Trig identities, used honestly

An identity holds for every input in the shared domain. That standard cuts both ways: no number of checks can prove one, but a single failed input kills one. That makes numerical testing the perfect weapon against the identities students invent.

The most tempting fake is distributing sine over a sum: $\sin(a+b) = \sin a + \sin b$. Test it: sin(30° + 60°) = sin 90° = 1, but sin 30° + sin 60° = 0.5 + 0.866 = 1.366. Dead. Sine is not a quantity multiplying a parenthesis; it is a function, and functions do not distribute. The same test executes $\sin 2\theta = 2\sin\theta$: at θ = π/2, the left side is sin π = 0 while the right side is 2. When in doubt, feed the claim one concrete angle before you trust it.

From the unit circle, where every point is (cos θ, sin θ) at distance 1 from the origin:

1. $\sin^2\theta + \cos^2\theta = 1$, for every θ, no exceptions.
2. Rearranged: $\sin^2\theta = 1 - \cos^2\theta$ and $\cos^2\theta = 1 - \sin^2\theta$.

Its everyday job is trading one function for the other. Given sin θ = 5/13 with θ in quadrant II: cos²θ = 1 − 25/169 = 144/169, so cos θ = ±12/13, and the quadrant decides the sign: cos θ = −12/13. The identity hands you the magnitude; the quadrant hands you the sign; forgetting the second half is how right-triangle arithmetic turns into wrong answers.

The real expansion rules for a sum of angles:

1. $\sin(a+b) = \sin a\cos b + \cos a\sin b$
2. $\cos(a+b) = \cos a\cos b - \sin a\sin b$ (note the minus; for a − b the signs flip)

Worked: sin 75° = sin(45° + 30°) = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4 ≈ 0.966. Setting b = a collapses the sum formulas into the double-angle family: $\sin 2\theta = 2\sin\theta\cos\theta$, and $\cos 2\theta$ in three interchangeable costumes: $\cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$. Pick the costume that matches the information you hold: given sin θ = 3/5, the third form gives cos 2θ = 1 − 2(9/25) = 7/25 with no need to find cos θ at all.

Identities earn their keep inside equations: they convert a mixed equation into one function of one angle. $\cos 2x + \sin x = 0$ becomes $1 - 2\sin^2 x + \sin x = 0$, a quadratic in sin x that factors to $(2\sin x + 1)(\sin x - 1) = 0$, giving sin x = −1/2 or sin x = 1: on [0, 2π), the solutions x = 7π/6, 11π/6, and π/2. Three of them, from both factors.

The discipline: factor, never divide. Solving $\sin 2x = \cos x$ by expanding to $2\sin x\cos x = \cos x$ and dividing by cos x throws away every solution where cos x = 0. Move everything to one side and factor instead: $\cos x(2\sin x - 1) = 0$, and the cos x = 0 branch returns π/2 and 3π/2 alongside the sin x = 1/2 pair π/6 and 5π/6. Four solutions, not two. Dividing by an expression that can be zero is how correct algebra produces incomplete answers.