Sections 5.1 – 5.3 Review

Trig Identities & Equations

Review: Simplifying with identities, verifying identities, and solving trig equations

Fundamental Identities Verifying Solving Equations

Section 5.1

Using Fundamental Identities

Reciprocal, Quotient, Pythagorean, Cofunction, Even/Odd

Reference

Fundamental Identities

Pythagorean Identities

\[\sin^2\theta + \cos^2\theta = 1 \qquad 1 + \tan^2\theta = \sec^2\theta \qquad 1 + \cot^2\theta = \csc^2\theta\]

Reciprocal Identities

\[\csc\theta = \frac{1}{\sin\theta} \qquad \sec\theta = \frac{1}{\cos\theta} \qquad \cot\theta = \frac{1}{\tan\theta}\]

Quotient Identities

\[\tan\theta = \frac{\sin\theta}{\cos\theta} \qquad \cot\theta = \frac{\cos\theta}{\sin\theta}\]

Even / Odd Identities

\[\cos(-\theta) = \cos\theta \qquad \sin(-\theta) = -\sin\theta \qquad \tan(-\theta) = -\tan\theta\]

Strategies

Simplifying Trig Expressions

1

Factor out a common term

2

Apply a Pythagorean identity

3

Rewrite in sine and cosine

4

Find a common denominator

5

Factor a difference of squares

6

Multiply by the conjugate

Goal

Reduce to fewer trig functions, eliminate fractions, or match a target form.

5.1 Practice

Simplify the expression

\[\csc x - \cos x \cot x\]

Rewrite in sine & cosine

\[\dfrac{1}{\sin x} - \cos x \cdot \dfrac{\cos x}{\sin x}\]

Combine over common denominator

\[\dfrac{1 - \cos^2 x}{\sin x}\]

Pythagorean identity

\[\dfrac{\sin^2 x}{\sin x}\]

Since \(\sin^2 x + \cos^2 x = 1\),
we have \(1 - \cos^2 x = \sin^2 x\)

Simplified

\[\sin x\]

5.1 Practice

Factor the expression

\[\sec^2 x + \tan x - 3\]

Pythagorean substitution

\[(1 + \tan^2 x) + \tan x - 3\]

Replace \(\sec^2 x\) with
\(1 + \tan^2 x\)

Simplify

\[\tan^2 x + \tan x - 2\]

Factor as a quadratic in tan x

\[(\tan x + 2)(\tan x - 1)\]

Think: \(u^2 + u - 2 = (u+2)(u-1)\)

Factored form

\[(\tan x + 2)(\tan x - 1)\]

Section 5.2

Verifying Trig Identities

Work one side at a time — never cross the equal sign

Guidelines

Verifying Identities

1

Work with the more complicated side

2

Convert everything to sine and cosine

3

Factor expressions

4

Find a common denominator

5

Multiply by the conjugate

6

Apply Pythagorean identities

✗

Never cross the equal sign — do not add, multiply, or square both sides

5.2 Practice

Verify the identity

\[\dfrac{1}{1 - \sin x} + \dfrac{1}{1 + \sin x} = 2\sec^2 x\]

Work the left side — common denominator

\[\dfrac{(1 + \sin x) + (1 - \sin x)}{(1 - \sin x)(1 + \sin x)}\]

Simplify numerator and denominator

\[\dfrac{2}{1 - \sin^2 x}\]

Numerator: \(\sin x\) cancels
Denominator: difference of squares

Pythagorean identity

\[\dfrac{2}{\cos^2 x}\]

Equals the right side ✓

\[2\sec^2 x \;\; \checkmark\]

5.2 Practice

Verify the identity

\[\dfrac{\csc^2 x - 1}{\csc^2 x} = \cos^2 x\]

Work the left side — split the fraction

\[\dfrac{\csc^2 x}{\csc^2 x} - \dfrac{1}{\csc^2 x}\]

Simplify each fraction

\[1 - \dfrac{1}{\csc^2 x}\]

Reciprocal identity

\[1 - \sin^2 x\]

Since \(\dfrac{1}{\csc x} = \sin x\)

Pythagorean identity

\[\cos^2 x\]

Equals the right side ✓

\[\cos^2 x \;\; \checkmark\]

Section 5.3

Solving Trig Equations

Restricted domains and general solutions

Key Ideas

Solving Trig Equations

1

Isolate the trig function (like solving for \(x\) in algebra)

2

Factor — never divide both sides by a trig function

3

Use Pythagorean identities to rewrite in one trig function

4

Use the unit circle for exact values

Restricted vs. General Solutions

Restricted \([0, 2\pi)\): list only angles in the interval. General: add \(+ 2\pi n\) (or \(+ \pi n\) for tan/cot).

Caution

Never divide both sides by a trig function — you'll lose solutions! Factor instead.

5.3 Practice

Solve in \([0, 2\pi)\)

\[2\sin^2\theta + \sin\theta - 1 = 0\]

Factor as a quadratic in sin θ

\[(2\sin\theta - 1)(\sin\theta + 1) = 0\]

Let \(u = \sin\theta\):
\(2u^2 + u - 1 = (2u-1)(u+1)\)

Set each factor equal to zero

\[2\sin\theta - 1 = 0 \qquad \text{or} \qquad \sin\theta + 1 = 0\]

Solve the first equation

\[\sin\theta = \dfrac{1}{2} \quad \Longrightarrow \quad \theta = \dfrac{\pi}{6},\; \dfrac{5\pi}{6}\]

Sine is \(\tfrac{1}{2}\)
in QI and QII

Solve the second equation

\[\sin\theta = -1 \quad \Longrightarrow \quad \theta = \dfrac{3\pi}{2}\]

Solution set

\[\theta = \dfrac{\pi}{6},\; \dfrac{5\pi}{6},\; \dfrac{3\pi}{2}\]

5.3 Practice

Solve for all solutions

\[\cos x \tan x = \cos x\]

Do NOT divide by cos x — factor instead!

\[\cos x \tan x - \cos x = 0\]

Factor out cos x

\[\cos x(\tan x - 1) = 0\]

First factor

\[\cos x = 0 \quad \Longrightarrow \quad x = \dfrac{\pi}{2} + \pi n\]

Second factor

\[\tan x = 1 \quad \Longrightarrow \quad x = \dfrac{\pi}{4} + \pi n\]

General solution

\[x = \dfrac{\pi}{2} + \pi n \quad \text{or} \quad x = \dfrac{\pi}{4} + \pi n\]

Both families are needed —
neither contains the other

5.3 Practice

Solve in \([0, 2\pi)\)

\[2\cos^2 x - 5\sin x + 1 = 0\]

Pythagorean substitution

\[2(1 - \sin^2 x) - 5\sin x + 1 = 0\]

Replace \(\cos^2 x\) with
\(1 - \sin^2 x\)

Expand and simplify

\[-2\sin^2 x - 5\sin x + 3 = 0 \quad \Longrightarrow \quad 2\sin^2 x + 5\sin x - 3 = 0\]

Factor

\[(2\sin x - 1)(\sin x + 3) = 0\]

Solve each factor

\[\sin x = \dfrac{1}{2} \;\Rightarrow\; x = \dfrac{\pi}{6},\, \dfrac{5\pi}{6} \qquad \sin x = -3 \;\Rightarrow\; \text{No solution}\]

\(\sin x = -3\) is impossible
since \(-1 \le \sin x \le 1\)

Solution set

\[x = \dfrac{\pi}{6},\; \dfrac{5\pi}{6}\]