AP Precalculus · Lesson 3

Trig identities and polar functions

Know Know it

The Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 comes from the unit circle. Dividing it by cos⁡2θ\cos^2\theta gives 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta. The reciprocal functions are sec⁡θ=1cos⁡θ\sec\theta = \frac{1}{\cos\theta}, csc⁡θ=1sin⁡θ\csc\theta = \frac{1}{\sin\theta} and cot⁡θ=cos⁡θsin⁡θ\cot\theta = \frac{\cos\theta}{\sin\theta}.

Sum identities: sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta and cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta. For a difference, change the sign between the two products. Double-angle identities: sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\theta and cos⁡2θ=cos⁡2θ−sin⁡2θ\cos 2\theta = \cos^2\theta - \sin^2\theta.

In polar coordinates a point is (r,θ)(r, \theta): go r units from the origin along the ray at angle θ\theta. If r is negative, go the other way. A polar function r=f(θ)r = f(\theta) is graphed by sweeping θ\theta and plotting each distance. The distance from the origin is |r|.

Moving between polar and rectangular coordinates
To getUse
x and y from (r,θ)(r, \theta)x=rcos⁡θx = r\cos\theta and y=rsin⁡θy = r\sin\theta
r from (x,y)(x, y)r=x2+y2r = \sqrt{x^2 + y^2}
θ\theta from (x,y)(x, y)tan⁡θ=yx\tan\theta = \frac{y}{x}, then check the quadrant of the point

Trap Trig functions do not split over a sum

sin⁡(α+β)\sin(\alpha + \beta) is not sin⁡α+sin⁡β\sin\alpha + \sin\beta in general. With α=π6\alpha = \frac{\pi}{6} and β=π3\beta = \frac{\pi}{3}, sin⁡π2=1\sin\frac{\pi}{2} = 1, but 12+32≈1.37\frac{1}{2} + \frac{\sqrt{3}}{2} \approx 1.37. Use the sum identity.

Worked example Easy

An angle θ\theta is in quadrant II and sin⁡θ=35\sin\theta = \frac{3}{5}. What is cos⁡θ\cos\theta?

  1. 45\frac{4}{5}
  2. −34-\frac{3}{4}
  3. −45-\frac{4}{5}Answer
  4. 25\frac{2}{5}

How to solve it

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

Why each choice is right or wrong

  • A. Incorrect. 45\frac{4}{5} has the right size, but cosine is negative in quadrant II, where x is negative.
  • B. Incorrect. −34-\frac{3}{4} is sin⁡θcos⁡θ\frac{\sin\theta}{\cos\theta}, the tangent of the angle, not its cosine.
  • C. Correct. cos⁡2θ=1−925=1625\cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25}, and cosine is negative in quadrant II, so cos⁡θ=−45\cos\theta = -\frac{4}{5}.
  • D. Incorrect. 25\frac{2}{5} subtracts sin⁡θ\sin\theta from 1, but the identity uses the squares, sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.

Explain Explain it

Task verbs: calculate, justify

To justify an identity, start from one side and rewrite it step by step until it becomes the other side, naming the identity you use at each step. Common moves: factor, replace sin⁡2θ+cos⁡2θ\sin^2\theta + \cos^2\theta with 1, and rewrite tan, sec, csc and cot in terms of sine and cosine.

To change (x,y)(x, y) to polar form, find r=x2+y2r = \sqrt{x^2 + y^2}, then pick the angle from the point's quadrant. Do not rely on arctan⁡yx\arctan\frac{y}{x} alone: it only returns angles between −π2-\frac{\pi}{2} and π2\frac{\pi}{2}.

Worked example Medium

Which polar coordinates (r,θ)(r, \theta), with r>0r > 0 and 0≤θ<2π0 \le \theta < 2\pi, name the rectangular point (−2,23)(-2, 2\sqrt{3})?

  1. (4,−π3)\left(4, -\frac{\pi}{3}\right)
  2. (16,2π3)\left(16, \frac{2\pi}{3}\right)
  3. (4,π3)\left(4, \frac{\pi}{3}\right)
  4. (4,2π3)\left(4, \frac{2\pi}{3}\right)Answer

How to solve it

  1. r=(−2)2+(23)2=16=4r = \sqrt{(-2)^2 + (2\sqrt{3})^2} = \sqrt{16} = 4.
  2. tan⁡θ=23−2=−3\tan\theta = \frac{2\sqrt{3}}{-2} = -\sqrt{3}, so the reference angle is π3\frac{\pi}{3}.
  3. The point is in quadrant II, so θ=π−π3=2π3\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}.

Why each choice is right or wrong

  • A. Incorrect. −π3-\frac{\pi}{3} is what arctan returns, but it points into quadrant IV and is outside 0≤θ<2π0 \le \theta < 2\pi.
  • B. Incorrect. 16 is x2+y2x^2 + y^2, which is r2r^2; the distance r is its square root, 4.
  • C. Incorrect. π3\frac{\pi}{3} points into quadrant I, but the point has a negative x-coordinate, so it is in quadrant II.
  • D. Correct. r=4+12=4r = \sqrt{4 + 12} = 4, and the point is in quadrant II with reference angle π3\frac{\pi}{3}, so θ=2π3\theta = \frac{2\pi}{3}.

Worked example Hard

Use π12=π3−π4\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4} to find the exact value of cos⁡π12\cos\frac{\pi}{12}.

  1. 6−24\frac{\sqrt{6} - \sqrt{2}}{4}
  2. 6+24\frac{\sqrt{6} + \sqrt{2}}{4}Answer
  3. 1−22\frac{1 - \sqrt{2}}{2}
  4. 24\frac{\sqrt{2}}{4}

How to solve it

  1. cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta.
  2. cos⁡π3cos⁡π4=12⋅22=24\cos\frac{\pi}{3}\cos\frac{\pi}{4} = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4}, and sin⁡π3sin⁡π4=32⋅22=64\sin\frac{\pi}{3}\sin\frac{\pi}{4} = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{6}}{4}.
  3. Add: cos⁡π12=6+24\cos\frac{\pi}{12} = \frac{\sqrt{6} + \sqrt{2}}{4}, about 0.966.

Why each choice is right or wrong

  • A. Incorrect. 6−24\frac{\sqrt{6} - \sqrt{2}}{4} uses a minus sign between the products, but for a difference of angles the cosine identity adds them.
  • B. Correct. cos⁡π3cos⁡π4+sin⁡π3sin⁡π4=24+64=6+24\cos\frac{\pi}{3}\cos\frac{\pi}{4} + \sin\frac{\pi}{3}\sin\frac{\pi}{4} = \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}.
  • C. Incorrect. 1−22\frac{1 - \sqrt{2}}{2} is cos⁡π3−cos⁡π4\cos\frac{\pi}{3} - \cos\frac{\pi}{4}, which splits cosine over a difference; it is even negative.
  • D. Incorrect. 24\frac{\sqrt{2}}{4} is only the first product, cos⁡π3cos⁡π4\cos\frac{\pi}{3}\cos\frac{\pi}{4}, and leaves out the product of the sines.

Justify Your turn

Show that cos⁡4θ−sin⁡4θ=cos⁡2θ\cos^4\theta - \sin^4\theta = \cos 2\theta for every angle θ\theta. Name each identity you use.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Factors the left side as a difference of squares, (cos⁡2θ−sin⁡2θ)(cos⁡2θ+sin⁡2θ)(\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta)
  • Replaces cos⁡2θ+sin⁡2θ\cos^2\theta + \sin^2\theta with 1 by the Pythagorean identity and cos⁡2θ−sin⁡2θ\cos^2\theta - \sin^2\theta with cos⁡2θ\cos 2\theta by the double-angle identity

Model answer

As a difference of squares, cos⁡4θ−sin⁡4θ=(cos⁡2θ−sin⁡2θ)(cos⁡2θ+sin⁡2θ)\cos^4\theta - \sin^4\theta = (\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta). By the Pythagorean identity the second factor is 1. By the double-angle identity the first factor is cos⁡2θ\cos 2\theta. So the left side equals cos⁡2θ⋅1=cos⁡2θ\cos 2\theta \cdot 1 = \cos 2\theta.

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