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3.12 - Equivalent Representations of Trigonometric Functions

Overview

Trig expressions can be written in many equivalent ways. An identity is an equation that is true for every input where both sides are defined, so you can swap one side for the other whenever it helps. This topic covers the Pythagorean identity and the sum, difference, and double-angle identities for sine and cosine, and how to use them to rewrite expressions, verify other identities, and solve equations.

The Pythagorean identity

Take the point P=(cos⁡θ,sin⁡θ)P = (\cos\theta, \sin\theta) where the terminal ray of θ\theta meets the unit circle. A vertical segment from PP down to the x-axis makes a right triangle. Its legs have lengths ∣cos⁡θ∣|\cos\theta| and ∣sin⁡θ∣|\sin\theta|, and its hypotenuse is a radius of the circle, so it has length 1.
3.12 - Equivalent Representations of Trigonometric Functions
The Pythagorean Theorem, a2+b2=c2a^2 + b^2 = c^2, gives
sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1
This holds for every angle. Squaring removes the signs, so it works in every quadrant, and it still works at angles like θ=0\theta = 0 where the triangle flattens out: 02+12=10^2 + 1^2 = 1. Remember that sin⁡2θ\sin^2\theta means (sin⁡θ)2(\sin\theta)^2.

Other forms

Dividing every term by cos⁡2θ\cos^2\theta (where cos⁡θ≠0\cos\theta \neq 0) gives a version with tangent and secant:
sin⁡2θcos⁡2θ+cos⁡2θcos⁡2θ=1cos⁡2θ⟹tan⁡2θ+1=sec⁡2θ\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta} \quad\Longrightarrow\quad \tan^2\theta + 1 = \sec^2\theta
Dividing by sin⁡2θ\sin^2\theta instead (where sin⁡θ≠0\sin\theta \neq 0) gives 1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta. Each identity can be rearranged, and the rearranged form is often the one you need.
Identity
Useful rearrangements
sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1
sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta and cos⁡2θ=1−sin⁡2θ\cos^2\theta = 1 - \sin^2\theta
1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta
tan⁡2θ=sec⁡2θ−1\tan^2\theta = \sec^2\theta - 1
1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta
cot⁡2θ=csc⁡2θ−1\cot^2\theta = \csc^2\theta - 1

Example 1: Finding cosine and tangent from sine

If sin⁡θ=−513\sin\theta = -\frac{5}{13} and π<θ<3π2\pi < \theta < \frac{3\pi}{2}, find cos⁡θ\cos\theta and tan⁡θ\tan\theta.
Step 1: Use the Pythagorean identity.
cos⁡2θ=1−sin⁡2θ=1−25169=144169\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{25}{169} = \frac{144}{169}
Step 2: Choose the sign. cos⁡θ=±1213\cos\theta = \pm\frac{12}{13}. The angle is in Quadrant III, where cosine is negative, so cos⁡θ=−1213\cos\theta = -\frac{12}{13}.
Step 3: Find tangent.
tan⁡θ=sin⁡θcos⁡θ=−5/13−12/13=512\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-5/13}{-12/13} = \frac{5}{12}
Answer: cos⁡θ=−1213\cos\theta = -\frac{12}{13} and tan⁡θ=512\tan\theta = \frac{5}{12}. The identity gives the size, and the quadrant gives the sign.

Connecting inverse trig functions

The Pythagorean identity also links the inverse trig functions. For example, for 0≤x≤10 \le x \le 1
arcsin⁡x=arccos⁡(1−x2)\arcsin x = \arccos\left(\sqrt{1 - x^2}\right)
Why it works: let θ=arcsin⁡x\theta = \arcsin x. Then sin⁡θ=x\sin\theta = x, and since 0≤x≤10 \le x \le 1, θ\theta is between 0 and π2\frac{\pi}{2}. By the Pythagorean identity, cos⁡θ=±1−x2\cos\theta = \pm\sqrt{1 - x^2}, and cosine is not negative on [0,π2]\left[0, \frac{\pi}{2}\right], so cos⁡θ=1−x2\cos\theta = \sqrt{1 - x^2}. That angle is in the output range of arccosine, so θ=arccos⁡(1−x2)\theta = \arccos\left(\sqrt{1 - x^2}\right).
Why the domain restriction matters: try x=−12x = -\frac{1}{2}. The left side is arcsin⁡(−12)=−π6\arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}, but the right side is arccos⁡(32)=π6\arccos\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}. They disagree, because arccosine never outputs a negative angle.

Sum and difference identities

These identities expand the sine or cosine of a sum of two angles:
sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\betacos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta
A way to remember them: the sine identity mixes the functions (sin cos, then cos sin) and keeps the plus sign. The cosine identity keeps like functions together (cos cos, then sin sin) and flips the sign.
Difference identities. Replace β\beta with −β-\beta. Cosine is even, cos⁡(−β)=cos⁡β\cos(-\beta) = \cos\beta, and sine is odd, sin⁡(−β)=−sin⁡β\sin(-\beta) = -\sin\beta, so
sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\betacos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta
Sine does not distribute: sin⁡(α+β)≠sin⁡α+sin⁡β\sin(\alpha + \beta) \neq \sin\alpha + \sin\beta. Check with α=β=π2\alpha = \beta = \frac{\pi}{2}: sin⁡π=0\sin\pi = 0, but sin⁡π2+sin⁡π2=2\sin\frac{\pi}{2} + \sin\frac{\pi}{2} = 2.

Example 2: An exact value

Find the exact value of sin⁡5π12\sin\frac{5\pi}{12}.
Step 1: Write the angle as a sum of special angles. 5π12=2π12+3π12=π6+π4\frac{5\pi}{12} = \frac{2\pi}{12} + \frac{3\pi}{12} = \frac{\pi}{6} + \frac{\pi}{4}.
Step 2: Apply the sum identity.
sin⁡(π6+π4)=sin⁡π6cos⁡π4+cos⁡π6sin⁡π4\sin\left(\frac{\pi}{6} + \frac{\pi}{4}\right) = \sin\frac{\pi}{6}\cos\frac{\pi}{4} + \cos\frac{\pi}{6}\sin\frac{\pi}{4}=12⋅22+32⋅22=2+64= \frac{1}{2}\cdot\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} = \frac{\sqrt{2} + \sqrt{6}}{4}
Answer: sin⁡5π12=2+64≈0.966\sin\frac{5\pi}{12} = \frac{\sqrt{2} + \sqrt{6}}{4} \approx 0.966, which matches a calculator.

Example 3: Simplifying with a difference identity

Simplify cos⁡(θ−π2)\cos\left(\theta - \frac{\pi}{2}\right).
cos⁡(θ−π2)=cos⁡θcos⁡π2+sin⁡θsin⁡π2=cos⁡θ⋅0+sin⁡θ⋅1=sin⁡θ\cos\left(\theta - \frac{\pi}{2}\right) = \cos\theta\cos\frac{\pi}{2} + \sin\theta\sin\frac{\pi}{2} = \cos\theta \cdot 0 + \sin\theta \cdot 1 = \sin\theta
This proves what the graphs showed: shifting the cosine graph right by π2\frac{\pi}{2} gives the sine graph.

Double-angle identities

Setting β=α\beta = \alpha in the sum identities gives the double-angle identities. For sine:
sin⁡2θ=sin⁡(θ+θ)=sin⁡θcos⁡θ+cos⁡θsin⁡θ=2sin⁡θcos⁡θ\sin 2\theta = \sin(\theta + \theta) = \sin\theta\cos\theta + \cos\theta\sin\theta = 2\sin\theta\cos\theta
For cosine:
cos⁡2θ=cos⁡2θ−sin⁡2θ\cos 2\theta = \cos^2\theta - \sin^2\theta
Replacing sin⁡2θ\sin^2\theta or cos⁡2θ\cos^2\theta with the Pythagorean identity gives two more forms of cos⁡2θ\cos 2\theta:
cos⁡2θ=2cos⁡2θ−1=1−2sin⁡2θ\cos 2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta
Pick the form that matches the problem. If the rest of the expression only has cosine, use 2cos⁡2θ−12\cos^2\theta - 1. If it only has sine, use 1−2sin⁡2θ1 - 2\sin^2\theta.

Example 4: Double angles from one value

If sin⁡θ=35\sin\theta = \frac{3}{5} and π2<θ<π\frac{\pi}{2} < \theta < \pi, find sin⁡2θ\sin 2\theta and cos⁡2θ\cos 2\theta.
Step 1: Find cos⁡θ\cos\theta. 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}.
Step 2: Apply the double-angle identities.
sin⁡2θ=2sin⁡θcos⁡θ=2(35)(−45)=−2425\sin 2\theta = 2\sin\theta\cos\theta = 2\left(\frac{3}{5}\right)\left(-\frac{4}{5}\right) = -\frac{24}{25}cos⁡2θ=1−2sin⁡2θ=1−2(925)=725\cos 2\theta = 1 - 2\sin^2\theta = 1 - 2\left(\frac{9}{25}\right) = \frac{7}{25}
Answer: sin⁡2θ=−2425\sin 2\theta = -\frac{24}{25} and cos⁡2θ=725\cos 2\theta = \frac{7}{25}. Check with the Pythagorean identity: (−2425)2+(725)2=576+49625=1\left(-\frac{24}{25}\right)^2 + \left(\frac{7}{25}\right)^2 = \frac{576 + 49}{625} = 1.
What the AP exam covers: the identities in the CED are the Pythagorean identity (and the forms you get by rearranging it) and the sum, difference, and double-angle identities for sine and cosine. Half-angle identities and the sum and difference identities for tangent aren't in the CED.

Verifying identities

To verify an identity, rewrite one side step by step until it matches the other side. Work on one side only, usually the more complicated one. Don't add, multiply, or cross-multiply across the equals sign, because that assumes the identity is already true.
Strategies that usually work:
  • Rewrite everything in terms of sine and cosine.
  • Look for a Pythagorean identity to substitute, like 1−cos⁡2θ=sin⁡2θ1 - \cos^2\theta = \sin^2\theta.
  • Expand double angles, or combine fractions over a common denominator.
  • Factor, then cancel common factors.

Example 5: A double-angle identity

Verify that sin⁡2θ1+cos⁡2θ=tan⁡θ\dfrac{\sin 2\theta}{1 + \cos 2\theta} = \tan\theta.
Start with the left side. The denominator has a 1+1 + in it, so use the form cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2\cos^2\theta - 1, which makes the 1s cancel:
sin⁡2θ1+cos⁡2θ=2sin⁡θcos⁡θ1+(2cos⁡2θ−1)=2sin⁡θcos⁡θ2cos⁡2θ=sin⁡θcos⁡θ=tan⁡θ\frac{\sin 2\theta}{1 + \cos 2\theta} = \frac{2\sin\theta\cos\theta}{1 + (2\cos^2\theta - 1)} = \frac{2\sin\theta\cos\theta}{2\cos^2\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta
The left side now matches the right side, so the identity is verified for every θ\theta where both sides are defined (where cos⁡θ≠0\cos\theta \neq 0).

Example 6: A Pythagorean identity

Verify that sec⁡θ−cos⁡θ=sin⁡θtan⁡θ\sec\theta - \cos\theta = \sin\theta\tan\theta.
sec⁡θ−cos⁡θ=1cos⁡θ−cos⁡2θcos⁡θ=1−cos⁡2θcos⁡θ=sin⁡2θcos⁡θ=sin⁡θ⋅sin⁡θcos⁡θ=sin⁡θtan⁡θ\sec\theta - \cos\theta = \frac{1}{\cos\theta} - \frac{\cos^2\theta}{\cos\theta} = \frac{1 - \cos^2\theta}{\cos\theta} = \frac{\sin^2\theta}{\cos\theta} = \sin\theta \cdot \frac{\sin\theta}{\cos\theta} = \sin\theta\tan\theta
The moves were: rewrite secant with cosine, combine over a common denominator, use 1−cos⁡2θ=sin⁡2θ1 - \cos^2\theta = \sin^2\theta, then split off a tangent.

Choosing a useful form

Sometimes rewriting a function makes its key features easy to read.

Example 7: Reading features from an equivalent form

(a) Find the maximum value and the period of f(θ)=4sin⁡θcos⁡θf(\theta) = 4\sin\theta\cos\theta.
4sin⁡θcos⁡θ=2(2sin⁡θcos⁡θ)=2sin⁡2θ4\sin\theta\cos\theta = 2(2\sin\theta\cos\theta) = 2\sin 2\theta
Now ff is a sinusoid with amplitude 2 and period 2π2=π\frac{2\pi}{2} = \pi. Its maximum value is 2, reached when 2θ=π22\theta = \frac{\pi}{2}, at θ=π4\theta = \frac{\pi}{4} (and every π\pi after that).
(b) Show that g(θ)=sin⁡θ+cos⁡θg(\theta) = \sin\theta + \cos\theta is a sinusoid with amplitude 2\sqrt{2}. Expand 2sin⁡(θ+π4)\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right) with the sum identity:
2sin⁡(θ+π4)=2(sin⁡θ⋅22+cos⁡θ⋅22)=sin⁡θ+cos⁡θ\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right) = \sqrt{2}\left(\sin\theta \cdot \frac{\sqrt{2}}{2} + \cos\theta \cdot \frac{\sqrt{2}}{2}\right) = \sin\theta + \cos\theta
So g(θ)=2sin⁡(θ+π4)g(\theta) = \sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right): amplitude 2\sqrt{2}, period 2π2\pi, and maximum value 2\sqrt{2} at θ=π4\theta = \frac{\pi}{4}. None of that is obvious from sin⁡θ+cos⁡θ\sin\theta + \cos\theta.

Solving equations with identities

An equation that mixes functions (like sin⁡θ\sin\theta and cos⁡2θ\cos^2\theta) or mixes angles (like 2θ2\theta and θ\theta) is hard to solve as written. Use an identity to rewrite it in terms of one function of one angle, then factor and solve.

Example 8: Using the Pythagorean identity

Solve 2cos⁡2θ+3sin⁡θ−3=02\cos^2\theta + 3\sin\theta - 3 = 0 for 0≤θ<2π0 \le \theta < 2\pi.
Step 1: Rewrite with only sine. Replace cos⁡2θ\cos^2\theta with 1−sin⁡2θ1 - \sin^2\theta:
2(1−sin⁡2θ)+3sin⁡θ−3=0⇒−2sin⁡2θ+3sin⁡θ−1=0⇒2sin⁡2θ−3sin⁡θ+1=02(1 - \sin^2\theta) + 3\sin\theta - 3 = 0 \quad\Rightarrow\quad -2\sin^2\theta + 3\sin\theta - 1 = 0 \quad\Rightarrow\quad 2\sin^2\theta - 3\sin\theta + 1 = 0
Step 2: Factor. This is a quadratic in sin⁡θ\sin\theta:
(2sin⁡θ−1)(sin⁡θ−1)=0(2\sin\theta - 1)(\sin\theta - 1) = 0
Step 3: Solve each factor. sin⁡θ=12\sin\theta = \frac{1}{2} gives θ=π6\theta = \frac{\pi}{6} or 5π6\frac{5\pi}{6}. sin⁡θ=1\sin\theta = 1 gives θ=π2\theta = \frac{\pi}{2}.
Answer: θ=π6,π2,5π6\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}.

Example 9: An equation and an inequality (AP-style)

For 0≤θ<2π0 \le \theta < 2\pi, (a) solve sin⁡2θ=3sin⁡θ\sin 2\theta = \sqrt{3}\sin\theta, and (b) solve sin⁡2θ>3sin⁡θ\sin 2\theta > \sqrt{3}\sin\theta.
(a) Step 1: Rewrite the double angle and move everything to one side.
2sin⁡θcos⁡θ−3sin⁡θ=02\sin\theta\cos\theta - \sqrt{3}\sin\theta = 0
Step 2: Factor.
sin⁡θ (2cos⁡θ−3)=0\sin\theta\,(2\cos\theta - \sqrt{3}) = 0
Don't divide both sides by sin⁡θ\sin\theta. That would throw away the solutions where sin⁡θ=0\sin\theta = 0.
Step 3: Solve each factor. sin⁡θ=0\sin\theta = 0 gives θ=0\theta = 0 or π\pi. cos⁡θ=32\cos\theta = \frac{\sqrt{3}}{2} gives θ=π6\theta = \frac{\pi}{6} or 11π6\frac{11\pi}{6}.
So the solutions are θ=0,π6,π,11π6\theta = 0, \frac{\pi}{6}, \pi, \frac{11\pi}{6}.
(b) The inequality is sin⁡θ (2cos⁡θ−3)>0\sin\theta\,(2\cos\theta - \sqrt{3}) > 0. The solutions from part (a) split [0,2π)[0, 2\pi) into intervals where the product keeps one sign. Check the sign of each factor:
Interval
sin⁡θ\sin\theta
2cos⁡θ−32\cos\theta - \sqrt{3}
Product
0<θ<π60 < \theta < \frac{\pi}{6}
++
++
++
π6<θ<π\frac{\pi}{6} < \theta < \pi
++
−-
−-
π<θ<11π6\pi < \theta < \frac{11\pi}{6}
−-
−-
++
11π6<θ<2π\frac{11\pi}{6} < \theta < 2\pi
−-
++
−-
Answer: 0<θ<π60 < \theta < \frac{\pi}{6} or π<θ<11π6\pi < \theta < \frac{11\pi}{6}. The solutions from part (a) are left out, because the product is 0 there, not positive. The graph confirms it: the shaded regions are exactly where y=sin⁡2θy = \sin 2\theta (black) is above y=3sin⁡θy = \sqrt{3}\sin\theta (red). The first region is only a thin sliver, because the two curves stay close together between 00 and π6\frac{\pi}{6}.
3.12 - Equivalent Representations of Trigonometric Functions

Common mistakes

  • Distributing sine or cosine. sin⁡(α+β)\sin(\alpha + \beta) is not sin⁡α+sin⁡β\sin\alpha + \sin\beta, and sin⁡2θ\sin 2\theta is not 2sin⁡θ2\sin\theta.
  • Sign errors in the cosine identities. The sum identity has a minus, cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta, and the difference identity has a plus.
  • Dividing by a trig function. Dividing both sides by sin⁡θ\sin\theta loses every solution where sin⁡θ=0\sin\theta = 0. Factor instead.
  • Ignoring the quadrant after a square root. cos⁡θ=±1−sin⁡2θ\cos\theta = \pm\sqrt{1 - \sin^2\theta}, and the quadrant decides the sign.
  • Working both sides of an identity at once. Transform one side until it matches the other.

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