Proving Trig Identities

A trig identity is an equation that’s true for every angle where both sides are defined, like $\sin^2\theta + \cos^2\theta = 1$. Proving identities trains you to see one expression in several different forms, a skill you’ll use to simplify problems throughout calculus and beyond.

Key ideas

Identity or equation?

An equation, like $\sin\theta = \tfrac{1}{2}$, is true only for some angles. An identity is true for all angles (where defined). Checking one angle can show something is not an identity, but it can’t prove that it is one.

The basic identities

Pythagorean identity. For a point on the unit circle, $x = \cos\theta$, $y = \sin\theta$, and $r = 1$. The Pythagorean theorem $x^2 + y^2 = r^2$ becomes:

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

The legs are $\cos\theta$ and $\sin\theta$, and the hypotenuse is $1$.

Here $\sin^2\theta$ means $(\sin\theta)^2$. Rearranged versions are just as useful: $\sin^2\theta = 1 - \cos^2\theta$ and $\cos^2\theta = 1 - \sin^2\theta$.

Quotient identity. Since $\tan\theta = \tfrac{y}{x}$, $\sin\theta = \tfrac{y}{r}$, and $\cos\theta = \tfrac{x}{r}$:

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

Reciprocal identities. $\csc\theta = \dfrac{1}{\sin\theta}$, $\sec\theta = \dfrac{1}{\cos\theta}$, $\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}$.

How to prove an identity

Write the left side (LS) and right side (RS) separately, and transform one side until it matches the other. Don’t move terms across the equals sign; you’re showing they’re equal, not assuming it.

Strategies:

1. Start with the more complicated side.
2. Rewrite everything in terms of $\sin$ and $\cos$.
3. Combine fractions with a common denominator.
4. Look for $\sin^2\theta + \cos^2\theta$, or a rearranged form, to replace.
5. Factor (common factors, difference of squares).

Worked examples

Example 1: Using the quotient identity

Prove $\tan\theta\cos\theta = \sin\theta$.

Solution.

$$\text{LS} = \tan\theta\cos\theta = \frac{\sin\theta}{\cos\theta} \times \cos\theta = \sin\theta = \text{RS}$$

Example 2: Using a rearranged Pythagorean identity

Prove $\dfrac{1 - \cos^2 x}{\sin x} = \sin x$.

Solution.

$$\text{LS} = \frac{1 - \cos^2 x}{\sin x} = \frac{\sin^2 x}{\sin x} = \sin x = \text{RS}$$

Example 3: Common denominators

Prove $\tan x + \dfrac{1}{\tan x} = \dfrac{1}{\sin x \cos x}$.

Solution. Write the left side in sine and cosine, and use the common denominator $\sin x \cos x$:

$$\begin{aligned} \text{LS} &= \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \\ &= \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} \\ &= \frac{1}{\sin x \cos x} \\ &= \text{RS} \end{aligned}$$

Example 4: Starting from the right side

Prove $\tan^2\theta = \sec^2\theta - 1$.

Solution. The right side is easier to work with:

$$\begin{aligned} \text{RS} &= \frac{1}{\cos^2\theta} - 1 \\ &= \frac{1 - \cos^2\theta}{\cos^2\theta} \\ &= \frac{\sin^2\theta}{\cos^2\theta} \\ &= \tan^2\theta \\ &= \text{LS} \end{aligned}$$

A quick numerical check (not a proof): at $\theta = 60^\circ$, $\tan^2 60^\circ = 3$ and $\sec^2 60^\circ - 1 = 4 - 1 = 3$. ✓

Common mistakes

Working on both sides as if it were an equation. Don’t add, subtract, or cross-multiply across the equals sign. Transform one side (or each side separately) until they match.

“Proving” with one angle. A numerical check is a good sanity test, but an identity has to work for every angle, so you need algebra.

Treating $\sin^2 x$ as $\sin(x^2)$. $\sin^2 x$ means $(\sin x)^2$.

Cancelling terms instead of factors. In $\dfrac{\sin x + \cos x}{\sin x}$, you can’t cancel the $\sin x$. Split it into $1 + \dfrac{\cos x}{\sin x}$ instead.

Writing $\sin x + \cos x = 1$. It’s the squares that add to $1$, not the ratios themselves.

Practice

1. (Warm-up) Verify that $\sin^2 60^\circ + \cos^2 60^\circ = 1$ using exact values.

Solution

$$\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{3}{4} + \frac{1}{4} = 1$$

2. (Warm-up) Prove $\dfrac{\sin\theta}{\tan\theta} = \cos\theta$.

Solution

$$\text{LS} = \sin\theta \div \frac{\sin\theta}{\cos\theta} = \sin\theta \times \frac{\cos\theta}{\sin\theta} = \cos\theta = \text{RS}$$

3. (Warm-up) Prove $\csc\theta\sin\theta = 1$.

Solution

$$\text{LS} = \frac{1}{\sin\theta} \times \sin\theta = 1 = \text{RS}$$

4. (Core) Prove $\dfrac{\sin^2 x + \cos^2 x}{\cos x} = \sec x$.

Solution

$$\text{LS} = \frac{1}{\cos x} = \sec x = \text{RS}$$

5. (Core) Prove $\cos x \tan x \csc x = 1$.

Solution

$$\text{LS} = \cos x \times \frac{\sin x}{\cos x} \times \frac{1}{\sin x} = 1 = \text{RS}$$

6. (Core) Prove $(1 - \sin^2\theta)(1 + \tan^2\theta) = 1$.

Solution

$$\begin{aligned} \text{LS} &= \cos^2\theta\left(1 + \frac{\sin^2\theta}{\cos^2\theta}\right) \\ &= \cos^2\theta + \sin^2\theta \\ &= 1 = \text{RS} \end{aligned}$$

7. (Core) Prove $\sin^4 x - \cos^4 x = \sin^2 x - \cos^2 x$.

Solution

Factor the left side as a difference of squares:

$$\begin{aligned} \text{LS} &= (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x) \\ &= (\sin^2 x - \cos^2 x)(1) \\ &= \text{RS} \end{aligned}$$

8. (Core) Prove $\dfrac{1}{1 - \sin x} + \dfrac{1}{1 + \sin x} = \dfrac{2}{\cos^2 x}$.

Solution

$$\begin{aligned} \text{LS} &= \frac{(1 + \sin x) + (1 - \sin x)}{(1 - \sin x)(1 + \sin x)} \\ &= \frac{2}{1 - \sin^2 x} \\ &= \frac{2}{\cos^2 x} = \text{RS} \end{aligned}$$

9. (Challenge) Show that $\sin x + \cos x = 1$ is not an identity.

Solution

One counterexample is enough. At $x = 45^\circ$:

$$\sin 45^\circ + \cos 45^\circ = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \approx 1.414 \ne 1$$

(It’s true for some angles, like $0^\circ$ and $90^\circ$, so it’s an equation, not an identity.)

10. (Challenge) Prove $\tan^2 x - \sin^2 x = \tan^2 x \sin^2 x$.

Solution

$$\begin{aligned} \text{LS} &= \frac{\sin^2 x}{\cos^2 x} - \sin^2 x \\ &= \frac{\sin^2 x - \sin^2 x\cos^2 x}{\cos^2 x} \\ &= \frac{\sin^2 x(1 - \cos^2 x)}{\cos^2 x} \\ &= \frac{\sin^2 x \cdot \sin^2 x}{\cos^2 x} \\ &= \tan^2 x \sin^2 x = \text{RS} \end{aligned}$$