Double-angle identities

About this lesson

Precalculus › Trigonometric identities and equations

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Double-angle identities

Warm-up

Three quick questions from earlier lessons.

Discover

Is sin⁡2x\sin 2x the same as 2sin⁡x2\sin x? Try x=π2x = \dfrac{\pi}{2}.

  1. No
  2. Yes

Key idea

Put A=B=xA = B = x in the sum identities.

sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x. cos⁡2x=cos⁡2x−sin⁡2x=2cos⁡2x−1=1−2sin⁡2x\cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x.

Worked example

sin⁡θ=35\sin\theta = \dfrac{3}{5}, θ\theta in quadrant I. Find sin⁡2θ\sin 2\theta and cos⁡2θ\cos 2\theta.

cos⁡θ=45\cos\theta = \dfrac{4}{5}
Pythagorean identity; positive in quadrant I.
sin⁡2θ=2⋅35⋅45=2425\sin 2\theta = 2\cdot\dfrac{3}{5}\cdot\dfrac{4}{5} = \dfrac{24}{25}
2 sin θ cos θ.
cos⁡2θ=1−2⋅925=725\cos 2\theta = 1 - 2\cdot\dfrac{9}{25} = \dfrac{7}{25}
1 − 2 sin²θ.

Your turn, guided

Your turn: cos⁡θ=13\cos\theta = \dfrac{1}{3}. Find cos⁡2θ\cos 2\theta.

cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2\cos^2\theta - 1
The form with cosine.
=−79= -\dfrac{7}{9}
2 · 1/9 − 1.

Practice ★

  1. cos⁡θ=37\cos\theta = \frac{3}{7}. Find the exact value of cos⁡2θ\cos 2\theta.
  2. cos⁡θ=17\cos\theta = \frac{1}{7}. Find the exact value of cos⁡2θ\cos 2\theta.
  3. cos⁡θ=25\cos\theta = \frac{2}{5}. Find the exact value of cos⁡2θ\cos 2\theta.

Watch out!

Practice ★★

  1. θ\theta is in quadrant I and sin⁡θ=1517\sin\theta = \frac{15}{17}. Find the exact value of sin⁡2θ\sin 2\theta.
  2. θ\theta is in quadrant I and sin⁡θ=35\sin\theta = \frac{3}{5}. Find the exact value of sin⁡2θ\sin 2\theta.
  1. Which expressions are equal to cos⁡2x\cos 2x? Tick all of them.
    1. cos⁡2x−sin⁡2x\cos^2 x - \sin^2 x
    2. 2cos⁡2x−12\cos^2 x - 1
    3. 1−2sin⁡2x1 - 2\sin^2 x
    4. 2cos⁡x2\cos x
    5. cos⁡2x+sin⁡2x\cos^2 x + \sin^2 x

Exit ticket

Three questions to check you have got it.

  1. cos⁡θ=37\cos\theta = \frac{3}{7}. Find the exact value of cos⁡2θ\cos 2\theta.
  2. cos⁡θ=17\cos\theta = \frac{1}{7}. Find the exact value of cos⁡2θ\cos 2\theta.
  3. cos⁡θ=25\cos\theta = \frac{2}{5}. Find the exact value of cos⁡2θ\cos 2\theta.

Summary card

Double-angle identities

  • sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x.
  • cos⁡2x=cos⁡2x−sin⁡2x=2cos⁡2x−1=1−2sin⁡2x\cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x.

Watch out: sin⁡2x≠2sin⁡x\sin 2x \ne 2\sin x (a sine is never above 1, but 2sin⁡x2\sin x can be).

End of the lesson

Next lesson: Solve equations with identities Back to the chapter

Answers: Double-angle identities

A

  1. −3149- \frac{31}{49}
  2. −4749- \frac{47}{49}
  3. −1725- \frac{17}{25}

B

  1. 240289\frac{240}{289}
  2. 2425\frac{24}{25}

C

  1. cos⁡2x−sin⁡2x\cos^2 x - \sin^2 x · 2cos⁡2x−12\cos^2 x - 1 · 1−2sin⁡2x1 - 2\sin^2 x

D

  1. −3149- \frac{31}{49}
  2. −4749- \frac{47}{49}
  3. −1725- \frac{17}{25}

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