Every trig identity, by category

The terminal side of \theta meets the unit circle at the point (\cos\theta, \sin\theta), and every point on the unit circle satisfies x^2 + y^2 = 1 by definition of the circle. Substitute x = \cos\theta and y = \sin\theta directly: \cos^2\theta + \sin^2\theta = 1. That's the whole proof. No triangle, no case split for which quadrant \theta lands in — the unit-circle equation itself doesn't care, so neither does this identity.

Place two points on the unit circle: P = (\cos\alpha, \sin\alpha) at angle \alpha from standard position, and Q = (\cos\beta, \sin\beta) at angle \beta. The angle between segments OP and OQ (where O is the origin) has measure \alpha - \beta. We'll compute the straight-line distance PQ two different ways and set the results equal.

Way 1 — the distance formula. PQ^2 = (\cos\alpha - \cos\beta)^2 + (\sin\alpha - \sin\beta)^2 Expand both squares: PQ^2 = \cos^2\alpha - 2\cos\alpha\cos\beta + \cos^2\beta + \sin^2\alpha - 2\sin\alpha\sin\beta + \sin^2\beta Regroup by point, and apply the Pythagorean identity to each point separately — \cos^2\alpha+\sin^2\alpha=1 since P is on the unit circle, and likewise for Q: PQ^2 = \underbrace{(\cos^2\alpha+\sin^2\alpha)}_{=\,1} + \underbrace{(\cos^2\beta+\sin^2\beta)}_{=\,1} - 2(\cos\alpha\cos\beta+\sin\alpha\sin\beta) = 2 - 2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)

Way 2 — the Law of Cosines. Triangle OPQ has two sides of length 1 (both P and Q are on the unit circle, so OP = OQ = 1) enclosing an angle of measure \alpha-\beta. The Law of Cosines gives PQ^2 = OP^2 + OQ^2 - 2\cdot OP\cdot OQ\cdot\cos(\alpha-\beta) = 1 + 1 - 2\cos(\alpha-\beta) = 2 - 2\cos(\alpha-\beta) This route treats \alpha-\beta as an actual interior angle of a triangle, which only makes literal sense for 0<\alpha-\beta<\pi — Way 1 makes no such assumption at all (it's pure coordinate algebra, valid for every real \alpha,\beta), which is what actually earns the theorem's "for any angles" claim; Way 2 is included because the Law-of-Cosines route is the more memorable picture, not because it alone would suffice.

Set the two expressions for PQ^2 equal — they measure the same segment: 2 - 2(\cos\alpha\cos\beta+\sin\alpha\sin\beta) = 2 - 2\cos(\alpha-\beta) Subtract 2 from both sides, then divide by -2: \cos\alpha\cos\beta + \sin\alpha\sin\beta = \cos(\alpha-\beta) which is the identity, read right to left. \blacksquare

Substitute \alpha=\beta=\theta into the sine sum formula \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta: \sin(2\theta) = \sin(\theta+\theta) = \sin\theta\cos\theta + \cos\theta\sin\theta = 2\sin\theta\cos\theta. The two middle terms are identical, so they combine by ordinary like-terms addition — not a new trig fact, just x + x = 2x wearing a trig costume.

Set \alpha=\beta=\theta in the cosine sum formula \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta: \cos(2\theta) = \cos(\theta+\theta) = \cos\theta\cos\theta - \sin\theta\sin\theta = \cos^2\theta - \sin^2\theta. That's the first form, and no further work is required — it's already a true statement about \cos(2\theta).

The other two forms come from a single substitution each, using the Pythagorean identity \sin^2\theta+\cos^2\theta=1. Replace \sin^2\theta with 1-\cos^2\theta: \cos(2\theta) = \cos^2\theta - \sin^2\theta = \cos^2\theta - (1-\cos^2\theta) = 2\cos^2\theta - 1. Or replace \cos^2\theta with 1-\sin^2\theta instead: \cos(2\theta) = \cos^2\theta - \sin^2\theta = (1-\sin^2\theta) - \sin^2\theta = 1 - 2\sin^2\theta. All three expressions equal \cos(2\theta); they're the same quantity written three ways, not three different facts.

By the quotient identity, \tan(2\theta) = \dfrac{\sin(2\theta)}{\cos(2\theta)}. Substitute the sine and cosine double-angle forms just proved: \tan(2\theta) = \frac{2\sin\theta\cos\theta}{\cos^2\theta - \sin^2\theta} Divide every term, top and bottom, by \cos^2\theta — valid only once \cos\theta\neq0, i.e. \theta\neq\frac{\pi}{2}+k\pi, since dividing by \cos^2\theta is what introduces \tan\theta in the first place: \tan(2\theta) = \frac{\dfrac{2\sin\theta\cos\theta}{\cos^2\theta}}{\dfrac{\cos^2\theta}{\cos^2\theta} - \dfrac{\sin^2\theta}{\cos^2\theta}} = \frac{2\tan\theta}{1-\tan^2\theta} Two separate restrictions survive into the final form, for two different reasons. \theta \neq \frac{\pi}{2}+k\pi is carried over from the division step above — \tan\theta itself doesn't exist there, so the right-hand side can't even be evaluated, even though \tan(2\theta) on the left is perfectly fine (e.g. \theta=\frac{\pi}{2} gives \tan(2\theta)=\tan\pi=0). Separately, \tan\theta \neq \pm 1 keeps the denominator 1-\tan^2\theta away from 0 — and that condition is exactly \theta = \frac{\pi}{4}+k\frac{\pi}{2}, precisely where 2\theta itself has no tangent, so both sides genuinely agree there is no value to assign.

Start from \cos(2\theta) = 1 - 2\sin^2\theta. Subtract 1 from both sides, then multiply by -1: \cos(2\theta) - 1 = -2\sin^2\theta \quad\Longrightarrow\quad 1 - \cos(2\theta) = 2\sin^2\theta Divide both sides by 2: \sin^2\theta = \frac{1-\cos(2\theta)}{2} Start instead from \cos(2\theta) = 2\cos^2\theta - 1. Add 1 to both sides: \cos(2\theta) + 1 = 2\cos^2\theta Divide both sides by 2: \cos^2\theta = \frac{1+\cos(2\theta)}{2}

Substitute \theta \to \dfrac{\theta}{2} into both power-reduction identities (so every 2\theta becomes \theta): \sin^2\frac{\theta}{2} = \frac{1-\cos\theta}{2}, \qquad \cos^2\frac{\theta}{2} = \frac{1+\cos\theta}{2} Square-rooting a squared quantity always produces a \pm — the square root property from Chapter 3 applies here exactly as it did to any equation of the form u^2 = k: \sin\frac{\theta}{2} = \pm\sqrt{\frac{1-\cos\theta}{2}}, \qquad \cos\frac{\theta}{2} = \pm\sqrt{\frac{1+\cos\theta}{2}} Dividing the two gives the tangent version, and it also simplifies to two radical-free forms, \tan\dfrac{\theta}{2} = \dfrac{1-\cos\theta}{\sin\theta} = \dfrac{\sin\theta}{1+\cos\theta}, which carry their own sign automatically — but the radical form above is the one that matches sine and cosine's pattern, and the one this course uses by default.

Write the cosine difference and sum formulas together: \cos(\alpha-\beta) = \cos\alpha\cos\beta+\sin\alpha\sin\beta, \qquad \cos(\alpha+\beta) = \cos\alpha\cos\beta-\sin\alpha\sin\beta Add them — the \sin\alpha\sin\beta terms cancel, leaving \cos\alpha\cos\beta doubled: \cos(\alpha-\beta)+\cos(\alpha+\beta) = 2\cos\alpha\cos\beta \implies \cos\alpha\cos\beta = \tfrac12\big[\cos(\alpha-\beta)+\cos(\alpha+\beta)\big] Subtract them instead — the \cos\alpha\cos\beta terms cancel, leaving \sin\alpha\sin\beta doubled: \cos(\alpha-\beta)-\cos(\alpha+\beta) = 2\sin\alpha\sin\beta \implies \sin\alpha\sin\beta = \tfrac12\big[\cos(\alpha-\beta)-\cos(\alpha+\beta)\big] The same add-and-cancel move on the sine sum and difference formulas, \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta and \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta, isolates the third identity above.