Double angle formulae

Covers AS91575

The double angle formulae

Putting $B = A$ into the compound formulae gives the double angle formulae:

$\sin 2A = 2\sin A \cos A$

$\cos 2A = \cos^2 A - \sin^2 A$

$\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$

Three forms of $\cos 2A$

Using $\sin^2 A + \cos^2 A = 1$ , the cosine formula can be written three equivalent ways:

$\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$

Choose the form that matches what you have — all-cosine ( $2\cos^2 A - 1$ ) or all-sine ( $1 - 2\sin^2 A$ ).
Rearranged, these give the power-reduction identities:

$\cos^2 A = \frac{1 + \cos 2A}{2}, \qquad \sin^2 A = \frac{1 - \cos 2A}{2}$

What to use them for

Simplify double-angle expressions, e.g. $2\sin 3x\cos 3x = \sin 6x$ (here $A = 3x$ ).
Find exact values, e.g. given $\sin A$ , work out $\sin 2A$ and $\cos 2A$ .
Prepare an equation for solving by turning $\sin 2x$ or $\cos 2x$ into single-angle functions.

Worked ExampleDouble angle from a given ratio

$\theta$ is acute with $\sin\theta = \tfrac{3}{5}$ . Find the exact values of $\sin 2\theta$ and $\cos 2\theta$ .

Step 1 — Find $\cos\theta$

$\theta$ is acute, so cosine is positive:

$\cos\theta = \frac{4}{5}$

Step 2 — Apply $\sin 2\theta = 2\sin\theta\cos\theta$

$\sin 2\theta = 2\cdot\frac{3}{5}\cdot\frac{4}{5} = \frac{24}{25}$

Step 3 — Apply $\cos 2\theta = 1 - 2\sin^2\theta$

$\cos 2\theta = 1 - 2\left(\frac{3}{5}\right)^2 = 1 - \frac{18}{25} = \frac{7}{25}$

$\boxed{\sin 2\theta = \frac{24}{25}, \quad \cos 2\theta = \frac{7}{25}}$

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Given $\cos A = \tfrac{2}{3}$ , find the exact value of $\cos 2A$ .

Merit

$x$ is in the second quadrant with $\sin x = \tfrac{5}{13}$ . Find the exact value of $\sin 2x$ .

Excellence

Prove that $\dfrac{1 - \cos 2x}{\sin 2x} = \tan x$ .

Double angle formulae

Covers AS91575

The double angle formulae

Putting $B = A$ into the compound formulae gives the double angle formulae:

$\sin 2A = 2\sin A \cos A$

$\cos 2A = \cos^2 A - \sin^2 A$

$\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$

Three forms of $\cos 2A$

Using $\sin^2 A + \cos^2 A = 1$ , the cosine formula can be written three equivalent ways:

$\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$

Choose the form that matches what you have — all-cosine ( $2\cos^2 A - 1$ ) or all-sine ( $1 - 2\sin^2 A$ ).
Rearranged, these give the power-reduction identities:

$\cos^2 A = \frac{1 + \cos 2A}{2}, \qquad \sin^2 A = \frac{1 - \cos 2A}{2}$

What to use them for

Simplify double-angle expressions, e.g. $2\sin 3x\cos 3x = \sin 6x$ (here $A = 3x$ ).
Find exact values, e.g. given $\sin A$ , work out $\sin 2A$ and $\cos 2A$ .
Prepare an equation for solving by turning $\sin 2x$ or $\cos 2x$ into single-angle functions.

Worked ExampleDouble angle from a given ratio

$\theta$ is acute with $\sin\theta = \tfrac{3}{5}$ . Find the exact values of $\sin 2\theta$ and $\cos 2\theta$ .

Step 1 — Find $\cos\theta$

$\theta$ is acute, so cosine is positive:

$\cos\theta = \frac{4}{5}$

Step 2 — Apply $\sin 2\theta = 2\sin\theta\cos\theta$

$\sin 2\theta = 2\cdot\frac{3}{5}\cdot\frac{4}{5} = \frac{24}{25}$

Step 3 — Apply $\cos 2\theta = 1 - 2\sin^2\theta$

$\cos 2\theta = 1 - 2\left(\frac{3}{5}\right)^2 = 1 - \frac{18}{25} = \frac{7}{25}$

$\boxed{\sin 2\theta = \frac{24}{25}, \quad \cos 2\theta = \frac{7}{25}}$

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Given $\cos A = \tfrac{2}{3}$ , find the exact value of $\cos 2A$ .

Merit

$x$ is in the second quadrant with $\sin x = \tfrac{5}{13}$ . Find the exact value of $\sin 2x$ .

Excellence

Prove that $\dfrac{1 - \cos 2x}{\sin 2x} = \tan x$ .

Double angle formulae

The double angle formulae

Three forms of cos⁡2A\cos 2Acos2A

What to use them for

Step 1 — Find cos⁡θ\cos\thetacosθ

Step 2 — Apply sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\thetasin2θ=2sinθcosθ

Step 3 — Apply cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2\sin^2\thetacos2θ=1−2sin2θ

Test yourself

Practice by grade

Double angle formulae

The double angle formulae

Three forms of cos⁡2A\cos 2Acos2A

What to use them for

Step 1 — Find cos⁡θ\cos\thetacosθ

Step 2 — Apply sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\thetasin2θ=2sinθcosθ

Step 3 — Apply cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2\sin^2\thetacos2θ=1−2sin2θ

Test yourself

Practice by grade

Three forms of $\cos 2A$

Step 1 — Find $\cos\theta$

Step 2 — Apply $\sin 2\theta = 2\sin\theta\cos\theta$

Step 3 — Apply $\cos 2\theta = 1 - 2\sin^2\theta$

Three forms of $\cos 2A$

Step 1 — Find $\cos\theta$

Step 2 — Apply $\sin 2\theta = 2\sin\theta\cos\theta$

Step 3 — Apply $\cos 2\theta = 1 - 2\sin^2\theta$