The Pythagorean identity and simplifying

About this lesson

Precalculus › Trigonometric identities and equations

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The Pythagorean identity and simplifying

Warm-up

Three quick questions from earlier lessons.

Discover

The point (cos⁡x,sin⁡x)(\cos x, \sin x) is on the unit circle, whose equation is X2+Y2=1X^2 + Y^2 = 1. What does that tell you?

  1. cos⁡2x+sin⁡2x=1\cos^2 x + \sin^2 x = 1
  2. cos⁡x+sin⁡x=1\cos x + \sin x = 1

Key idea

An identity is an equation that is true for every value of the variable.

sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1, so cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x and sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x. Also tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{\sin x}{\cos x}. Use them to find one value from another (the quadrant gives the sign) and to simplify.

Worked example

cos⁡θ=−35\cos\theta = -\dfrac{3}{5} and θ\theta is in quadrant III. Find sin⁡θ\sin\theta.

sin⁡2θ=1−925=1625\sin^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}
The Pythagorean identity.
sin⁡θ=±45\sin\theta = \pm\dfrac{4}{5}
Two square roots are possible.
sin⁡θ=−45\sin\theta = -\dfrac{4}{5}
Sine is negative in quadrant III.

Your turn, guided

Your turn: sin⁡θ=513\sin\theta = \dfrac{5}{13} and θ\theta is in quadrant I.

cos⁡2θ=144169\cos^2\theta = \dfrac{144}{169}
1 − 25/169.
cos⁡θ=1213\cos\theta = \dfrac{12}{13}
Positive in quadrant I.

Practice ★

  1. sin⁡θ=15\sin\theta = \frac{1}{5} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.
  2. sin⁡θ=38\sin\theta = \frac{3}{8} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.
  3. sin⁡θ=16\sin\theta = \frac{1}{6} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.
  4. sin⁡θ=13\sin\theta = \frac{1}{3} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.

Watch out!

Practice ★★

  1. Which expression is equal to 1−cos⁡2xsin⁡x\dfrac{1 - \cos^2 x}{\sin x} (where sin⁡x≠0\sin x \ne 0)?
    1. sin⁡x\sin x
    2. cos⁡x\cos x
    3. 1−cos⁡x1 - \cos x
  1. Which equation is true for every xx?
    1. sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1
    2. sin⁡x+cos⁡x=1\sin x + \cos x = 1
    3. sin⁡2x−cos⁡2x=1\sin^2 x - \cos^2 x = 1
    4. tan⁡x=cos⁡xsin⁡x\tan x = \dfrac{\cos x}{\sin x}

Exit ticket

Three questions to check you have got it.

  1. sin⁡θ=15\sin\theta = \frac{1}{5} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.
  2. sin⁡θ=38\sin\theta = \frac{3}{8} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.
  3. sin⁡θ=16\sin\theta = \frac{1}{6} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.

Summary card

The Pythagorean identity

  • sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 for every xx.
  • From one value: square, subtract from 1, take the root, choose the sign by quadrant.

Watch out: the identity needs the squares.

End of the lesson

Next lesson: Solve sin x = c, cos x = c and tan x = c Back to the chapter

Answers: The Pythagorean identity and simplifying

A

  1. −265- \frac{2 \sqrt{6}}{5}
  2. −558- \frac{\sqrt{55}}{8}
  3. −356- \frac{\sqrt{35}}{6}
  4. −223- \frac{2 \sqrt{2}}{3}

B

  1. sin⁡x\sin x

C

  1. sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1

D

  1. −265- \frac{2 \sqrt{6}}{5}
  2. −558- \frac{\sqrt{55}}{8}
  3. −356- \frac{\sqrt{35}}{6}

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