The Trig Identities Worth Memorising

An identity is an equation that holds for every value of the variable where both sides make sense. That is a stronger claim than an equation such as $\sin \theta =\frac{1}{2}$, which is true only at particular angles. Because an identity is true everywhere, it is really a statement that two expressions are two spellings of the same thing.

The reason there are so many of them is that sine and cosine are related by a single constraint — the circle — and every consequence of that constraint can be written several ways. This is also why they collapse into so few independent facts. Almost the entire standard list descends from ${\sin }^2\theta +{\cos }^2\theta =1$ and the two angle-sum formulas.

Proving an identity is therefore not a search. It is a translation problem: take the expression you have, rewrite it in the vocabulary of the expression you want, and the two will meet. The tactics below are just a sensible order for those rewrites.

Recognising an identity problem is easy when the instruction says “prove”. The useful cues are the ones where an identity is the hidden step:

• An expression contains both ${\sin }^2$ and ${\cos }^2$. One of them can almost always be eliminated, and doing so usually simplifies everything downstream.
• A trig equation mixes $\sin 2\theta$ with $\sin \theta$, or $\cos 2\theta$ with $\cos \theta$. Two different angles cannot be solved together, so a double-angle formula has to reduce them to one.
• An integral contains ${\sin }^{2}x$ or ${\cos }^{2}x$. Neither is a standard antiderivative in that form, so power reduction is the first move rather than an optional tidy-up.
• You need an exact value at a non-standard angle such as $75^{\circ }$. Splitting it as $45^{\circ }+30^{\circ }$ turns an unknown into two known values.
• A denominator contains $1+\cos \theta$ or $1-\sin \theta$. That shape exists to be multiplied by its conjugate.

All of these assume you can already read exact values off the circle. If $\cos \frac{5\pi }{6}$ is not immediate, start with the unit circle you actually need first — the identities are much harder to see when every value has to be looked up.

Two of the three core identities were derived on the unit circle page. Here is the third, the one most people take on trust. It costs six lines and it generates the entire angle-sum, double-angle, and half-angle family.

Put two points on the unit circle, $P$ at angle $A$ and $Q$ at angle $B$, and compute the square of the distance between them:

Expand, then group the squared terms into two applications of the Pythagorean identity:

Now rotate the whole picture by $-B$. Rotation does not change distances, and it moves $Q$ onto $(1,0)$ and $P$ onto the point at angle $A-B$. Computing the same distance in the new position:

Two expressions for the same length must agree. Setting them equal, cancelling the 2s and dividing by $-2$:

Everything else falls out of this. Replacing $B$ with $-B$ and using $\cos (-B)=\cos B$, $\sin (-B)=-\sin B$ gives the sum formula. Writing $\sin (A+B)$ as $\cos \left(\frac{\pi }{2}-A-B\right)$ and applying the result again gives the sine version. Setting $B=A$ gives the double-angle formulas. Rearranging those gives power reduction, and halving the angle there gives the half-angle formulas. The reference sheet below has each of those one-step derivations attached to the formula it produces.

One more thing has to be in place before any of this is usable. Step two of the procedure below says “convert to sine and cosine”, which assumes you can. Four of the six functions are defined as combinations of the other two, and these are definitions rather than results:

The pairing of secant with cosine rather than with sine looks perverse and is worth noticing deliberately, because reaching for the wrong one is a silent error that survives several lines before anything looks odd.

It is also worth testing a claimed identity numerically before committing to a proof. Substitute $\theta =\frac{\pi }{6}$, where every value is known exactly, and evaluate both sides. If they disagree, either you have copied the problem wrongly or it is not an identity, and either way you have saved yourself twenty minutes. If they agree, try $\theta =\frac{\pi }{4}$ as well — one agreement can be a coincidence, two rarely are. This is a check, not a proof: an identity has to hold everywhere, so no finite number of successful substitutions establishes it.

To prove a given identity, work in this order:

1. Pick the messier side and stay on it. Simplifying is easier than complicating, and touching both sides at once is not a valid proof structure.
2. Convert to sine and cosine. Four of the six functions are just quotients. Removing them halves the number of things you are tracking.
3. Combine fractions over a common denominator. This is the step that most often produces a numerator the Pythagorean identity can collapse.
4. Hunt for a Pythagorean substitution. Any $1-{\sin }^2\theta$, $1-{\cos }^2\theta$, or ${\sec }^2\theta -1$ is an invitation.
5. Multiply by a conjugate when a denominator resists. $1+\sin \theta$ times $1-\sin \theta$ is ${\cos }^2\theta$, which is usually the breakthrough.
6. Match the target exactly. If the right-hand side says $2\sec \theta$, convert your $2/\cos \theta$ rather than leaving it.

The first is a single Pythagorean substitution. The second is a factoring problem wearing a trigonometric costume. The third stacks two fractions and needs every tactic on the list.

Cross-multiplying across the equals sign

Treating the identity as an equation and multiplying both sides by a denominator is circular: it uses the equality as a premise in its own proof. If you want to work from both ends, run two independent columns that each reduce to the same expression, and say so explicitly. That is a valid structure; a single manipulated equation is not.

Reaching for the wrong form of cos 2θ

There are three, and choosing badly turns a two-line proof into a page. Look at the target: if it contains only sines, use $1-2{\sin }^2\theta$; if only cosines, use $2{\cos }^2\theta -1$; if both, use ${\cos }^2\theta -{\sin }^2\theta$. Let the destination pick the form.

Distributing a function over addition

$\sin (A+B)$ is not $\sin A+\sin B$, and ${\sin }^2\theta$ is not $\sin ({\theta }^2)$. One numerical check kills the first: $\sin 90^{\circ }=1$ but $\sin 45^{\circ }+\sin 45^{\circ }=\sqrt{2}\approx 1.414$. Sine is a function, not a multiplier, and the angle-sum formula exists precisely because the naive distribution fails.

Dropping the plus-or-minus on a half angle

$\sin \frac{\theta }{2}=\pm \sqrt{\frac{1-\cos \theta }{2}}$ carries a sign that the formula cannot determine. You decide it from the quadrant of $\theta /2$, not the quadrant of $\theta$. For $\theta =300^{\circ }$, the half angle is $150^{\circ }$, which is in quadrant II, so the sine is positive even though $\theta$ itself is in quadrant IV.

Use $\text{sqrt()}$ for roots and $\text{^}$ for powers. Each of these takes at most two identity applications.

The second example above was really a difference of two squares in disguise, and that pattern recurs constantly once expressions contain fourth powers of sine and cosine. Equations that arrive as $2{\sin }^2\theta -\sin \theta -1=0$ are quadratics in $\sin \theta$, so the quadratic formula finishes them before the circle takes over to find every angle. If a proof stalls halfway, the trigonometry solver will show which identity it applied and why that side was the one to start from.