1. What an identity is
An equation that is true for every value of the variable for which both sides are defined — not just for special values that happen to satisfy it.
Compare two superficially similar statements:
$$ \sin\theta = \tfrac{1}{2} \qquad\text{vs.}\qquad \sin^2\theta + \cos^2\theta = 1 $$The first is a conditional equation. It pins $\theta$ down: only certain angles ($\theta = 30°, 150°, 390°, \ldots$) satisfy it. Pick a different $\theta$ and the statement is simply false.
The second is an identity. Plug in any angle — $0$, $\pi/7$, a billion radians — and both sides agree. The equality isn't a puzzle to solve; it's a structural fact about the trig functions themselves.
Identities are the rewrite rules of trigonometry. Need to integrate $\sin^2\theta$? Rewrite it with a double-angle identity. Stuck simplifying $\frac{\sin\theta}{1+\cos\theta}$? An identity collapses it to $\tan(\theta/2)$. They're not facts to memorize as trivia — they're transformations you reach for when an expression won't budge.
The basic reciprocal & quotient identities
Before getting to the heavy hitters, three families follow directly from the definitions of the six trig functions:
$$ \begin{aligned} \csc\theta &= \frac{1}{\sin\theta}, & \sec\theta &= \frac{1}{\cos\theta}, & \cot\theta &= \frac{1}{\tan\theta} \\[2pt] \tan\theta &= \frac{\sin\theta}{\cos\theta}, & \cot\theta &= \frac{\cos\theta}{\sin\theta} \end{aligned} $$These aren't theorems — they're how the functions are defined. Treat them as renaming rules and use them freely.
2. The Pythagorean identities
One identity dominates trigonometry, and the other two are its shadow. The master identity is:
$$ \boxed{\;\sin^2\theta + \cos^2\theta = 1\;} $$The proof is the most direct in mathematics. On the unit circle, every point $(x, y)$ satisfies $x^2 + y^2 = 1$ by definition. But by the unit-circle definition of sine and cosine, that point is $(\cos\theta, \sin\theta)$. Substitute and you have the identity. The Pythagorean theorem and this identity are the same fact wearing different clothes.
The two derived identities
Divide the master identity by $\cos^2\theta$ (assuming $\cos\theta \neq 0$):
$$ \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta} \;\Longrightarrow\; \tan^2\theta + 1 = \sec^2\theta $$Divide instead by $\sin^2\theta$ (assuming $\sin\theta \neq 0$):
$$ \frac{\sin^2\theta}{\sin^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \frac{1}{\sin^2\theta} \;\Longrightarrow\; 1 + \cot^2\theta = \csc^2\theta $$So the three Pythagorean identities are:
$$ \begin{aligned} \sin^2\theta + \cos^2\theta &= 1 \\ 1 + \tan^2\theta &= \sec^2\theta \\ 1 + \cot^2\theta &= \csc^2\theta \end{aligned} $$You don't have to memorize the second two as independent facts. Memorize the first and the trick of dividing — the others fall out in seconds.
3. Angle-sum and angle-difference identities
What is $\sin(\alpha + \beta)$? A natural guess is $\sin\alpha + \sin\beta$. That guess is wrong, and seeing why matters: trig functions are not linear. They mix the two angles in a specific way.
The correct identities are:
$$ \begin{aligned} \sin(\alpha + \beta) &= \sin\alpha \cos\beta + \cos\alpha \sin\beta \\ \sin(\alpha - \beta) &= \sin\alpha \cos\beta - \cos\alpha \sin\beta \\ \cos(\alpha + \beta) &= \cos\alpha \cos\beta - \sin\alpha \sin\beta \\ \cos(\alpha - \beta) &= \cos\alpha \cos\beta + \sin\alpha \sin\beta \end{aligned} $$Two patterns are worth burning in:
- Sine mixes — each term has one sine and one cosine. The sign follows the input: $+$ becomes $+$, $-$ becomes $-$.
- Cosine separates — cosine times cosine, then sine times sine. The sign flips: $+$ in the input becomes $-$ in the formula, and vice versa.
For $\sin$, the sign on the right matches the sign on the left. For $\cos$, the sign on the right is the opposite. "Sine same, cosine opposite" is the mnemonic. Get this backwards and every derivation downstream is poisoned.
For tangent the corresponding identities (derived from sin/cos by long division) are:
$$ \tan(\alpha \pm \beta) = \frac{\tan\alpha \pm \tan\beta}{1 \mp \tan\alpha \tan\beta} $$Same sign-swap discipline applies: the numerator follows the input, the denominator flips.
4. Double-angle and half-angle identities
Set $\beta = \alpha$ in the sum identities and the dependence on two angles collapses to one. These are useful so often that they get their own names.
Double-angle
From $\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$ with $\beta = \alpha$:
$$ \sin(2\theta) = 2\sin\theta\cos\theta $$From $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$ with $\beta = \alpha$:
$$ \cos(2\theta) = \cos^2\theta - \sin^2\theta $$That last one has two cousins that are often more useful, both produced by substituting the Pythagorean identity. Using $\sin^2\theta = 1 - \cos^2\theta$:
$$ \cos(2\theta) = 2\cos^2\theta - 1 $$And using $\cos^2\theta = 1 - \sin^2\theta$:
$$ \cos(2\theta) = 1 - 2\sin^2\theta $$Three equivalent forms — pick whichever leaves you with the variable you want to see. Need a formula in terms of $\sin\theta$ only? Use the third. In terms of $\cos\theta$ only? Use the second.
For tangent:
$$ \tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta} $$Half-angle
The double-angle identities run in both directions. Solving the $\cos(2\theta)$ forms for $\sin^2\theta$ and $\cos^2\theta$ gives the half-angle (or "power-reduction") identities:
$$ \sin^2\theta = \frac{1 - \cos(2\theta)}{2}, \qquad \cos^2\theta = \frac{1 + \cos(2\theta)}{2} $$Replace $2\theta$ with $\theta$ and $\theta$ with $\theta/2$ and you get the more recognizable shape:
$$ \sin\tfrac{\theta}{2} = \pm\sqrt{\tfrac{1 - \cos\theta}{2}}, \qquad \cos\tfrac{\theta}{2} = \pm\sqrt{\tfrac{1 + \cos\theta}{2}} $$The $\pm$ is genuine — the correct sign depends on which quadrant $\theta/2$ sits in. The squared versions above are the workhorses in calculus, where they turn $\sin^2\theta$ and $\cos^2\theta$ into things that integrate cleanly.
5. Using identities — strategy for proofs
A trig "identity proof" looks like an equation and asks you to show the two sides really are equal. The danger is treating it like a regular equation and doing the same operation to both sides — that proves nothing, since you'd be assuming what you're trying to show.
The standard discipline:
- Pick the more complicated side — more terms, more functions, more nesting. That's the side with room to simplify.
- Convert everything to sines and cosines when stuck. Almost every identity can be expressed in $\sin$ and $\cos$, and reducing to a common alphabet often makes the structure visible.
- Apply known identities — Pythagorean, reciprocal, quotient — one rewrite at a time, watching for the shape of the other side.
- Don't touch the other side. You're transforming the messy expression into the clean one. The other side is the target, not a participant.
A short walk-through
Show that $\dfrac{1 - \cos^2\theta}{\sin\theta} = \sin\theta$.
The left side is the more complicated one. Recognize $1 - \cos^2\theta$ from the Pythagorean identity:
$$ \frac{1 - \cos^2\theta}{\sin\theta} = \frac{\sin^2\theta}{\sin\theta} = \sin\theta $$Done. Two rewrites and the left side has been carried over to the right. We never wrote "$\sin\theta = \sin\theta$" or applied an operation to both sides — we showed that one expression is the other.
Working "down" from the complicated side is a one-way street: if you reach the target, you're done. If you get stuck, the failed work doesn't accidentally smuggle in an unjustified step. The discipline forces honest derivations.
6. Common pitfalls
It's $\cos\alpha\cos\beta \,{\color{var(--danger)}\mathbf{-}}\, \sin\alpha\sin\beta$ — minus, not plus. Sine matches the input sign; cosine flips it. Mixing this up is the single most common source of wrong answers when angle-sum identities show up in proofs and calculus problems.
The notation $\sin^2\theta$ is shorthand for $(\sin\theta)^2$ — square the value, not the angle. $\sin(\theta^2)$ would mean "feed $\theta^2$ into the sine function," which is an entirely different beast. The convention is unfortunate but universal.
An identity is "true for every $\theta$ where both sides are defined." $1 + \tan^2\theta = \sec^2\theta$ is genuinely true wherever $\tan\theta$ exists — but neither side exists at $\theta = \pi/2$. If your proof step requires evaluating a tangent or secant at $\pi/2$, the identity isn't broken; you've just walked off the legal domain.
Squaring both sides, multiplying both sides by an expression, or "moving" a term across the equals sign while proving an identity assumes the equation already holds. That's circular. Transform one side at a time, leave the other alone.
7. Worked examples
Each example is a clean application of one of the identity families above. Work through the steps yourself before opening the solution — the goal is to recognize which identity unlocks the expression, not just to reach the answer.
Example 1 · Verify $\sec^2\theta - \tan^2\theta = 1$
Start from the Pythagorean identity in its tangent–secant form:
$$ 1 + \tan^2\theta = \sec^2\theta $$Subtract $\tan^2\theta$ from both sides (we're rearranging a known identity, not "operating on both sides of an unknown one"):
$$ 1 = \sec^2\theta - \tan^2\theta $$Which is the claim. ✓
Example 2 · Compute $\sin 75°$ exactly using $\sin(45° + 30°)$
Apply the angle-sum identity for sine:
$$ \sin 75° = \sin(45° + 30°) = \sin 45°\cos 30° + \cos 45°\sin 30° $$Plug in the unit-circle values $\sin 45° = \cos 45° = \tfrac{\sqrt{2}}{2}$, $\sin 30° = \tfrac{1}{2}$, $\cos 30° = \tfrac{\sqrt{3}}{2}$:
$$ \sin 75° = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4} $$A clean exact value for a "non-standard" angle, built from two standard ones.
Example 3 · Simplify $\dfrac{\sin\theta}{1 - \cos\theta} + \dfrac{1 - \cos\theta}{\sin\theta}$
Combine over a common denominator:
$$ \frac{\sin^2\theta + (1 - \cos\theta)^2}{\sin\theta(1 - \cos\theta)} $$Expand the squared term in the numerator:
$$ \sin^2\theta + 1 - 2\cos\theta + \cos^2\theta $$Apply the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$:
$$ 1 + 1 - 2\cos\theta = 2(1 - \cos\theta) $$Substitute back:
$$ \frac{2(1 - \cos\theta)}{\sin\theta(1 - \cos\theta)} = \frac{2}{\sin\theta} = 2\csc\theta $$A two-term mess collapses to a single cosecant.
Example 4 · If $\cos\theta = \tfrac{3}{5}$ and $\theta$ is in the first quadrant, find $\sin\theta$
The Pythagorean identity is the right tool:
$$ \sin^2\theta = 1 - \cos^2\theta = 1 - \tfrac{9}{25} = \tfrac{16}{25} $$So $\sin\theta = \pm\tfrac{4}{5}$. Quadrant I means sine is positive, so:
$$ \sin\theta = \tfrac{4}{5} $$The quadrant information is doing real work — without it the answer would be ambiguous.
Example 5 · Verify $\sin 60° = \sin(2 \cdot 30°)$ using the double-angle identity
Apply $\sin(2\theta) = 2\sin\theta\cos\theta$ with $\theta = 30°$:
$$ \sin 60° = 2\sin 30°\cos 30° = 2 \cdot \tfrac{1}{2} \cdot \tfrac{\sqrt{3}}{2} = \tfrac{\sqrt{3}}{2} $$And from the unit circle directly, $\sin 60° = \tfrac{\sqrt{3}}{2}$. ✓
The identity isn't telling us anything we couldn't have read off the unit circle here — but it's the same machinery that lets us compute, say, $\sin(2 \cdot 17°)$ without a table.