Double Angle Formulas

Nothing New to Memorize

The double angle formulas are not a fresh set of rules. They are the addition formulas with both angles the same.

Set B=AB = A in sin⁡(A+B)\sin(A + B) and you get sin⁡(A+A)=sin⁡2A\sin(A + A) = \sin 2A. That is all a “double angle” is.

The yellow ray turns at angle θ\theta. The green ray turns at 2θ2\theta, sweeping twice as fast. The two bars show sin⁡2θ\sin 2\theta and 2sin⁡θcos⁡θ2\sin\theta\cos\theta locked together, equal at every instant. That equality is the formula.


Deriving Sine

Start from the addition formula and let the two angles be equal:

sin⁡2θ=sin⁡(θ+θ)=sin⁡θcos⁡θ+cos⁡θsin⁡θ=2sin⁡θcos⁡θ\begin{aligned} \sin 2\theta &= \sin(\theta + \theta) \\[0.5em] &= \sin\theta\cos\theta + \cos\theta\sin\theta \\[0.5em] &= 2\sin\theta\cos\theta \end{aligned}

sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\theta


Deriving Cosine

Same move with the cosine addition formula:

cos⁡2θ=cos⁡(θ+θ)=cos⁡θcos⁡θ−sin⁡θsin⁡θ=cos⁡2θ−sin⁡2θ\begin{aligned} \cos 2\theta &= \cos(\theta + \theta) \\[0.5em] &= \cos\theta\cos\theta - \sin\theta\sin\theta \\[0.5em] &= \cos^2\theta - \sin^2\theta \end{aligned}

cos⁡2θ=cos⁡2θ−sin⁡2θ\cos 2\theta = \cos^2\theta - \sin^2\theta


Cosine Has Three Faces

This is where cosine gets useful. Using the Pythagorean identity cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1, we can rewrite cos⁡2θ\cos 2\theta in two more ways.

Replace cos⁡2θ\cos^2\theta with 1−sin⁡2θ1 - \sin^2\theta:

cos⁡2θ=(1−sin⁡2θ)−sin⁡2θ=1−2sin⁡2θ\cos 2\theta = (1 - \sin^2\theta) - \sin^2\theta = 1 - 2\sin^2\theta

Replace sin⁡2θ\sin^2\theta with 1−cos⁡2θ1 - \cos^2\theta:

cos⁡2θ=cos⁡2θ−(1−cos⁡2θ)=2cos⁡2θ−1\cos 2\theta = \cos^2\theta - (1 - \cos^2\theta) = 2\cos^2\theta - 1

So all three of these are the same thing:

FormUse it when
cos⁡2θ−sin⁡2θ\cos^2\theta - \sin^2\thetayou know both
1−2sin⁡2θ1 - 2\sin^2\thetayou only know sin⁡θ\sin\theta
2cos⁡2θ−12\cos^2\theta - 1you only know cos⁡θ\cos\theta

Pick the version that matches what you already have.


Tangent

Setting B=AB = A in the tangent formula gives:

tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}


Don’t Fall for the Shortcut

Doubling the angle is not doubling the sine:

sin⁡2θ≠2sin⁡θ\sin 2\theta \neq 2\sin\theta

The green bar in the animation is sin⁡2θ\sin 2\theta, but the true value is 2sin⁡θcos⁡θ2\sin\theta\cos\theta. That extra cos⁡θ\cos\theta factor is the whole point.


Example

Suppose sin⁡θ=35\sin\theta = \frac{3}{5} and θ\theta is acute. Find sin⁡2θ\sin 2\theta.

First get cos⁡θ\cos\theta from the Pythagorean identity: cos⁡θ=45\cos\theta = \frac{4}{5}.

Then:

sin⁡2θ=2sin⁡θcos⁡θ=2⋅35⋅45=2425\sin 2\theta = 2\sin\theta\cos\theta = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25}


The Set to Know

FormulaResult
sin⁡2θ\sin 2\theta2sin⁡θcos⁡θ2\sin\theta\cos\theta
cos⁡2θ\cos 2\thetacos⁡2θ−sin⁡2θ=1−2sin⁡2θ=2cos⁡2θ−1\cos^2\theta - \sin^2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1
tan⁡2θ\tan 2\theta2tan⁡θ1−tan⁡2θ\dfrac{2\tan\theta}{1 - \tan^2\theta}