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)=sin2A\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 sin2θ\sin 2\theta and 2sinθcosθ2\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:

sin2θ=sin(θ+θ)=sinθcosθ+cosθsinθ=2sinθcosθ\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}

sin2θ=2sinθcosθ\sin 2\theta = 2\sin\theta\cos\theta


Deriving Cosine

Same move with the cosine addition formula:

cos2θ=cos(θ+θ)=cosθcosθsinθsinθ=cos2θsin2θ\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}

cos2θ=cos2θsin2θ\cos 2\theta = \cos^2\theta - \sin^2\theta


Cosine Has Three Faces

This is where cosine gets useful. Using the Pythagorean identity cos2θ+sin2θ=1\cos^2\theta + \sin^2\theta = 1, we can rewrite cos2θ\cos 2\theta in two more ways.

Replace cos2θ\cos^2\theta with 1sin2θ1 - \sin^2\theta:

cos2θ=(1sin2θ)sin2θ=12sin2θ\cos 2\theta = (1 - \sin^2\theta) - \sin^2\theta = 1 - 2\sin^2\theta

Replace sin2θ\sin^2\theta with 1cos2θ1 - \cos^2\theta:

cos2θ=cos2θ(1cos2θ)=2cos2θ1\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
cos2θsin2θ\cos^2\theta - \sin^2\thetayou know both
12sin2θ1 - 2\sin^2\thetayou only know sinθ\sin\theta
2cos2θ12\cos^2\theta - 1you only know cosθ\cos\theta

Pick the version that matches what you already have.


Tangent

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

tan2θ=2tanθ1tan2θ\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:

sin2θ2sinθ\sin 2\theta \neq 2\sin\theta

The green bar in the animation is sin2θ\sin 2\theta, but the true value is 2sinθcosθ2\sin\theta\cos\theta. That extra cosθ\cos\theta factor is the whole point.


Example

Suppose sinθ=35\sin\theta = \frac{3}{5} and θ\theta is acute. Find sin2θ\sin 2\theta.

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

Then:

sin2θ=2sinθcosθ=23545=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
sin2θ\sin 2\theta2sinθcosθ2\sin\theta\cos\theta
cos2θ\cos 2\thetacos2θsin2θ=12sin2θ=2cos2θ1\cos^2\theta - \sin^2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1
tan2θ\tan 2\theta2tanθ1tan2θ\dfrac{2\tan\theta}{1 - \tan^2\theta}