Double-Angle Identities Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Double-Angle Identities.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

Formulas expressing sin⁡(2θ)\sin(2\theta), cos⁡(2θ)\cos(2\theta), and tan⁡(2θ)\tan(2\theta) in terms of single-angle trig functions.

What if both angles in the sum formula are the same? Setting A=B=θA = B = \theta in the sum identities gives you the double-angle formulas. They answer: if you know the trig values for an angle, what are the trig values for twice that angle? The cosine double-angle formula is especially versatile because it has three equivalent forms, each useful in different situations—pick whichever one simplifies your problem.

Read the full concept explanation →

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: sin⁡2θ\sin 2\theta, cos⁡2θ\cos 2\theta, tan⁡2θ\tan 2\theta written from single-angle values, with cosine offering three interchangeable forms.

Common stuck point: The procedure for double-angle identities is the easy part; the trap is writing sin⁡2θ=2sin⁡θ\sin 2\theta=2\sin\theta. Asking "Is the angle exactly twice another, so I can express it from single-angle trig values?" first is what keeps a correct-looking calculation from being attached to the wrong concept.

Sense of Study hint: Ask: Is the angle exactly twice another, so I can express it from single-angle trig values?

Worked Examples

Example 1

easy
If cos⁡(θ)=35\cos(\theta) = \frac{3}{5}, find cos⁡(2θ)\cos(2\theta) using the double-angle formula.

Answer

cos⁡(2θ)=−725\cos(2\theta) = -\frac{7}{25}

First step

1
Use the double-angle formula: cos⁡(2θ)=2cos⁡2(θ)−1\cos(2\theta) = 2\cos^2(\theta) - 1.

Full solution

  1. 2
    Substitute: cos⁡(2θ)=2(35)2−1=2⋅925−1\cos(2\theta) = 2\left(\frac{3}{5}\right)^2 - 1 = 2 \cdot \frac{9}{25} - 1.
  2. 3
    =1825−1=1825−2525=−725= \frac{18}{25} - 1 = \frac{18}{25} - \frac{25}{25} = -\frac{7}{25}.
The double-angle formula for cosine has three equivalent forms: cos⁡(2θ)=cos⁡2θ−sin⁡2θ=2cos⁡2θ−1=1−2sin⁡2θ\cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta. Choose the form that uses the information you have — here we used 2cos⁡2θ−12\cos^2\theta - 1 since we knew cosine.

Example 2

medium
Find sin⁡(2θ)\sin(2\theta) given that tan⁡(θ)=512\tan(\theta) = \frac{5}{12} and θ\theta is in Quadrant I.

Example 3

medium
Solve cos⁡(2x)=cos⁡x\cos(2x) = \cos x on [0,2π)[0, 2\pi).

Example 4

medium
Use power-reduction to write cos⁡2x\cos^2 x in terms of cos⁡(2x)\cos(2x).

Example 5

hard
Prove that sin⁡(2θ)1+cos⁡(2θ)=tan⁡θ\dfrac{\sin(2\theta)}{1 + \cos(2\theta)} = \tan\theta.

Example 6

hard
Show that sin⁡(3θ)=3sin⁡θ−4sin⁡3θ\sin(3\theta) = 3\sin\theta - 4\sin^3\theta using sin⁡(3θ)=sin⁡(2θ+θ)\sin(3\theta) = \sin(2\theta + \theta).

Example 7

challenge
Find the maximum value of f(θ)=sin⁡θcos⁡θ+cos⁡(2θ)f(\theta) = \sin\theta\cos\theta + \cos(2\theta) on [0,2π)[0, 2\pi).

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

medium
Simplify cos⁡2(x)−sin⁡2(x)\cos^2(x) - \sin^2(x) using a double-angle identity.

Example 2

hard
Solve sin⁡(2x)=cos⁡(x)\sin(2x) = \cos(x) for x∈[0,2π)x \in [0, 2\pi).

Example 3

easy
State the double-angle formula for sin⁡(2θ)\sin(2\theta).

Example 4

easy
State one form of the double-angle formula for cos⁡(2θ)\cos(2\theta).

Example 5

easy
Give the form of cos⁡(2θ)\cos(2\theta) in terms of cos⁡2θ\cos^2\theta only.

Example 6

easy
Give the form of cos⁡(2θ)\cos(2\theta) in terms of sin⁡2θ\sin^2\theta only.

Example 7

easy
If sin⁡θ=35\sin\theta = \frac35 and cos⁡θ=45\cos\theta = \frac45, find sin⁡(2θ)\sin(2\theta).

Example 8

easy
State the double-angle formula for tan⁡(2θ)\tan(2\theta).

Example 9

easy
If cos⁡θ=35\cos\theta = \frac35, use a double-angle form to find cos⁡(2θ)\cos(2\theta).

Example 10

easy
Is sin⁡(2θ)=2sin⁡θ\sin(2\theta) = 2\sin\theta correct? Answer yes or no.

Example 11

medium
If sin⁡θ=513\sin\theta = \frac{5}{13} and θ\theta is in Quadrant I, find cos⁡(2θ)\cos(2\theta).

Example 12

medium
If cos⁡θ=−35\cos\theta = -\frac35 and θ\theta is in Quadrant II, find sin⁡(2θ)\sin(2\theta).

Example 13

medium
Simplify 2sin⁡15°cos⁡15°2\sin 15°\cos 15°.

Example 14

medium
Simplify cos⁡225°−sin⁡225°\cos^2 25° - \sin^2 25°.

Example 15

medium
Simplify 1−2sin⁡230°1 - 2\sin^2 30°.

Example 16

medium
If tan⁡θ=13\tan\theta = \frac13, find tan⁡(2θ)\tan(2\theta).

Example 17

medium
Express sin⁡(4θ)\sin(4\theta) using a double-angle formula once.

Example 18

medium
If sin⁡θ=45\sin\theta = \frac{4}{5} and θ\theta is in Quadrant I, find cos⁡(2θ)\cos(2\theta).

Example 19

medium
Simplify 2tan⁡22.5°1−tan⁡222.5°\frac{2\tan 22.5°}{1 - \tan^2 22.5°}.

Example 20

challenge
If cos⁡θ=45\cos\theta = \frac45 and θ\theta is in Quadrant IV, find sin⁡(2θ)\sin(2\theta) and cos⁡(2θ)\cos(2\theta).

Example 21

challenge
Prove the power-reduction formula sin⁡2θ=1−cos⁡(2θ)2\sin^2\theta = \frac{1 - \cos(2\theta)}{2}.

Example 22

challenge
If sin⁡(2θ)=12\sin(2\theta) = \frac12, find the value of sin⁡θcos⁡θ\sin\theta\cos\theta.

Example 23

easy
Compute 2sin⁡45°cos⁡45°2\sin 45°\cos 45°.

Example 24

easy
If sin⁡θ=12\sin\theta = \tfrac{1}{2} and cos⁡θ=32\cos\theta = \tfrac{\sqrt{3}}{2}, find sin⁡(2θ)\sin(2\theta).

Example 25

easy
Simplify 2cos⁡260°−12\cos^2 60° - 1.

Example 26

easy
Find tan⁡(2⋅0°)\tan(2 \cdot 0°) and use the formula to verify it equals 00.

Example 27

easy
Simplify 1−2sin⁡245°1 - 2\sin^2 45°.

Example 28

medium
If sin⁡θ=−35\sin\theta = -\tfrac{3}{5} and θ\theta is in Quadrant III, find sin⁡(2θ)\sin(2\theta).

Example 29

medium
If cos⁡θ=725\cos\theta = \tfrac{7}{25} and θ\theta is in Quadrant IV, find cos⁡(2θ)\cos(2\theta).

Example 30

medium
Simplify sin⁡(2θ)sin⁡θ\dfrac{\sin(2\theta)}{\sin\theta}.

Example 31

medium
Write cos⁡4θ\cos 4\theta in terms of cos⁡2θ\cos 2\theta using a double-angle formula.

Example 32

medium
If tan⁡θ=−12\tan\theta = -\tfrac{1}{2}, find tan⁡(2θ)\tan(2\theta).

Example 33

medium
Express sin⁡θcos⁡θ\sin\theta\cos\theta in terms of sin⁡(2θ)\sin(2\theta).

Example 34

medium
Simplify 1−cos⁡(2x)sin⁡(2x)\dfrac{1 - \cos(2x)}{\sin(2x)}.

Example 35

medium
If cos⁡θ=1213\cos\theta = \tfrac{12}{13} and θ\theta is in Quadrant I, find tan⁡(2θ)\tan(2\theta).

Example 36

hard
Solve sin⁡(2x)+sin⁡x=0\sin(2x) + \sin x = 0 on [0,2π)[0, 2\pi).

Example 37

hard
If sin⁡θ+cos⁡θ=12\sin\theta + \cos\theta = \tfrac{1}{2}, find sin⁡(2θ)\sin(2\theta).

Example 38

hard
Solve cos⁡(2x)=sin⁡x\cos(2x) = \sin x on [0,2π)[0, 2\pi).

Example 39

hard
Find the exact value of sin⁡22.5°\sin 22.5° using a half-angle approach derived from the cosine double-angle identity.

Example 40

challenge
If sin⁡(2θ)=35\sin(2\theta) = \tfrac{3}{5} and θ∈(0,π/4)\theta \in (0, \pi/4), find sin⁡θ−cos⁡θ\sin\theta - \cos\theta.
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Background Knowledge

These ideas may be useful before you work through the harder examples.

trig identities sum difference