Trig Identities Solver

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

$\csc\theta = \frac{1}{\sin\theta}, \quad \sec\theta = \frac{1}{\cos\theta}, \quad \cot\theta = \frac{\cos\theta}{\sin\theta}, \quad \tan\theta = \frac{\sin\theta}{\cos\theta}$

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

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

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

Double angle:

$\sin 2\theta = 2\sin\theta\cos\theta, \qquad \cos 2\theta = \cos^2\theta - \sin^2\theta = 1 - 2\sin^2\theta$

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 $\sin$ and $\cos$. This alone finishes most textbook problems.
3. Combine fractions over a common denominator.
4. Look for a Pythagorean pattern: any $1 - \sin^2\theta$, $\sec^2\theta - 1$ or $1 + \cot^2\theta$ collapses at once.
5. Multiply by a conjugate when $1 \pm \cos\theta$ sits in a denominator, since $(1-\cos\theta)(1+\cos\theta) = \sin^2\theta$.
6. Factor out common terms and differences of squares.

The conditions that come with it

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

• 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: $\sin(A + B) \neq \sin A + \sin B$, and $\cos 2\theta \neq 2\cos\theta$. Use the sum and double-angle formulas instead.
• Cancelling a factor that can be zero: dividing through by $\cos\theta$ quietly discards every angle where $\cos\theta = 0$.
• Dropping the $\pm$ in half-angle work: $\sin\frac{\theta}{2} = \pm\sqrt{\frac{1-\cos\theta}{2}}$, and the sign is decided by the quadrant of $\frac{\theta}{2}$.
• Ignoring the domain: an identity is only ever claimed where both sides are defined.