)) + (1 + cos(x))/sin(x) = 2csc(x)
Simplify (1 - cos^2 x)(1 + cot^2 x)
Verify (1 - cos(2x))/sin(2x) = tan(x)
Simplify sec(x) - sec(x)sin^2(x)

The Core Trig Identities

A trigonometric identity is an equation that holds for every angle where both sides are defined — unlike a trig equation, which is true only at particular angles.

Reciprocal and quotient identities

csc⁡θ=1sin⁡θ,sec⁡θ=1cos⁡θ,cot⁡θ=cos⁡θsin⁡θ,tan⁡θ=sin⁡θcos⁡θ\csc\theta = \frac{1}{\sin\theta}, \quad \sec\theta = \frac{1}{\cos\theta}, \quad \cot\theta = \frac{\cos\theta}{\sin\theta}, \quad \tan\theta = \frac{\sin\theta}{\cos\theta}

Pythagorean identities — all three follow from x2+y2=1x^2 + y^2 = 1 on the unit circle:

sin⁡2θ+cos⁡2θ=1,1+tan⁡2θ=sec⁡2θ,1+cot⁡2θ=csc⁡2θ\sin^2\theta + \cos^2\theta = 1, \quad 1 + \tan^2\theta = \sec^2\theta, \quad 1 + \cot^2\theta = \csc^2\theta

Even and odd: cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta, while sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta and tan⁡(−θ)=−tan⁡θ\tan(-\theta) = -\tan\theta.

Double angle:

sin⁡2θ=2sin⁡θcos⁡θ,cos⁡2θ=cos⁡2θ−sin⁡2θ=1−2sin⁡2θ\sin 2\theta = 2\sin\theta\cos\theta, \qquad \cos 2\theta = \cos^2\theta - \sin^2\theta = 1 - 2\sin^2\theta

How to Prove a Trig Identity

The rules of the game

A proof must turn one side into the other, or reduce each side independently to the same expression. You may not cross-multiply or square both sides — that assumes the very statement you are proving.

A reliable order of attack

  1. Start on the messier side, the one with more terms, fractions or distinct functions.
  2. Rewrite everything in sin⁡\sin and cos⁡\cos. This alone finishes most textbook problems.
  3. Combine fractions over a common denominator.
  4. Look for a Pythagorean pattern: any 1−sin⁡2θ1 - \sin^2\theta, sec⁡2θ−1\sec^2\theta - 1 or 1+cot⁡2θ1 + \cot^2\theta collapses at once.
  5. Multiply by a conjugate when 1±cos⁡θ1 \pm \cos\theta sits in a denominator, since (1−cos⁡θ)(1+cos⁡θ)=sin⁡2θ(1-\cos\theta)(1+\cos\theta) = \sin^2\theta.
  6. Factor out common terms and differences of squares.

The conditions that come with it

Every identity carries a domain restriction. tan⁡θ=sin⁡θ/cos⁡θ\tan\theta = \sin\theta/\cos\theta fails at θ=π2+nπ\theta = \frac{\pi}{2} + n\pi, and anything containing cot⁡θ\cot\theta or csc⁡θ\csc\theta excludes sin⁡θ=0\sin\theta = 0. A complete proof names those excluded angles.

Common Mistakes to Avoid

  • Working both sides at once: manipulating the equation as though it were already true proves nothing. Keep the two sides apart.
  • Distributing the function name: sin⁡(A+B)≠sin⁡A+sin⁡B\sin(A + B) \neq \sin A + \sin B, and cos⁡2θ≠2cos⁡θ\cos 2\theta \neq 2\cos\theta. Use the sum and double-angle formulas instead.
  • Cancelling a factor that can be zero: dividing through by cos⁡θ\cos\theta quietly discards every angle where cos⁡θ=0\cos\theta = 0.
  • Dropping the ±\pm in half-angle work: sin⁡θ2=±1−cos⁡θ2\sin\frac{\theta}{2} = \pm\sqrt{\frac{1-\cos\theta}{2}}, and the sign is decided by the quadrant of θ2\frac{\theta}{2}.
  • Ignoring the domain: an identity is only ever claimed where both sides are defined.

示例题目

Step 1: Combine the left side over the common denominator sin⁡θ(1+cos⁡θ)\sin\theta(1+\cos\theta): sin⁡2θ+(1+cos⁡θ)2sin⁡θ(1+cos⁡θ)\dfrac{\sin^2\theta + (1+\cos\theta)^2}{\sin\theta(1+\cos\theta)}
Step 2: Expand the numerator: sin⁡2θ+1+2cos⁡θ+cos⁡2θ\sin^2\theta + 1 + 2\cos\theta + \cos^2\theta
Step 3: Apply sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1: the numerator becomes 2+2cos⁡θ=2(1+cos⁡θ)2 + 2\cos\theta = 2(1+\cos\theta)
Step 4: Cancel the factor 1+cos⁡θ1+\cos\theta: 2(1+cos⁡θ)sin⁡θ(1+cos⁡θ)=2sin⁡θ=2csc⁡θ\dfrac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)} = \dfrac{2}{\sin\theta} = 2\csc\theta
Step 5: Valid for all θ\theta with sin⁡θ≠0\sin\theta \neq 0 and cos⁡θ≠−1\cos\theta \neq -1
Answer: Both sides equal 2csc⁡θ2\csc\theta, so the identity holds.

Step 1: Use the Pythagorean identity on the first factor: 1−cos⁡2x=sin⁡2x1 - \cos^2 x = \sin^2 x
Step 2: Use the Pythagorean identity on the second factor: 1+cot⁡2x=csc⁡2x1 + \cot^2 x = \csc^2 x
Step 3: Multiply: sin⁡2x⋅csc⁡2x=sin⁡2x⋅1sin⁡2x\sin^2 x \cdot \csc^2 x = \sin^2 x \cdot \dfrac{1}{\sin^2 x}
Step 4: The sin⁡2x\sin^2 x terms cancel, leaving 11 (for sin⁡x≠0\sin x \neq 0)
Answer: 11

Step 1: Rewrite the numerator with cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2 x: 1−(1−2sin⁡2x)=2sin⁡2x1 - (1 - 2\sin^2 x) = 2\sin^2 x
Step 2: Rewrite the denominator with sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x
Step 3: Form the quotient: 2sin⁡2x2sin⁡xcos⁡x\dfrac{2\sin^2 x}{2\sin x\cos x}
Step 4: Cancel 2sin⁡x2\sin x: sin⁡xcos⁡x=tan⁡x\dfrac{\sin x}{\cos x} = \tan x
Step 5: Requires sin⁡2x≠0\sin 2x \neq 0, i.e. x≠nπ2x \neq \frac{n\pi}{2}
Answer: The identity holds, with both sides equal to tan⁡x\tan x.

常见问题

An identity is true for every angle in its domain, so proving it means showing the two sides are the same expression. A trig equation is true only for specific angles, so solving it means finding those angles. You verify identities; you solve equations.

The three Pythagorean identities, the quotient and reciprocal definitions, and the sum, difference and double-angle formulas cover almost every problem. Everything else — half-angle, product-to-sum, cofunction — can be re-derived from those in a line or two.

No. Cross-multiplying treats the equation as already true, which is exactly what you are being asked to establish. Transform one side only, or simplify each side separately until they meet at the same expression.

Usually because the expression was not converted to sine and cosine early enough, or a common denominator was never formed. Rewrite everything in terms of sin and cos, combine the fractions, then scan for a Pythagorean pattern to collapse.

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