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Step 14: Trigonometry and precalculus

Trig identities and equations

Step 14 · Trigonometry and precalculus topics

Use the Pythagorean identity, the sum and double-angle formulas, and factoring to simplify expressions and to find every solution of a trig equation in a given range.

The Pythagorean identity

Every point on the unit circle has x² + y² = 1, and the point is (cos θ, sin θ). So for every angle:

sin⁡2θ+cos⁡2θ=1\sin^{2}\theta + \cos^{2}\theta = 1

Example: Cosine from sine

sin⁡θ=35\sin\theta = \frac{3}{5} and θ is in quadrant II. Find cos θ and tan θ.

  1. cos⁡2θ=1−925=1625\cos^{2}\theta = 1 - \frac{9}{25} = \frac{16}{25}.
  2. In quadrant II cosine is negative, so cos⁡θ=−45\cos\theta = -\frac{4}{5}.
  3. tan⁡θ=sin⁡θcos⁡θ=−34\tan\theta = \frac{\sin\theta}{\cos\theta} = -\frac{3}{4}.

Answercos⁡θ=−45\cos\theta = -\frac{4}{5} and tan⁡θ=−34\tan\theta = -\frac{3}{4}

Example: Simplifying

Simplify 1−cos⁡2xsin⁡x\frac{1 - \cos^{2}x}{\sin x}.

  1. 1 − cos²x is sin²x.
  2. sin²x ÷ sin x is sin x.

Answersin x

Sum and double-angle formulas

The sine of a sum is not the sum of the sines. Instead:

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡Bcos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\sin(A + B) = \sin A\cos B + \cos A\sin B \qquad \cos(A + B) = \cos A\cos B - \sin A\sin B

Example: An exact value

Find sin 75° exactly, using 75° = 45° + 30°.

  1. sin⁡45∘cos⁡30∘+cos⁡45∘sin⁡30∘\sin 45^{\circ}\cos 30^{\circ} + \cos 45^{\circ}\sin 30^{\circ}
  2. =22⋅32+22⋅12= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}

Answer6+24\frac{\sqrt{6} + \sqrt{2}}{4}

Putting A = B gives the double-angle formulas.

sin⁡2A=2sin⁡Acos⁡Acos⁡2A=cos⁡2A−sin⁡2A=1−2sin⁡2A\sin 2A = 2\sin A\cos A \qquad \cos 2A = \cos^{2}A - \sin^{2}A = 1 - 2\sin^{2}A

Example: A double angle

sin⁡A=35\sin A = \frac{3}{5} and cos⁡A=45\cos A = \frac{4}{5}. Find sin 2A.

  1. sin⁡2A=2⋅35⋅45\sin 2A = 2 \cdot \frac{3}{5} \cdot \frac{4}{5}

Answer2425\frac{24}{25}

Solving trig equations

To solve a trig equation, get the trig function on its own, find the reference angle, then list every angle in the range where the function has the right sign. There is usually more than one solution.

Example: Two solutions

Solve 2 sin x = 1 for 0 ≤ x < 2π.

  1. sin⁡x=12\sin x = \frac{1}{2}.
  2. The reference angle is π6\frac{\pi}{6}, and sine is positive in quadrants I and II.

Answerx=π6x = \frac{\pi}{6} or x=5π6x = \frac{5\pi}{6}

Example: A quadratic in disguise

Solve 2cos²x − cos x − 1 = 0 for 0 ≤ x < 2π.

  1. Factor: (2 cos x + 1)(cos x − 1) = 0.
  2. cos⁡x=−12\cos x = -\frac{1}{2} gives 2π3\frac{2\pi}{3} and 4π3\frac{4\pi}{3}; cos x = 1 gives 0.

Answerx=0,2π3,4π3x = 0, \frac{2\pi}{3}, \frac{4\pi}{3}

Example: Using an identity

Solve sin 2x = sin x for 0 ≤ x < 2π.

  1. 2 sin x cos x − sin x = 0, so sin x(2 cos x − 1) = 0.
  2. sin x = 0 gives 0 and π; cos⁡x=12\cos x = \frac{1}{2} gives π3\frac{\pi}{3} and 5π3\frac{5\pi}{3}.

Answerx=0,π3,π,5π3x = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}

Try it yourself

Work each one on paper first, then open the answer to check.

  1. cos⁡θ=513\cos\theta = \frac{5}{13} and θ is in quadrant IV. Find sin θ.

    Show answer

    −1213-\frac{12}{13}

  2. Simplify cos x · tan x.

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    sin x

  3. Find cos 15° exactly, using 15° = 45° − 30°.

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    6+24\frac{\sqrt{6} + \sqrt{2}}{4}

  4. sin A = 0.6 and cos A = 0.8. Find cos 2A.

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    0.28

  5. Solve 2 cos x = √3 for 0 ≤ x < 2π.

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    x=π6,11π6x = \frac{\pi}{6}, \frac{11\pi}{6}

  6. Solve tan x = 1 for 0 ≤ x < 2π.

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    x=π4,5π4x = \frac{\pi}{4}, \frac{5\pi}{4}

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