Topic 3.12B

Equivalent Representations of Trigonometric Functions

Part A connected $\sin \theta$ and $\cos \theta$ through the Pythagorean identity. Part B answers a new question: what happens to sine and cosine when you add two angles? The answer unlocks exact values for angles like $5\pi /12$ , proves a fact from 3.5, and gives the double-angle identities.

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01
Choose the form of $\cos 2\theta$ that matches the rest of the equation: $1 - 2\sin ^{2}\theta$ with sines, $2\cos ^{2}\theta - 1$ with cosines.

Practice 33 problems Next: Chapter 3.13A Trigonometry and Polar Coordinates

Sine and Cosine Do Not Distribute

It is tempting to think $\sin (\alpha + \beta ) = \sin \alpha + \sin \beta$ . Test it with $\alpha = \pi /6$ and $\beta = \pi /3$ : $\sin (\pi /6 + \pi /3) = \sin (\pi /2) = 1$ , but $\sin (\pi /6) + \sin (\pi /3) = 1/2 + \sqrt{3}/2 \approx 1.366$ . They are not equal, and the graph shows they are different functions.

$y = \sin (\theta + \pi /3)$ is a phase shift of sine; $y = \sin \theta + \sin (\pi /3)$ is a vertical shift. Different graphs.

The correct rules are the sum and difference identities:

\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta

\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta

CONCEPT

Reading the signs

In the sine identity the sign in the middle MATCHES the sign in the angle: $\sin (\alpha - \beta ) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$ .

In the cosine identity the sign FLIPS: $\cos (\alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$ , and $\cos (\alpha - \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$ .

Sine mixes: $\sin \cdot \cos$ and $\cos \cdot \sin$ . Cosine keeps pairs together: $\cos \cdot \cos$ and $\sin \cdot \sin$ .

Worked example

Example 1. Find the exact values of $\sin (5\pi /12)$ and $\cos (7\pi /12)$ .

01
Write each angle as a sum of special angles. $5\pi /12 = 3\pi /12 + 2\pi /12 = \pi /4 + \pi /6$ , and $7\pi /12 = 4\pi /12 + 3\pi /12 = \pi /3 + \pi /4$ .
02
$\sin (\pi /4 + \pi /6) = \sin (\pi /4)\cos (\pi /6) + \cos (\pi /4)\sin (\pi /6) = (\sqrt{2}/2)(\sqrt{3}/2) + (\sqrt{2}/2)(1/2) = (\sqrt{6} + \sqrt{2})/4 \approx 0.966$ .
03
$\cos (\pi /3 + \pi /4) = \cos (\pi /3)\cos (\pi /4) - \sin (\pi /3)\sin (\pi /4) = (1/2)(\sqrt{2}/2) - (\sqrt{3}/2)(\sqrt{2}/2) = (\sqrt{2} - \sqrt{6})/4 \approx -0.259$ .
04
Sense check: $7\pi /12$ is a little past $\pi /2$ , in Quadrant II, so its cosine should be a small negative number ✓.

COMMON MISTAKE

" $\cos (\pi /3 + \pi /4) = \cos (\pi /3)\cos (\pi /4) + \sin (\pi /3)\sin (\pi /4)$ ."

The cosine identity flips the sign. With + you get $(\sqrt{2} + \sqrt{6})/4 \approx 0.966$ , a positive number — impossible for an angle in Quadrant II. A quick quadrant check catches a wrong sign every time.

Quick check

Find $\sin\left(\frac{5\pi}{12}\right)$ exactly.

Rewriting Shifted Functions

Sum identities also turn a phase shift into simpler terms. Remember from 3.5 that $\cos \theta = \sin (\theta + \pi /2)$ — we saw it on the graph. Now we can prove it: $\sin (\theta + \pi /2) = \sin \theta \cos (\pi /2) + \cos \theta \sin (\pi /2) = \sin \theta \cdot 0 + \cos \theta \cdot 1 = \cos \theta$ ✓.

Worked example

Example 2. Simplify $\cos (\theta + 3\pi /2)$ and $\sin (\theta - \pi )$ .

01
$\cos (\theta + 3\pi /2) = \cos \theta \cos (3\pi /2) - \sin \theta \sin (3\pi /2)$ . Since $\cos (3\pi /2) = 0$ and $\sin (3\pi /2) = -1$ , this is $0 - \sin \theta (-1) = \sin \theta$ .
02
$\sin (\theta - \pi ) = \sin \theta \cos \pi - \cos \theta \sin \pi = \sin \theta (-1) - \cos \theta (0) = -\sin \theta$ .
03
Graph check: shifting sine right by $\pi$ turns it upside down, so $-\sin \theta$ makes sense (3.6B).

Double-Angle Identities

Let $\beta = \alpha$ in the sum identities. $\sin (\alpha + \alpha ) = \sin \alpha \cos \alpha + \cos \alpha \sin \alpha = 2 \sin \alpha \cos \alpha$ , and $\cos (\alpha + \alpha ) = \cos ^{2} \alpha - \sin ^{2} \alpha$ . Replacing $\sin ^{2} \alpha$ or $\cos ^{2} \alpha$ with the Pythagorean identity (Part A) gives two more forms of $\cos 2\alpha$ .

\sin 2\theta=2\sin\theta\cos\theta\qquad \cos 2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1=1-2\sin^2\theta

COMMON MISTAKE

" $\sin$ $2\theta = 2 \sin \theta$ ."

Doubling the angle does not double the output. At $\theta = \pi /2$ : $\sin$ $2\theta = \sin \pi = 0$ , but $2 \sin (\pi /2) = 2$ . Sine can never be bigger than 1, so $2 \sin \theta$ cannot always equal a sine value.

Worked example

Example 3. $\cos \theta = -4/5$ and $\theta$ is in Quadrant III. Find $\sin 2\theta$ and $\cos 2\theta$ .

01
First find $\sin \theta$ (Part A): $\sin ^{2} \theta = 1 - 16/25 = 9/25$ , so $\sin \theta = \pm 3/5$ . In Quadrant III sine is negative: $\sin \theta = -3/5$ .
02
$\sin$ $2\theta = 2 \sin \theta \cos \theta = 2(-3/5)(-4/5) = 24/25$ .
03
$\cos$ $2\theta = \cos ^{2} \theta - \sin ^{2} \theta = 16/25 - 9/25 = 7/25$ .
04
Sense check: $\theta$ is between $\pi$ and $3\pi /2$ , so $2\theta$ is between $2\pi$ and $3\pi$ — the same position as an angle between 0 and $\pi$ , where sine is positive ✓. And $(24/25)^{2} + (7/25)^{2} = 1$ ✓.

KEY RULE

Choose the form of $\cos 2\theta$ that matches the rest of the equation: $1 - 2\sin ^{2}\theta$ with sines, $2\cos ^{2}\theta - 1$ with cosines.

Quick check

If $\cos\theta = \frac{3}{5}$ , what is $\cos 2\theta$ ?

Solving Equations with Double Angles

If an equation contains both $2\theta$ and $\theta$ , use a double-angle identity so every term uses $\theta$ . Then factor.

Worked example

Example 4. Solve $\sin$ $2\theta = \sqrt{3} \cos \theta$ for $0 \le \theta < 2\pi$ .

01
Replace $\sin 2\theta$ : $2 \sin \theta \cos \theta = \sqrt{3} \cos \theta$ . Move everything to one side: $2 \sin \theta \cos \theta - \sqrt{3} \cos \theta = 0$ .
02
Factor out $\cos \theta$ — do NOT divide by it (3.10A): $\cos \theta$ $(2 \sin \theta - \sqrt{3}) = 0$ .
03
$\cos \theta = 0$ gives $\theta = \pi /2$ and $3\pi /2$ . $\sin \theta = \sqrt{3}/2$ gives $\theta = \pi /3$ and $2\pi /3$ .
04
Solutions: $\pi /3$ , $\pi /2$ , $2\pi /3$ , and $3\pi /2$ . Dividing by $\cos \theta$ would have lost $\pi /2$ and $3\pi /2$ .

$y = \sin 2\theta$ and $y = \sqrt{3} \cos \theta$ meet four times in $[0, 2\pi )$ .

Worked example

Example 5. Solve $\cos$ $2\theta + 5 \cos \theta + 3 = 0$ for $0 \le \theta < 2\pi$ .

01
The other term uses $\cos \theta$ , so choose $\cos$ $2\theta = 2 \cos ^{2} \theta - 1$ : $2 \cos ^{2} \theta - 1 + 5 \cos \theta + 3 = 0$ .
02
Simplify: $2 \cos ^{2} \theta + 5 \cos \theta + 2 = 0$ , which factors as $(2 \cos \theta + 1)(\cos \theta + 2) = 0$ .
03
$\cos \theta = -1/2$ gives $\theta = 2\pi /3$ and $4\pi /3$ . $\cos \theta = -2$ has NO solution, because cosine is always between −1 and 1. Reject it.
04
Solutions: $\theta = 2\pi /3$ and $4\pi /3$ .

Common slips

" $\cos (\pi /3 + \pi /4) = \cos (\pi /3)\cos (\pi /4) + \sin (\pi /3)\sin (\pi /4)$ ."

The cosine identity flips the sign. With + you get $(\sqrt{2} + \sqrt{6})/4 \approx 0.966$ , a positive number — impossible for an angle in Quadrant ii. A quick quadrant check catches a wrong sign every time.
" $\sin$ $2\theta = 2 \sin \theta$ ."

Doubling the angle does not double the output. At $\theta = \pi /2$ : $\sin$ $2\theta = \sin \pi = 0$ , but $2 \sin (\pi /2) = 2$ . Sine can never be bigger than 1, so $2 \sin \theta$ cannot always equal a sine value.

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6 lines to re-read the night before.

01
$\sin (\alpha + \beta ) \ne \sin \alpha + \sin \beta$ . Use $\sin (\alpha \pm \beta ) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta$ and $\cos (\alpha \pm \beta ) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta$ .
02
Split angles like $5\pi /12$ and $7\pi /12$ into sums of special angles to find exact values, then check the sign with the quadrant.
03
Sum identities turn shifts into simpler forms, for example $\sin (\theta + \pi /2) = \cos \theta$ and $\sin (\theta - \pi ) = -\sin \theta$ .
04
$\sin$ $2\theta = 2 \sin \theta \cos \theta$ ; $\cos$ $2\theta = \cos ^{2}\theta - \sin ^{2}\theta = 2\cos ^{2}\theta - 1 = 1 - 2\sin ^{2}\theta$ .
05
In equations, rewrite so every term uses the same angle, factor, and reject impossible values like $\cos \theta = -2$ .
06
Coming up in 3.13: describing points with a distance and an angle — polar coordinates.

Practice 33 problems Next: Chapter 3.13A Trigonometry and Polar Coordinates

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Equivalent Representations of Trigonometric Functions

Sine and Cosine Do Not Distribute

Rewriting Shifted Functions

Double-Angle Identities

Solving Equations with Double Angles

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