Trigonometric Identities: Core List, Proof Ideas, and Example

Trigonometric identities are formulas involving $\sin$, $\cos$, $\tan$, and related functions that are true for every angle where both sides are defined. If you are looking for the standard trig identities used in algebra, precalculus, and early calculus, the core list is reciprocal, quotient, Pythagorean, even-odd, cofunction, sum-and-difference, double-angle, and half-angle identities.

The fastest way to make them stick is to group them by purpose. Some rewrite one trig function in terms of another, some connect $\sin \theta$ and $\cos \theta$, and some change the angle from $\theta$ to $2\theta$ or $\theta/2$.

What makes an equation a trigonometric identity?

An identity is true for every angle in its domain. For example,

$$\sin^2 \theta + \cos^2 \theta = 1$$

is an identity because it holds for every $\theta$.

By contrast,

$$\sin \theta = \frac{1}{2}$$

is not an identity. It is true only for specific angles.

The domain condition matters. For example,

$$\tan \theta = \frac{\sin \theta}{\cos \theta}$$

is true only when $\cos \theta \neq 0$.

Core trigonometric identities list

Reciprocal identities

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

Each formula requires the denominator to be nonzero.

Quotient identities

$$\tan \theta = \frac{\sin \theta}{\cos \theta}, \qquad \cot \theta = \frac{\cos \theta}{\sin \theta}$$

These are often the first step in simplification problems because they rewrite everything in terms of $\sin$ and $\cos$.

Pythagorean identities

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

The first identity is the source of the other two.

Even-odd identities

$$\sin(-\theta) = -\sin \theta, \qquad \cos(-\theta) = \cos \theta, \qquad \tan(-\theta) = -\tan \theta$$

The same pattern extends to the reciprocal functions: $\csc$ and $\cot$ are odd, while $\sec$ is even.

Cofunction identities

$$\sin\left(\frac{\pi}{2} - \theta\right) = \cos \theta$$ $$\cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta$$ $$\tan\left(\frac{\pi}{2} - \theta\right) = \cot \theta$$

These come from complementary angles.

Sum and difference identities

$$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$$ $$\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$$ $$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$$ $$\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$$ $$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$$ $$\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$$

For the tangent formulas, the denominator must be nonzero.

Double-angle identities

Set $\alpha = \beta = \theta$ in the angle-sum formulas.

$$\sin(2\theta) = 2 \sin \theta \cos \theta$$ $$\cos(2\theta) = \cos^2 \theta - \sin^2 \theta$$ $$\cos(2\theta) = 2\cos^2 \theta - 1$$ $$\cos(2\theta) = 1 - 2\sin^2 \theta$$ $$\tan(2\theta) = \frac{2\tan \theta}{1 - \tan^2 \theta}$$

The tangent version also needs $1 - \tan^2 \theta \neq 0$.

Half-angle identities

These come from rearranging the double-angle formulas.

$$\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$$ $$\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$$

For an angle written as $\theta/2$, the square-root forms are

$$\sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \theta}{2}}$$ $$\cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \theta}{2}}$$

The sign depends on the quadrant of $\theta/2$, so the $\pm$ cannot be dropped blindly.

Where the main trig identities come from

The unit circle gives the first Pythagorean identity

On the unit circle, the point at angle $\theta$ is $(\cos \theta, \sin \theta)$. Because every point on that circle satisfies $x^2 + y^2 = 1$, substituting $x = \cos \theta$ and $y = \sin \theta$ gives

$$\cos^2 \theta + \sin^2 \theta = 1$$

That is the basic Pythagorean identity.

The other Pythagorean identities come from division

If $\cos \theta \neq 0$, divide

$$\sin^2 \theta + \cos^2 \theta = 1$$

by $\cos^2 \theta$:

$$\frac{\sin^2 \theta}{\cos^2 \theta} + 1 = \frac{1}{\cos^2 \theta}$$ $$\tan^2 \theta + 1 = \sec^2 \theta$$

If $\sin \theta \neq 0$, dividing by $\sin^2 \theta$ gives

$$1 + \cot^2 \theta = \csc^2 \theta$$

Double-angle identities come from the angle-sum formulas

Start with

$$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$$

and set $\alpha = \beta = \theta$:

$$\sin(2\theta) = 2\sin \theta \cos \theta$$

The cosine and tangent double-angle identities are derived the same way.

Worked example: simplify a double-angle expression

Simplify

$$\frac{1 - \cos(2\theta)}{\sin(2\theta)}$$

for angles where the original expression is defined.

Use the double-angle identities:

$$1 - \cos(2\theta) = 1 - \left(1 - 2\sin^2 \theta\right) = 2\sin^2 \theta$$

and

$$\sin(2\theta) = 2\sin \theta \cos \theta$$

Now substitute:

$$\frac{1 - \cos(2\theta)}{\sin(2\theta)} = \frac{2\sin^2 \theta}{2\sin \theta \cos \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta$$

This conclusion is valid only where the original denominator is nonzero, so $\sin(2\theta) \neq 0$. That condition matters because canceling a factor can hide values that were excluded at the start.

Common mistakes with trigonometric identities

Ignoring domain restrictions is the mistake that causes the most trouble. Dividing by $\sin \theta$ or $\cos \theta$ is valid only when that quantity is not zero.

Another common error is dropping the $\pm$ in half-angle formulas. The square root alone does not determine the sign of the trig value.

Students also mix up $\sin^2 \theta$ and $\sin(\theta^2)$. The notation $\sin^2 \theta$ means $(\sin \theta)^2$.

When trig identities are used

Trig identities show up whenever you need to rewrite an expression into a more useful form. That includes simplifying homework problems, proving two expressions are equal, solving trig equations, and preparing for calculus topics such as integration.

In practice, many problems become easier once everything is rewritten in terms of $\sin \theta$ and $\cos \theta$.

Try a similar problem

Simplify

$$\frac{\sin(2\theta)}{1 + \cos(2\theta)}$$

using double-angle identities, and keep the domain condition from the original expression in view. If you want another step after that, compare your result with $\tan \theta$.