Unit 3 · Topic 3.12

Equivalent Representations of Trigonometric Functions

The Pythagorean identity sin² x + cos² x = 1 and its forms with tan , sec , cot and csc ; the sum and difference identities for sine and cosine; and the double-angle identities sin (2x) = 2sin xcos x and cos (2x) = cos² x − sin² x = 2cos² x − 1 = 1 − 2sin² x .

Practice 1: Procedural and Symbolic Fluency

Part 1 · Hook

Why this matters

Two speakers play the same note, one a little behind the other. Your ear hears a single wave, louder or softer depending on the delay. The math behind that is a trigonometric identity: a sum of two sine waves can be rewritten as one. Identities let you trade a messy expression for a simpler one with the same graph.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is $\sin^2\frac{\pi}{6} + \cos^2\frac{\pi}{6}$ ?

$1$
$\frac{1}{2}$
$\frac{\sqrt{3} + 1}{2}$
$2$

Show the answer

$\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$ $= \frac{1}{4} + \frac{3}{4}$ $= 1$ : the point is on the unit circle.

Correct: $1$ :
$\frac{1}{2}$ :
$\frac{\sqrt{3} + 1}{2}$ :
$2$ :

2. Which equals $\csc x$ ?

$\frac{1}{\sin x}$
$\frac{1}{\cos x}$
$\frac{\cos x}{\sin x}$
$\sin^{-1} x$

Show the answer

Cosecant is the reciprocal of sine.

Correct: $\frac{1}{\sin x}$ :
$\frac{1}{\cos x}$ :
$\frac{\cos x}{\sin x}$ :
$\sin^{-1} x$ :

Part 4 · See it

See it first

Graphs of cos squared x and sin squared x (dashed) and of cos(2x) (solid) for x from 0 to 2 pi. At every x the solid height equals the first dashed height minus the second. — At every $x$ , the height of $\cos(2x)$ equals $\cos^2 x$ minus $\sin^2 x$ : the double-angle identity as a picture. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

A point $(\cos x, \sin x)$ is on the unit circle $\sin^2 x + \cos^2 x = 1$ , the Pythagorean identity
Divide it by $\cos^2 x$ or by $\sin^2 x$ $\tan^2 x + 1 = \sec^2 x$ and $1 + \cot^2 x = \csc^2 x$
Add two angles $\sin(a + b)$ $= \sin a\cos b$ $+ \cos a\sin b$ and $\cos(a + b)$ $= \cos a\cos b - \sin a\sin b$
Set $b = a$ in the sum identities $\sin(2a) = 2\sin a\cos a$ and $\cos(2a)$ $= \cos^2 a - \sin^2 a$

Part 6 · Key ideas

Key ideas

An identity is true for every input where both sides are defined; you use it to swap one form for another.
$\sin(a + b)$ is not $\sin a + \sin b$ ; use the sum identity.
$\cos(2x)$ has three forms: $\cos^2 x - \sin^2 x$ , $2\cos^2 x - 1$ and $1 - 2\sin^2 x$ . Pick the one that matches the rest of the expression.
To rewrite as a single trigonometric function, write everything in sines and cosines, simplify, and stop at one function.

Part 7 · Misconception

A common mistake

The wrong idea: $\sin(2x) = 2\sin x$ , because the $2$ can come out front.

What actually happens: The $2$ is part of the angle. $\sin(2x) = 2\sin x\cos x$ ; for example $\sin\pi = 0$ while $2\sin\frac{\pi}{2} = 2$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of $f(x) = 1 - 2\sin^2 x$

1. The graph shows $f(x) = 1 - 2\sin^2 x$ . Which function has the same graph?

$y = \cos(2x)$
$y = \cos^2 x$
$y = \sin(2x)$
$y = 2\cos x$

Show the answer

One form of the double-angle identity is $\cos(2x) = 1 - 2\sin^2 x$ . The graph agrees: it starts at $1$ , reaches $-1$ at $x = \frac{\pi}{2}$ and repeats every $\pi$ .

Correct: $y = \cos(2x)$ : Right: $1 - 2\sin^2 x$ is one of the three forms of $\cos(2x)$ .
$y = \cos^2 x$ : $\cos^2 x$ is never negative, but the graph goes down to $-1$ .
$y = \sin(2x)$ : $\sin(2x)$ starts at $0$ ; the graph starts at $1$ .
$y = 2\cos x$ : $2\cos x$ reaches $2$ and has period $2\pi$ ; the graph stays between $-1$ and $1$ and has period $\pi$ .

2. For the function in the graph, find the exact value of $f\left(\frac{\pi}{6}\right)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$f\left(\frac{\pi}{6}\right)$ $= 1 - 2\left(\frac{1}{2}\right)^2$ $= 1 - \frac{1}{2}$ $= \frac{1}{2}$ . Check with the identity: $\cos\left(2\cdot\frac{\pi}{6}\right)$ $= \cos\frac{\pi}{3}$ $= \frac{1}{2}$ .

Answer: $\frac{1}{2}$ ≈ 0.500

3. For the function in the graph, find every solution of $f(x) = 0$ on $[0, 2\pi)$ .

$\frac{\pi}{4}$ , $\frac{3\pi}{4}$ , $\frac{5\pi}{4}$ and $\frac{7\pi}{4}$
$\frac{\pi}{4}$ and $\frac{3\pi}{4}$
$\frac{\pi}{2}$ and $\frac{3\pi}{2}$
$\frac{\pi}{4}$ , $\frac{3\pi}{4}$ , $\frac{5\pi}{4}$ , $\frac{7\pi}{4}$ and $\frac{9\pi}{4}$

Show the answer

$f(x) = \cos(2x) = 0$ . With $u = 2x$ in $[0, 4\pi)$ : $u$ $= \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$ , so $x$ $= \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ . The graph crosses the axis four times.

Correct: $\frac{\pi}{4}$ , $\frac{3\pi}{4}$ , $\frac{5\pi}{4}$ and $\frac{7\pi}{4}$ : Right: two periods of length $\pi$ , two zeros each.
$\frac{\pi}{4}$ and $\frac{3\pi}{4}$ : That is one period; $[0, 2\pi)$ holds two periods of $f$ .
$\frac{\pi}{2}$ and $\frac{3\pi}{2}$ : These are values of $2x$ ; divide by $2$ .
$\frac{\pi}{4}$ , $\frac{3\pi}{4}$ , $\frac{5\pi}{4}$ , $\frac{7\pi}{4}$ and $\frac{9\pi}{4}$ : $\frac{9\pi}{4}$ is outside $[0, 2\pi)$ .

Table

Two angles $\alpha$ and $\beta$ in Quadrant I

Sine and cosine of two angles α and β, each between 0 and π/2
Angle	Sine	Cosine
$\alpha$	$\frac{3}{5}$	$\frac{4}{5}$
$\beta$	$\frac{12}{13}$	$\frac{5}{13}$

4. Using the table, find the exact value of $\sin(\alpha + \beta)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\sin(\alpha + \beta)$ $= \sin\alpha\cos\beta$ $+ \cos\alpha\sin\beta$ $= \frac{3}{5}\cdot\frac{5}{13}$ $+ \frac{4}{5}\cdot\frac{12}{13}$ $= \frac{15}{65}$ $+ \frac{48}{65}$ $= \frac{63}{65}$ .

Answer: $\frac{63}{65}$ ≈ 0.969

5. Rewrite $\frac{\sin(2x)}{2\cos^2 x}$ as a single trigonometric function of $x$ .

Type an expression for $\frac{\sin(2x)}{2\cos^2 x} =$ .

Show the answer

Use $\sin(2x) = 2\sin x\cos x$ : $\frac{2\sin x\cos x}{2\cos^2 x}$ $= \frac{\sin x}{\cos x}$ $= \tan x$ . Stop only when one trigonometric function is left.

Answer: $\frac{\sin(2x)}{2\cos^2 x} =$ $\tan(x)$

6. Find every solution of $\sin(2x) = \sin x$ on $[0, 2\pi)$ .

$0$ , $\frac{\pi}{3}$ , $\pi$ and $\frac{5\pi}{3}$
$\frac{\pi}{3}$ and $\frac{5\pi}{3}$
$0$ and $\pi$
$0$ , $\frac{\pi}{3}$ , $\frac{2\pi}{3}$ and $\pi$

Show the answer

$2\sin x\cos x = \sin x$ , so $2\sin x\cos x - \sin x = 0$ and $\sin x(2\cos x - 1) = 0$ . Either $\sin x = 0$ ( $x = 0, \pi$ ) or $\cos x = \frac{1}{2}$ ( $x = \frac{\pi}{3}, \frac{5\pi}{3}$ ).

Correct: $0$ , $\frac{\pi}{3}$ , $\pi$ and $\frac{5\pi}{3}$ : Right: factor out $\sin x$ and solve both factors.
$\frac{\pi}{3}$ and $\frac{5\pi}{3}$ : Dividing by $\sin x$ loses $x = 0$ and $x = \pi$ , where $\sin x = 0$ .
$0$ and $\pi$ : These come from $\sin x = 0$ alone; the factor $2\cos x - 1$ gives two more.
$0$ , $\frac{\pi}{3}$ , $\frac{2\pi}{3}$ and $\pi$ : $\cos\frac{2\pi}{3} = -\frac{1}{2}$ ; the second angle with cosine $\frac{1}{2}$ is $\frac{5\pi}{3}$ .

Part 9 · Summary

Summary

The Pythagorean identity

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

and its forms with

\tan

\sec

\cot

and

\csc

; the sum and difference identities for sine and cosine; and the double-angle identities

\sin(2x) = 2\sin x\cos x

and

\cos(2x)

= \cos^2 x - \sin^2 x

= 2\cos^2 x - 1

= 1 - 2\sin^2 x

. Use them to find exact values, to rewrite expressions as one function, and to solve equations.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The graph of f(x)=1−2sin2⁡xf(x) = 1 - 2\sin^2 x

Two angles α\alpha and β\beta in Quadrant I

Summary

What comes next

Connections

The graph of $f(x) = 1 - 2\sin^2 x$

Two angles $\alpha$ and $\beta$ in Quadrant I