Pythagorean and Double-Angle Identities

\(\sin^2x+\cos^2x=1\)

\(\sin2x=2\sin x\cos x\)

\(\cos2x=\cos^2x-\sin^2x=2\cos^2x-1=1-2\sin^2x\)

Choose the cosine double-angle form that leaves one trigonometric ratio. Determine signs from the quadrant; a square root alone does not determine the sign.

Example

If \(\cos x=-\dfrac35\) and \(\dfrac\pi2<x<\pi\), then

\(\begin{aligned}\sin x&=\sqrt{1-\dfrac9{25}}=\dfrac45\\\sin2x&=2\left(\dfrac45\right)\left(-\dfrac35\right)=-\dfrac{24}{25}\\\cos2x&=2\left(\dfrac9{25}\right)-1=-\dfrac7{25}.\end{aligned}\)

The argument must match: \(\sin^2(3x)+\cos^2(3x)=1\). An identity holds wherever both sides are defined.