Chapter 9 — Trigonometric Identities and Equations
Identities are the algebra of trigonometry. Mastering them lets you simplify, rewrite, and solve trigonometric expressions and equations. This chapter builds from verifying known identities to discovering powerful formulas for sums, differences, double angles, and half angles — then applies them to solve trig equations.
Table of Contents
1 — Verifying Trigonometric Identities
2 — Sum and Difference Identities
- 2.1 Cosine Sum and Difference
- 2.2 Sine Sum and Difference
- 2.3 Tangent Sum and Difference
- 2.4 Cofunction Identities
3 — Double-Angle, Half-Angle, and Reduction Formulas
4 — Sum-to-Product and Product-to-Sum Formulas
5 — Solving Trigonometric Equations
Glossary
| Term | Definition |
|---|---|
| Identity | An equation true for all values in the domain of the variable |
| Conditional equation | An equation true only for particular values of the variable |
| Verify | Show both sides of an identity are equivalent by algebraic manipulation |
| Double-angle formula | Identity expressing $\sin 2\theta$, $\cos 2\theta$, $\tan 2\theta$ in terms of $\theta$ |
| Half-angle formula | Identity expressing $\sin(\theta/2)$, $\cos(\theta/2)$ in terms of $\theta$ |
| Power-reduction | Rewriting $\sin^2\theta$ or $\cos^2\theta$ in terms of first-power cosine |
1 — Verifying Trigonometric Identities
1.1 Fundamental Identities Review
Reciprocal Identities:
\[\csc\theta = \frac{1}{\sin\theta}, \qquad \sec\theta = \frac{1}{\cos\theta}, \qquad \cot\theta = \frac{1}{\tan\theta}\]Quotient Identities:
\[\tan\theta = \frac{\sin\theta}{\cos\theta}, \qquad \cot\theta = \frac{\cos\theta}{\sin\theta}\]Pythagorean Identities:
\(\sin^2\theta + \cos^2\theta = 1\) \(1 + \tan^2\theta = \sec^2\theta\) \(1 + \cot^2\theta = \csc^2\theta\)
Even-Odd Identities:
\(\cos(-\theta) = \cos\theta, \qquad \sec(-\theta) = \sec\theta\) \(\sin(-\theta) = -\sin\theta, \qquad \csc(-\theta) = -\csc\theta\) \(\tan(-\theta) = -\tan\theta, \qquad \cot(-\theta) = -\cot\theta\)
1.2 Strategies for Verification
Verification Playbook:
- Work with one side only — usually the more complex side.
- Convert everything to sines and cosines.
- Factor where possible.
- Combine fractions using a common denominator.
- Multiply by a conjugate to clear $1 \pm \sin\theta$ or $1 \pm \cos\theta$.
- Use Pythagorean substitutions ($\sin^2\theta = 1 - \cos^2\theta$, etc.).
- Split a fraction: $\dfrac{A + B}{C} = \dfrac{A}{C} + \dfrac{B}{C}$.
Never treat an identity like an equation — don’t add/subtract the same thing to both sides. You must transform one side into the other independently.
1.3 Worked Verifications
Verify: $\dfrac{\sin\theta}{1 - \cos\theta} = \dfrac{1 + \cos\theta}{\sin\theta}$
LHS: Multiply numerator and denominator by the conjugate $(1 + \cos\theta)$:
\(\frac{\sin\theta}{1 - \cos\theta} \cdot \frac{1 + \cos\theta}{1 + \cos\theta} = \frac{\sin\theta(1 + \cos\theta)}{1 - \cos^2\theta} = \frac{\sin\theta(1 + \cos\theta)}{\sin^2\theta} = \frac{1 + \cos\theta}{\sin\theta} = \text{RHS} \checkmark\)
Verify: $\sec\theta - \cos\theta = \sin\theta\tan\theta$
LHS: $\dfrac{1}{\cos\theta} - \cos\theta = \dfrac{1 - \cos^2\theta}{\cos\theta} = \dfrac{\sin^2\theta}{\cos\theta} = \sin\theta \cdot \dfrac{\sin\theta}{\cos\theta} = \sin\theta\tan\theta = \text{RHS} \checkmark$
Verify: $\dfrac{\tan\theta}{\sec\theta} = \sin\theta$
LHS: $\dfrac{\sin\theta/\cos\theta}{1/\cos\theta} = \sin\theta = \text{RHS} \checkmark$
2 — Sum and Difference Identities
2.1 Cosine Sum and Difference
\(\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta\)
Find the exact value of $\cos 75°$.
$\cos 75° = \cos(45° + 30°) = \cos 45°\cos 30° - \sin 45°\sin 30°$
$= \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2} = \dfrac{\sqrt{6} - \sqrt{2}}{4}$
2.2 Sine Sum and Difference
\(\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta\)
Find the exact value of $\sin\dfrac{7\pi}{12}$.
$\sin\dfrac{7\pi}{12} = \sin!\left(\dfrac{\pi}{3} + \dfrac{\pi}{4}\right) = \sin\dfrac{\pi}{3}\cos\dfrac{\pi}{4} + \cos\dfrac{\pi}{3}\sin\dfrac{\pi}{4}$
$= \dfrac{\sqrt{3}}{2} \cdot \dfrac{\sqrt{2}}{2} + \dfrac{1}{2} \cdot \dfrac{\sqrt{2}}{2} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$
2.3 Tangent Sum and Difference
\(\tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}\)
2.4 Cofunction Identities
These follow directly from the sum/difference formulas:
\(\sec\!\left(\frac{\pi}{2} - \theta\right) = \csc\theta \qquad \csc\!\left(\frac{\pi}{2} - \theta\right) = \sec\theta\)
3 — Double-Angle, Half-Angle, and Reduction Formulas
3.1 Double-Angle Formulas
Sine:
\[\sin 2\theta = 2\sin\theta\cos\theta\]Cosine (three equivalent forms):
\[\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta\]Tangent:
\(\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}\)
Given $\sin\theta = \dfrac{3}{5}$ with $\theta$ in Q I, find $\sin 2\theta$, $\cos 2\theta$, $\tan 2\theta$.
$\cos\theta = \dfrac{4}{5}$ (Q I positive).
$\sin 2\theta = 2 \cdot \dfrac{3}{5} \cdot \dfrac{4}{5} = \dfrac{24}{25}$
$\cos 2\theta = \left(\dfrac{4}{5}\right)^2 - \left(\dfrac{3}{5}\right)^2 = \dfrac{16 - 9}{25} = \dfrac{7}{25}$
$\tan 2\theta = \dfrac{24/25}{7/25} = \dfrac{24}{7}$
3.2 Half-Angle Formulas
The $\pm$ sign depends on the quadrant in which $\theta/2$ lies.
Find the exact value of $\cos 15°$.
$\cos 15° = \cos\dfrac{30°}{2} = \sqrt{\dfrac{1 + \cos 30°}{2}} = \sqrt{\dfrac{1 + \frac{\sqrt{3}}{2}}{2}} = \sqrt{\dfrac{2 + \sqrt{3}}{4}} = \dfrac{\sqrt{2 + \sqrt{3}}}{2}$
(Positive because $15°$ is in Q I.)
3.3 Power-Reduction Formulas
Derived from the double-angle cosine formulas:
\(\tan^2\theta = \frac{1 - \cos 2\theta}{1 + \cos 2\theta}\)
Power-reduction is essential in calculus for integrating even powers of trig functions.
4 — Sum-to-Product and Product-to-Sum Formulas
4.1 Product-to-Sum Formulas
\(\sin\alpha\cos\beta = \frac{1}{2}[\sin(\alpha + \beta) + \sin(\alpha - \beta)]\)
Write $\sin 3x\cos 5x$ as a sum.
$\sin 3x\cos 5x = \dfrac{1}{2}[\sin(3x + 5x) + \sin(3x - 5x)] = \dfrac{1}{2}[\sin 8x + \sin(-2x)] = \dfrac{1}{2}[\sin 8x - \sin 2x]$
4.2 Sum-to-Product Formulas
\(\cos\alpha - \cos\beta = -2\sin\!\left(\frac{\alpha + \beta}{2}\right)\sin\!\left(\frac{\alpha - \beta}{2}\right)\)
5 — Solving Trigonometric Equations
5.1 Linear Trig Equations
A linear trig equation has a single trig function to the first power.
Strategy: Isolate the trig function, then use known angles (unit circle, inverse trig).
Solve $2\sin\theta - 1 = 0$ on $[0, 2\pi)$.
$\sin\theta = \dfrac{1}{2}$
$\theta = \dfrac{\pi}{6}$ (Q I) or $\theta = \dfrac{5\pi}{6}$ (Q II)
General solution: $\theta = \dfrac{\pi}{6} + 2n\pi$ or $\theta = \dfrac{5\pi}{6} + 2n\pi$, $n \in \mathbb{Z}$
5.2 Quadratic-Type Trig Equations
If the equation looks like $a\sin^2\theta + b\sin\theta + c = 0$, treat $\sin\theta$ (or $\cos\theta$) as a variable and factor or use the quadratic formula.
Solve $2\cos^2\theta + \cos\theta - 1 = 0$ on $[0, 2\pi)$.
Let $u = \cos\theta$:
$2u^2 + u - 1 = 0 \Rightarrow (2u - 1)(u + 1) = 0$
$u = \dfrac{1}{2}$ or $u = -1$
$\cos\theta = \dfrac{1}{2} \Rightarrow \theta = \dfrac{\pi}{3}, \dfrac{5\pi}{3}$
$\cos\theta = -1 \Rightarrow \theta = \pi$
Solutions: $\theta = \dfrac{\pi}{3}, \pi, \dfrac{5\pi}{3}$
5.3 Equations Using Identities
Sometimes you need to apply an identity to reduce an equation to one trig function.
Solve $\sin 2\theta = \cos\theta$ on $[0, 2\pi)$.
$2\sin\theta\cos\theta = \cos\theta$
$2\sin\theta\cos\theta - \cos\theta = 0$
$\cos\theta(2\sin\theta - 1) = 0$
$\cos\theta = 0 \Rightarrow \theta = \dfrac{\pi}{2}, \dfrac{3\pi}{2}$
$\sin\theta = \dfrac{1}{2} \Rightarrow \theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}$
Solutions: $\theta = \dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}$
Never divide both sides by a trig function — you lose solutions where that function equals zero. Always factor instead.
5.4 Equations with Multiple Angles
Solve $\sin 3\theta = \dfrac{\sqrt{3}}{2}$ on $[0, 2\pi)$.
Let $u = 3\theta$, so $\sin u = \dfrac{\sqrt{3}}{2}$.
$u = \dfrac{\pi}{3} + 2n\pi$ or $u = \dfrac{2\pi}{3} + 2n\pi$
Since $\theta \in [0, 2\pi)$, we need $u \in [0, 6\pi)$. So $n = 0, 1, 2$:
$3\theta = \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \dfrac{7\pi}{3}, \dfrac{8\pi}{3}, \dfrac{13\pi}{3}, \dfrac{14\pi}{3}$
$\theta = \dfrac{\pi}{9}, \dfrac{2\pi}{9}, \dfrac{7\pi}{9}, \dfrac{8\pi}{9}, \dfrac{13\pi}{9}, \dfrac{14\pi}{9}$
Key Takeaways
- Verification = transform one side to match the other. Never cross the equals sign.
- Sum/Difference formulas let you find exact values for non-standard angles (e.g., $15°, 75°, 105°$).
- Double-angle formulas: $\sin 2\theta = 2\sin\theta\cos\theta$; cosine has three forms.
- Half-angle formulas use $\pm$ sign based on the quadrant of $\theta/2$.
- Power-reduction eliminates squares: $\sin^2\theta = (1 - \cos 2\theta)/2$.
- Product-to-sum and sum-to-product convert between forms — crucial in signal processing and Fourier analysis.
- Solving trig equations: Factor (never divide by a trig function), use the unit circle for all solutions, and account for multiple-angle arguments by expanding the solution interval.
Practice Questions
Q1. Verify: $\csc\theta - \sin\theta = \cos\theta\cot\theta$.
Answer:
LHS $= \dfrac{1}{\sin\theta} - \sin\theta = \dfrac{1 - \sin^2\theta}{\sin\theta} = \dfrac{\cos^2\theta}{\sin\theta} = \cos\theta \cdot \dfrac{\cos\theta}{\sin\theta} = \cos\theta\cot\theta = \text{RHS}$ $\checkmark$
Q2. Find the exact value of $\sin 105°$.
Answer:
$\sin 105° = \sin(60° + 45°) = \sin 60°\cos 45° + \cos 60°\sin 45°$
$= \dfrac{\sqrt{3}}{2} \cdot \dfrac{\sqrt{2}}{2} + \dfrac{1}{2} \cdot \dfrac{\sqrt{2}}{2} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$
Q3. Given $\cos\theta = -\dfrac{12}{13}$ with $\theta$ in Q III, find $\sin 2\theta$.
Answer:
Q III: $\sin\theta < 0$. $\sin\theta = -\sqrt{1 - \frac{144}{169}} = -\dfrac{5}{13}$.
$\sin 2\theta = 2\sin\theta\cos\theta = 2!\left(-\dfrac{5}{13}\right)!\left(-\dfrac{12}{13}\right) = \dfrac{120}{169}$
Q4. Find the exact value of $\tan\dfrac{\pi}{8}$ using a half-angle formula.
Answer:
$\tan\dfrac{\pi}{8} = \tan\dfrac{\pi/4}{2} = \dfrac{1 - \cos(\pi/4)}{\sin(\pi/4)} = \dfrac{1 - \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \dfrac{2 - \sqrt{2}}{\sqrt{2}} = \dfrac{(2 - \sqrt{2})\sqrt{2}}{2} = \dfrac{2\sqrt{2} - 2}{2} = \sqrt{2} - 1$
Q5. Solve $2\sin^2\theta - \sin\theta - 1 = 0$ on $[0, 2\pi)$.
Answer:
Let $u = \sin\theta$: $2u^2 - u - 1 = 0 \Rightarrow (2u + 1)(u - 1) = 0$
$u = -\dfrac{1}{2}$ or $u = 1$.
$\sin\theta = -\dfrac{1}{2} \Rightarrow \theta = \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$
$\sin\theta = 1 \Rightarrow \theta = \dfrac{\pi}{2}$
Solutions: $\theta = \dfrac{\pi}{2}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$
Q6. Solve $\cos 2\theta = \sin\theta$ on $[0, 2\pi)$.
Answer:
$1 - 2\sin^2\theta = \sin\theta$
$2\sin^2\theta + \sin\theta - 1 = 0$
$(2\sin\theta - 1)(\sin\theta + 1) = 0$
$\sin\theta = \dfrac{1}{2} \Rightarrow \theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}$
$\sin\theta = -1 \Rightarrow \theta = \dfrac{3\pi}{2}$
Solutions: $\theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}$