Precalculus · Chapter 7

Analytic Trigonometry

SECTION 7.1

Trigonometric Identities

An equation is a statement that two mathematical expressions are equal. An identity is an equation that is true for all values of the variable(s). A trigonometric identity is an identity involving trigonometric functions.

Equation vs. Identity

An equation such as \( x + 2 = 5 \) is true only for specific values (here \( x = 3 \)). An identity such as \( \sin^2 t + \cos^2 t = 1 \) holds for every value of the variable. To show an equation is not an identity, it is enough to find a single value of the variable for which it fails.

Fundamental Trigonometric Identities

These identities, established in Chapters 5 and 6, are the building blocks for all algebraic work in this chapter.

— Reciprocal Identities

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

— Pythagorean Identities

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

— Even-Odd Identities

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

— Cofunction Identities

\[ \sin\!\left(\tfrac{\pi}{2} - x\right) = \cos x \qquad \tan\!\left(\tfrac{\pi}{2} - x\right) = \cot x \qquad \sec\!\left(\tfrac{\pi}{2} - x\right) = \csc x \] \[ \cos\!\left(\tfrac{\pi}{2} - x\right) = \sin x \qquad \cot\!\left(\tfrac{\pi}{2} - x\right) = \tan x \qquad \csc\!\left(\tfrac{\pi}{2} - x\right) = \sec x \]

Instructor Note

Memorize the three Pythagorean identities cold — almost every simplification and proof in this chapter starts by swapping one of them in. The two non-obvious ones, \( \tan^2 x + 1 = \sec^2 x \) and \( 1 + \cot^2 x = \csc^2 x \), come straight from dividing \( \sin^2 x + \cos^2 x = 1 \) by \( \cos^2 x \) and \( \sin^2 x \) respectively. If you ever forget them, re-derive them that way.

Simplifying Trigonometric Expressions

Identities let us rewrite the same expression in different forms. We use the same algebraic tools as before — factoring, common denominators, special products — combined with the fundamental identities. A common strategy is to rewrite everything in terms of sine and cosine.

EXAMPLE 1

Simplify \( \cos t + \tan t \sin t \).

Solution

Rewrite \( \tan t \) using a reciprocal identity: \( \cos t + \left(\dfrac{\sin t}{\cos t}\right)\sin t \).

Combine over a common denominator: \( \dfrac{\cos^2 t + \sin^2 t}{\cos t} \).

Apply the Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \): \( \dfrac{1}{\cos t} \).

Apply a reciprocal identity: \( \sec t \). \[ \cos t + \tan t \sin t = \sec t \]

EXAMPLE 2

Simplify \( \dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{1 + \sin\theta} \).

Solution

Combine over the common denominator \( \cos\theta\,(1+\sin\theta) \): \( \dfrac{\sin\theta\,(1+\sin\theta) + \cos^2\theta}{\cos\theta\,(1+\sin\theta)} \).

Distribute in the numerator: \( \dfrac{\sin\theta + \sin^2\theta + \cos^2\theta}{\cos\theta\,(1+\sin\theta)} \).

Apply \( \sin^2\theta + \cos^2\theta = 1 \): \( \dfrac{\sin\theta + 1}{\cos\theta\,(1+\sin\theta)} \).

Cancel the common factor \( 1+\sin\theta \) and use a reciprocal identity: \( \dfrac{1}{\cos\theta} = \sec\theta \). \[ \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{1 + \sin\theta} = \sec\theta \]

Proving Trigonometric Identities

To prove an equation is an identity, transform one side into the other through a sequence of steps, each of which is itself an identity. To disprove it, exhibit one value of the variable for which the two sides differ. For example, \( \sin x + \cos x = 1 \) fails at \( x = \pi/4 \), where the left side equals \( \sqrt{2} \neq 1 \).

— Guidelines for Proving Identities

1. **Start with one side** (usually the more complicated one) and aim to transform it into the other. 2. **Use known identities** along with algebra: common denominators, factoring, special products. 3. **Convert to sines and cosines** if you get stuck.

Instructor Note

Warning that costs students marks every year: you may not perform the same operation on both sides of an identity you are trying to prove — that assumes the result. Only reversible work on a single side counts. Squaring is the classic trap: \( \sin x = -\sin x \) is false, but squaring both sides gives a true statement.

EXAMPLE 3

Verify that \( \cos\theta\,(\sec\theta - \cos\theta) = \sin^2\theta \).

Solution

Start with the more complicated left-hand side: \( \cos\theta\,(\sec\theta - \cos\theta) \).

Reciprocal identity \( \sec\theta = \dfrac{1}{\cos\theta} \): \( \cos\theta\!\left(\dfrac{1}{\cos\theta} - \cos\theta\right) \).

Expand: \( 1 - \cos^2\theta \).

Pythagorean identity: \( \sin^2\theta \) = RHS. ∎

EXAMPLE 4

Verify that \( 2\tan x\sec x = \dfrac{1}{1 - \sin x} - \dfrac{1}{1 + \sin x} \).

Solution

Work on the right-hand side.

Common denominator: \( \dfrac{(1+\sin x) - (1-\sin x)}{(1-\sin x)(1+\sin x)} \).

Simplify the numerator: \( \dfrac{2\sin x}{1 - \sin^2 x} \).

Pythagorean identity in the denominator: \( \dfrac{2\sin x}{\cos^2 x} \).

Split and factor: \( 2\,\dfrac{\sin x}{\cos x}\!\left(\dfrac{1}{\cos x}\right) \).

Reciprocal identities: \( 2\tan x\sec x \) = LHS. ∎

EXAMPLE 5

Verify that \( \dfrac{\cos\theta}{1 - \sin\theta} = \sec\theta + \tan\theta \).

Solution

Start with the left-hand side and multiply numerator and denominator by \( 1 + \sin\theta \).

\( \dfrac{\cos\theta}{1-\sin\theta}\cdot\dfrac{1+\sin\theta}{1+\sin\theta} \).

Expand the denominator (difference of squares): \( \dfrac{\cos\theta\,(1+\sin\theta)}{1 - \sin^2\theta} \).

Pythagorean identity: \( \dfrac{\cos\theta\,(1+\sin\theta)}{\cos^2\theta} \).

Cancel \( \cos\theta \): \( \dfrac{1+\sin\theta}{\cos\theta} \).

Separate into two fractions: \( \dfrac{1}{\cos\theta} + \dfrac{\sin\theta}{\cos\theta} \).

Reciprocal identities: \( \sec\theta + \tan\theta \) = RHS. ∎ > 📝 ADMIN NOTE: The "multiply by the conjugate" move in Example 5 is the single most useful trick for identities with \( 1 \pm \sin\theta \) or \( 1 \pm \cos\theta \) in a denominator. Multiplying by \( 1+\sin\theta \) creates \( 1-\sin^2\theta = \cos^2\theta \), which almost always unlocks the problem.

EXAMPLE 6

Verify that \( \dfrac{1 + \cos\theta}{\cos\theta} = \dfrac{\tan^2\theta}{\sec\theta - 1} \).

Solution

Reduce each side independently to the same expression.

LHS: \( \dfrac{1+\cos\theta}{\cos\theta} = \dfrac{1}{\cos\theta} + \dfrac{\cos\theta}{\cos\theta} = \sec\theta + 1 \).

RHS: \( \dfrac{\tan^2\theta}{\sec\theta - 1} = \dfrac{\sec^2\theta - 1}{\sec\theta - 1} = \dfrac{(\sec\theta-1)(\sec\theta+1)}{\sec\theta - 1} = \sec\theta + 1 \).

Both sides equal \( \sec\theta + 1 \), so the equation is an identity. ∎

EXAMPLE 7

Substitute \( \sin\theta \) for \( x \) in \( \sqrt{1 - x^2} \) and simplify, assuming \( 0 \le \theta \le \pi/2 \).

Solution

Substitute \( x = \sin\theta \): \( \sqrt{1 - \sin^2\theta} \).

Pythagorean identity: \( \sqrt{\cos^2\theta} \).

Take the square root: \( \cos\theta \) (valid because \( \cos\theta \ge 0 \) on this interval). \[ \sqrt{1 - x^2} = \cos\theta \] > 📝 ADMIN NOTE: Trig substitution like Example 7 is exactly what you'll use in calculus to integrate expressions with \( \sqrt{a^2 - x^2} \), \( \sqrt{a^2 + x^2} \), and \( \sqrt{x^2 - a^2} \). Tie each radical to the matching Pythagorean identity now and it will pay off later. Figure 7.1.1 — Graphs of \( y = \cos\theta(\sec\theta - \cos\theta) \) and \( y = \sin^2\theta \) coincide, confirming the Example 3 identity. (See the source course, Ch. 7.) ---

SECTION 7.2

Addition and Subtraction Formulas

These formulas express trigonometric functions of a sum or difference of two angles in terms of the functions of the individual angles.

Addition and Subtraction Formulas

The cosine subtraction formula is proved with the distance formula on the unit circle; the rest follow from it together with the even-odd and cofunction identities.

— Formulas for Sine

\[ \sin(s + t) = \sin s\cos t + \cos s\sin t \] \[ \sin(s - t) = \sin s\cos t - \cos s\sin t \]

— Formulas for Cosine

\[ \cos(s + t) = \cos s\cos t - \sin s\sin t \] \[ \cos(s - t) = \cos s\cos t + \sin s\sin t \]

— Formulas for Tangent

\[ \tan(s + t) = \frac{\tan s + \tan t}{1 - \tan s\tan t} \qquad \tan(s - t) = \frac{\tan s - \tan t}{1 + \tan s\tan t} \]

Instructor Note

Watch the signs — they "flip" between the formula and its name. Cosine of a sum uses a minus; cosine of a difference uses a plus. Sine keeps the same sign as the operation. Mixing these up is the most common error on this section's problems.

EXAMPLE 1

Find the exact value of (a) \( \cos 75° \) and (b) \( \cos\dfrac{\pi}{12} \).

Solution

EXAMPLE 2

Find the exact value of \( \sin 20°\cos 40° + \cos 20°\sin 40° \).

Solution

This matches the right side of the sine addition formula with \( s = 20° \), \( t = 40° \). \[ \sin 20°\cos 40° + \cos 20°\sin 40° = \sin(20° + 40°) = \sin 60° = \frac{\sqrt{3}}{2} \]

EXAMPLE 3

Prove \( \cos\!\left(\dfrac{\pi}{2} - u\right) = \sin u \).

Solution

Subtraction formula for cosine: \( \cos\dfrac{\pi}{2}\cos u + \sin\dfrac{\pi}{2}\sin u \).

Substitute \( \cos\dfrac{\pi}{2} = 0 \), \( \sin\dfrac{\pi}{2} = 1 \): \( 0\cdot\cos u + 1\cdot\sin u = \sin u \). ∎

EXAMPLE 4

Verify that \( \dfrac{1 + \tan x}{1 - \tan x} = \tan\!\left(\dfrac{\pi}{4} + x\right) \).

Solution

Start from the right-hand side with the tangent addition formula.

\( \tan\!\left(\dfrac{\pi}{4} + x\right) = \dfrac{\tan\frac{\pi}{4} + \tan x}{1 - \tan\frac{\pi}{4}\tan x} \).

Substitute \( \tan\dfrac{\pi}{4} = 1 \): \( \dfrac{1 + \tan x}{1 - \tan x} \) = LHS. ∎

EXAMPLE 5

If \( f(x) = \sin x \), show that \[ \frac{f(x + h) - f(x)}{h} = \sin x\!\left(\frac{\cos h - 1}{h}\right) + \cos x\!\left(\frac{\sin h}{h}\right). \]

Solution

Definition of \( f \): \( \dfrac{\sin(x + h) - \sin x}{h} \).

Sine addition formula: \( \dfrac{\sin x\cos h + \cos x\sin h - \sin x}{h} \).

Factor \( \sin x \): \( \dfrac{\sin x\,(\cos h - 1) + \cos x\sin h}{h} \).

Separate the fraction: \( \sin x\!\left(\dfrac{\cos h - 1}{h}\right) + \cos x\!\left(\dfrac{\sin h}{h}\right) \). ∎

Evaluating Expressions Involving Inverse Trig Functions

By sketching reference triangles for the inverse-function angles, sums like \( \sin(\cos^{-1}x + \tan^{-1}y) \) can be rewritten as purely algebraic expressions.

EXAMPLE 6

Write \( \sin(\cos^{-1}x + \tan^{-1}y) \) as an algebraic expression in \( x \) and \( y \), with \( -1 \le x \le 1 \).

Solution

Let \( \theta = \cos^{-1}x \) and \( \phi = \tan^{-1}y \), so \( \cos\theta = x \) and \( \tan\phi = y \). From reference triangles: \[ \sin\theta = \sqrt{1 - x^2}, \quad \cos\phi = \frac{1}{\sqrt{1 + y^2}}, \quad \sin\phi = \frac{y}{\sqrt{1 + y^2}} \]

Sine addition formula: \( \sin\theta\cos\phi + \cos\theta\sin\phi \).

Substitute: \( \sqrt{1 - x^2}\cdot\dfrac{1}{\sqrt{1+y^2}} + x\cdot\dfrac{y}{\sqrt{1+y^2}} \).

Factor: \( \dfrac{1}{\sqrt{1+y^2}}\left(\sqrt{1-x^2} + xy\right) \).

EXAMPLE 7

Evaluate \( \sin(\theta + \phi) \) where \( \sin\theta = \tfrac{12}{13} \) with \( \theta \) in Quadrant II, and \( \tan\phi = \tfrac{3}{4} \) with \( \phi \) in Quadrant III.

Solution

For \( \theta \) in QII: \( \cos\theta = -\tfrac{5}{13} \) (negative in QII).

For \( \phi \) in QIII: \( \sin\phi = -\tfrac{3}{5} \), \( \cos\phi = -\tfrac{4}{5} \) (both negative in QIII).

Sine addition formula: \( \sin\theta\cos\phi + \cos\theta\sin\phi = \left(\tfrac{12}{13}\right)\!\left(-\tfrac{4}{5}\right) + \left(-\tfrac{5}{13}\right)\!\left(-\tfrac{3}{5}\right) \).

Simplify: \( -\tfrac{48}{65} + \tfrac{15}{65} = -\tfrac{33}{65} \). > 📝 ADMIN NOTE: For Examples 6 and 7, always sketch the reference triangle and fix the signs from the quadrant before plugging into a formula. Get the quadrant signs right first and the arithmetic takes care of itself.

Expressions of the Form \( A\sin x + B\cos x \)

Any linear combination of \( \sin x \) and \( \cos x \) collapses to a single sine function with a phase shift — the key to reading off amplitude, period, and phase.

— Sums of Sines and Cosines

If \( A \) and \( B \) are real numbers, then \[ A\sin x + B\cos x = k\,\sin(x + \phi) \] where \( k = \sqrt{A^2 + B^2} \) and \( \phi \) satisfies \[ \cos\phi = \frac{A}{\sqrt{A^2 + B^2}}, \qquad \sin\phi = \frac{B}{\sqrt{A^2 + B^2}}. \]

EXAMPLE 8

Express \( 3\sin x + 4\cos x \) in the form \( k\sin(x + \phi) \).

Solution

Amplitude: \( k = \sqrt{3^2 + 4^2} = 5 \).

Phase: \( \sin\phi = \tfrac{4}{5} \), \( \cos\phi = \tfrac{3}{5} \), so \( \phi \) is in Quadrant I and \( \phi = \sin^{-1}\tfrac{4}{5} \approx 53.1° \).

Result: \( 3\sin x + 4\cos x \approx 5\sin(x + 53.1°) \).

EXAMPLE 9

Write \( f(x) = \sin 2x + \sqrt{3}\cos 2x \) in the form \( k\sin(2x + \phi) \) and describe its graph.

Solution

Here \( A = 1 \), \( B = \sqrt{3} \), so \( k = \sqrt{1 + 3} = 2 \).

\( \cos\phi = \tfrac{1}{2} \), \( \sin\phi = \tfrac{\sqrt{3}}{2} \) → \( \phi = \dfrac{\pi}{3} \).

\( f(x) = 2\sin\!\left(2x + \dfrac{\pi}{3}\right) = 2\sin 2\!\left(x + \dfrac{\pi}{6}\right) \).

Read off: amplitude \( 2 \), period \( \dfrac{2\pi}{2} = \pi \), phase shift \( -\dfrac{\pi}{6} \). Figure 7.2.1 — Unit-circle construction (points \( P_0, P_1, Q_0, Q_1 \)) used to prove the cosine addition formula via equal chords. (See the source course, Ch. 7.) Figure 7.2.2 — Graph of \( y = 2\sin 2\!\left(x + \tfrac{\pi}{3}\right) \): a sine curve of amplitude 2, period \( \pi \), phase shift \( \pi/3 \). (See the source course, Ch. 7.) ---

SECTION 7.3

Double-Angle, Half-Angle, and Product-Sum Formulas

All formulas in this section are consequences of the addition formulas from Section 7.2.

Double-Angle Formulas

These give the trigonometric functions at \( 2x \) in terms of the functions at \( x \). The three forms of \( \cos 2x \) are interchangeable via the Pythagorean identity.

— Double-Angle Formulas

\[ \sin 2x = 2\sin x\cos x \] \[ \cos 2x = \cos^2 x - \sin^2 x = 1 - 2\sin^2 x = 2\cos^2 x - 1 \] \[ \tan 2x = \frac{2\tan x}{1 - \tan^2 x} \]

EXAMPLE 1

If \( \cos x = -\tfrac{2}{3} \) and \( x \) is in Quadrant II, find \( \cos 2x \) and \( \sin 2x \).

Solution

\( \cos 2x = 2\cos^2 x - 1 = 2\left(-\tfrac{2}{3}\right)^2 - 1 = \tfrac{8}{9} - 1 = -\tfrac{1}{9} \).

Find \( \sin x \): in QII it is positive, \( \sin x = \sqrt{1 - \tfrac{4}{9}} = \tfrac{\sqrt{5}}{3} \).

\( \sin 2x = 2\sin x\cos x = 2\left(\tfrac{\sqrt{5}}{3}\right)\!\left(-\tfrac{2}{3}\right) = -\tfrac{4\sqrt{5}}{9} \).

EXAMPLE 2

Write \( \cos 3x \) in terms of \( \cos x \).

Solution

\( \cos 3x = \cos(2x + x) = \cos 2x\cos x - \sin 2x\sin x \).

Substitute double-angle forms: \( (2\cos^2 x - 1)\cos x - (2\sin x\cos x)\sin x \).

Expand: \( 2\cos^3 x - \cos x - 2\sin^2 x\cos x \).

Replace \( \sin^2 x = 1 - \cos^2 x \): \( 2\cos^3 x - \cos x - 2\cos x(1 - \cos^2 x) \).

Simplify: \( 4\cos^3 x - 3\cos x \). > 📝 ADMIN NOTE: Example 2 is your first taste of the Tchebycheff (Chebyshev) polynomials: \( \cos nx \) can always be written as a degree-\( n \) polynomial in \( \cos x \). These show up constantly in signal processing and numerical methods — remember where you first saw them.

EXAMPLE 3

Verify that \( \dfrac{\sin 3x}{\sin x\cos x} = 4\cos x - \sec x \).

Solution

Start with the left-hand side.

Write \( \sin 3x = \sin(x + 2x) = \sin x\cos 2x + \cos x\sin 2x \).

Double-angle forms: \( \dfrac{\sin x(2\cos^2 x - 1) + \cos x(2\sin x\cos x)}{\sin x\cos x} \).

Separate the fraction: \( \dfrac{\sin x(2\cos^2 x - 1)}{\sin x\cos x} + \dfrac{\cos x(2\sin x\cos x)}{\sin x\cos x} \).

Cancel: \( \dfrac{2\cos^2 x - 1}{\cos x} + 2\cos x \).

Separate again: \( 2\cos x - \dfrac{1}{\cos x} + 2\cos x \).

Reciprocal identity: \( 4\cos x - \sec x \) = RHS. ∎

Formulas for Lowering Powers

These rewrite even powers of sine and cosine in terms of the first power of cosine — essential for integration in calculus.

— Formulas for Lowering Powers

\[ \sin^2 x = \frac{1 - \cos 2x}{2} \qquad \cos^2 x = \frac{1 + \cos 2x}{2} \qquad \tan^2 x = \frac{1 - \cos 2x}{1 + \cos 2x} \]

EXAMPLE 4

Express \( \sin^2 x\cos^2 x \) in terms of the first power of cosine.

Solution

Apply the lowering formulas: \( \left(\dfrac{1 - \cos 2x}{2}\right)\!\left(\dfrac{1 + \cos 2x}{2}\right) \).

Multiply (difference of squares): \( \dfrac{1 - \cos^2 2x}{4} \).

Lower \( \cos^2 2x = \dfrac{1 + \cos 4x}{2} \): \( \dfrac{1}{4} - \dfrac{1}{4}\cdot\dfrac{1 + \cos 4x}{2} \).

Simplify: \( \dfrac{1}{8}(1 - \cos 4x) \). \[ \sin^2 x\cos^2 x = \tfrac{1}{8}(1 - \cos 4x) \]

Half-Angle Formulas

These relate the functions at \( \tfrac{u}{2} \) to the functions at \( u \). The \( \pm \) sign is chosen by the quadrant of \( \tfrac{u}{2} \); the tangent forms carry no sign ambiguity.

— Half-Angle Formulas

\[ \sin\frac{u}{2} = \pm\sqrt{\frac{1 - \cos u}{2}} \qquad \cos\frac{u}{2} = \pm\sqrt{\frac{1 + \cos u}{2}} \] \[ \tan\frac{u}{2} = \frac{1 - \cos u}{\sin u} = \frac{\sin u}{1 + \cos u} \]

EXAMPLE 5

Find the exact value of \( \sin 22.5° \).

Solution

Note \( 22.5° = \tfrac{45°}{2} \); choose \( + \) since \( 22.5° \) is in QI.

\( \sin\dfrac{45°}{2} = \sqrt{\dfrac{1 - \cos 45°}{2}} \).

Substitute \( \cos 45° = \dfrac{\sqrt 2}{2} \): \( \sqrt{\dfrac{1 - \sqrt 2/2}{2}} \).

Common denominator inside: \( \sqrt{\dfrac{2 - \sqrt 2}{4}} \).

Simplify: \( \dfrac{1}{2}\sqrt{2 - \sqrt 2} \).

EXAMPLE 6

Find \( \tan\dfrac{u}{2} \) if \( \sin u = \tfrac{2}{5} \) and \( u \) is in Quadrant II.

Solution

Find \( \cos u \): negative in QII, \( \cos u = -\sqrt{1 - \tfrac{4}{25}} = -\dfrac{\sqrt{21}}{5} \).

Use \( \tan\dfrac{u}{2} = \dfrac{1 - \cos u}{\sin u} = \dfrac{1 + \sqrt{21}/5}{2/5} \).

Simplify: \( \dfrac{5 + \sqrt{21}}{2} \). > 📝 ADMIN NOTE: Prefer the tangent half-angle forms \( \dfrac{1-\cos u}{\sin u} \) or \( \dfrac{\sin u}{1+\cos u} \) over the square-root version whenever you already know \( \sin u \) and \( \cos u \) — they sidestep the \( \pm \) sign decision entirely.

EXAMPLE 7

Write \( \sin(2\cos^{-1}x) \) as an algebraic expression in \( x \), with \( -1 \le x \le 1 \).

Solution

Let \( \theta = \cos^{-1}x \), so \( \cos\theta = x \) and (from the reference triangle) \( \sin\theta = \sqrt{1 - x^2} \).

Double-angle formula: \( \sin 2\theta = 2\sin\theta\cos\theta \).

Substitute: \( 2x\sqrt{1 - x^2} \).

EXAMPLE 8

Evaluate \( \sin 2\theta \) where \( \cos\theta = -\tfrac{2}{5} \) with \( \theta \) in Quadrant II.

Solution

Reference triangle: \( x = -2 \), \( r = 5 \), so \( y = \sqrt{25 - 4} = \sqrt{21} \) (positive in QII), giving \( \sin\theta = \dfrac{\sqrt{21}}{5} \).

Double-angle formula: \( \sin 2\theta = 2\sin\theta\cos\theta = 2\!\left(\dfrac{\sqrt{21}}{5}\right)\!\left(-\dfrac{2}{5}\right) = -\dfrac{4\sqrt{21}}{25} \).

Product-Sum Formulas

The Product-to-Sum formulas turn products of sines/cosines into sums; the Sum-to-Product formulas do the reverse. Both come directly from the addition and subtraction formulas.

— Product-to-Sum Formulas

\[ \sin u\cos v = \tfrac{1}{2}\big[\sin(u + v) + \sin(u - v)\big] \] \[ \cos u\sin v = \tfrac{1}{2}\big[\sin(u + v) - \sin(u - v)\big] \] \[ \cos u\cos v = \tfrac{1}{2}\big[\cos(u - v) + \cos(u + v)\big] \] \[ \sin u\sin v = \tfrac{1}{2}\big[\cos(u - v) - \cos(u + v)\big] \]

— Sum-to-Product Formulas

\[ \sin x + \sin y = 2\sin\frac{x + y}{2}\cos\frac{x - y}{2} \] \[ \sin x - \sin y = 2\cos\frac{x + y}{2}\sin\frac{x - y}{2} \] \[ \cos x + \cos y = 2\cos\frac{x + y}{2}\cos\frac{x - y}{2} \] \[ \cos x - \cos y = -2\sin\frac{x + y}{2}\sin\frac{x - y}{2} \]

EXAMPLE 9

Express \( \sin 3x\sin 5x \) as a sum.

Solution

Product-to-sum (the \( \sin u\sin v \) form), \( u = 3x \), \( v = 5x \): \( \tfrac{1}{2}\big[\cos(3x - 5x) - \cos(3x + 5x)\big] \).

Simplify, using \( \cos(-2x) = \cos 2x \): \( \tfrac{1}{2}\cos 2x - \tfrac{1}{2}\cos 8x \).

EXAMPLE 10

Write \( \sin 7x + \sin 3x \) as a product.

Solution

Sum-to-product: \( 2\sin\dfrac{7x + 3x}{2}\cos\dfrac{7x - 3x}{2} \).

Simplify: \( 2\sin 5x\cos 2x \).

EXAMPLE 11

Verify that \( \dfrac{\sin 3x + \sin x}{\cos 3x + \cos x} = \tan 2x \).

Solution

Sum-to-product on numerator and denominator: \( \dfrac{2\sin 2x\cos x}{2\cos 2x\cos x} \).

Cancel \( 2\cos x \): \( \dfrac{\sin 2x}{\cos 2x} = \tan 2x \) = RHS. ∎ Figure 7.3.1 — Reference triangle for \( \theta = \cos^{-1}x \) (legs \( x \) and \( \sqrt{1-x^2} \), hypotenuse 1) used in Example 7. (See the source course, Ch. 7.) ---

SECTION 7.4

Basic Trigonometric Equations

An equation containing trigonometric functions is a trigonometric equation. Solving one means finding all values of the variable that make it true.

Basic Trigonometric Equations

Solving any trigonometric equation reduces to solving a basic equation of the form \( T(\theta) = c \), where \( T \) is a trigonometric function and \( c \) is a constant. Because the functions are periodic, a solvable basic equation has infinitely many solutions: find them in one period, then add integer multiples of the period.

— General Solution Pattern

If \( \theta_0 \) is a solution of \( \sin\theta = c \) or \( \cos\theta = c \), all solutions add multiples of the period \( 2\pi \): \( \theta = \theta_0 + 2k\pi \). For \( \tan\theta = c \) the period is \( \pi \): \( \theta = \theta_0 + k\pi \). Here \( k \) is any integer.

EXAMPLE 1

Solve \( \sin\theta = \tfrac{1}{2} \).

Solution

In \( [0, 2\pi) \), sine is \( \tfrac{1}{2} \) in QI and QII: \( \theta = \dfrac{\pi}{6} \) and \( \theta = \dfrac{5\pi}{6} \).

Add multiples of \( 2\pi \): \( \theta = \dfrac{\pi}{6} + 2k\pi \) and \( \theta = \dfrac{5\pi}{6} + 2k\pi \), \( k \) any integer.

EXAMPLE 2

Solve \( \cos\theta = -\dfrac{\sqrt 2}{2} \) and list eight specific solutions.

Solution

In \( [0, 2\pi) \), cosine is \( -\dfrac{\sqrt 2}{2} \) in QII and QIII: \( \theta = \dfrac{3\pi}{4} \) and \( \theta = \dfrac{5\pi}{4} \).

General solution: \( \theta = \dfrac{3\pi}{4} + 2k\pi \) and \( \theta = \dfrac{5\pi}{4} + 2k\pi \).

Eight specific solutions (taking \( k = -1, 0, 1, 2 \)): \[ -\tfrac{5\pi}{4},\ -\tfrac{3\pi}{4},\ \tfrac{3\pi}{4},\ \tfrac{5\pi}{4},\ \tfrac{11\pi}{4},\ \tfrac{13\pi}{4},\ \tfrac{19\pi}{4},\ \tfrac{21\pi}{4} \]

EXAMPLE 3

Solve \( \cos\theta = 0.65 \).

Solution

One solution: \( \theta = \cos^{-1}(0.65) \approx 0.86 \) (radian mode).

Cosine is positive in QI and QIV, so the other solution in one period is \( \theta \approx 2\pi - 0.86 \approx 5.42 \).

General solution: \( \theta \approx 0.86 + 2k\pi \) and \( \theta \approx 5.42 + 2k\pi \).

EXAMPLE 4

Solve \( \tan\theta = 2 \).

Solution

One solution: \( \theta = \tan^{-1}(2) \approx 1.12 \), the only solution in \( (-\pi/2, \pi/2) \).

Tangent has period \( \pi \): \( \theta \approx 1.12 + k\pi \), \( k \) any integer. > 📝 ADMIN NOTE: Don't forget the period difference: sine and cosine repeat every \( 2\pi \), but tangent repeats every \( \pi \). Using \( 2k\pi \) on a tangent equation will make you miss half the solutions — a frequent exam mistake.

EXAMPLE 5

Find all solutions of (a) \( 2\sin\theta - 1 = 0 \) and (b) \( \tan^2\theta - 3 = 0 \).

Solution

Solving Trigonometric Equations by Factoring

Move all terms to one side, factor, and apply the Zero-Product Property: if \( AB = 0 \), then \( A = 0 \) or \( B = 0 \).

EXAMPLE 6

Solve \( 2\cos^2\theta - 7\cos\theta + 3 = 0 \).

Solution

Factor as a quadratic in \( \cos\theta \): \( (2\cos\theta - 1)(\cos\theta - 3) = 0 \).

Set each factor to zero: \( \cos\theta = \tfrac{1}{2} \) or \( \cos\theta = 3 \).

\( \cos\theta = 3 \) has no solution (range of cosine is \( [-1,1] \)).

\( \cos\theta = \tfrac{1}{2} \) gives \( \theta = \dfrac{\pi}{3} + 2k\pi \) and \( \theta = \dfrac{5\pi}{3} + 2k\pi \).

EXAMPLE 7

Solve \( 5\sin\theta\cos\theta + 4\cos\theta = 0 \).

Solution

Factor out \( \cos\theta \): \( \cos\theta(5\sin\theta + 4) = 0 \).

First factor: \( \cos\theta = 0 \) → \( \theta = \dfrac{\pi}{2}, \dfrac{3\pi}{2} \) in one period.

Second factor: \( \sin\theta = -0.8 \) → \( \theta = \sin^{-1}(-0.8) \approx -0.93 \), so within \( [0, 2\pi) \), \( \theta \approx \pi + 0.93 \approx 4.07 \) and \( \theta \approx 2\pi - 0.93 \approx 5.35 \).

Add \( 2k\pi \) to each solution for the general solution. > 📝 ADMIN NOTE: Critical habit: never divide both sides by \( \cos\theta \) (or any trig factor) to "simplify." You'll throw away the solutions where that factor is zero. Always move everything to one side and factor instead, as in Example 7. Figure 7.4.1 — Unit-circle solutions of \( \sin\theta = \tfrac12 \) at \( \theta = \pi/6 \) and \( 5\pi/6 \). (See the source course, Ch. 7.) ---

SECTION 7.5

More Trigonometric Equations

Here we solve trigonometric equations that first require an identity to simplify, and equations involving functions of multiples of an angle.

Solving Trigonometric Equations by Using Identities

When an equation mixes different functions or powers, use an identity to rewrite it in terms of a single function, then factor and solve.

EXAMPLE 1

Solve \( 1 + \sin\theta = 2\cos^2\theta \).

Solution

Pythagorean identity \( \cos^2\theta = 1 - \sin^2\theta \): \( 1 + \sin\theta = 2(1 - \sin^2\theta) \).

Move all terms to one side: \( 2\sin^2\theta + \sin\theta - 1 = 0 \).

Factor: \( (2\sin\theta - 1)(\sin\theta + 1) = 0 \).

Solve each: \( \sin\theta = \tfrac{1}{2} \) → \( \theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6} \); and \( \sin\theta = -1 \) → \( \theta = \dfrac{3\pi}{2} \).

General solution: \( \theta = \dfrac{\pi}{6} + 2k\pi,\ \dfrac{5\pi}{6} + 2k\pi,\ \dfrac{3\pi}{2} + 2k\pi \).

EXAMPLE 2

Solve \( \sin 2\theta - \cos\theta = 0 \).

Solution

Double-angle formula \( \sin 2\theta = 2\sin\theta\cos\theta \): \( 2\sin\theta\cos\theta - \cos\theta = 0 \).

Factor out \( \cos\theta \): \( \cos\theta(2\sin\theta - 1) = 0 \).

\( \cos\theta = 0 \) → \( \theta = \dfrac{\pi}{2}, \dfrac{3\pi}{2} \); \( \sin\theta = \tfrac{1}{2} \) → \( \theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6} \).

General solution: add \( 2k\pi \) to each of \( \dfrac{\pi}{2},\ \dfrac{3\pi}{2},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6} \).

EXAMPLE 3

Solve \( \cos\theta + 1 = \sin\theta \) on \( [0, 2\pi) \).

Solution

Square both sides: \( \cos^2\theta + 2\cos\theta + 1 = \sin^2\theta \).

Pythagorean identity \( \sin^2\theta = 1 - \cos^2\theta \): \( \cos^2\theta + 2\cos\theta + 1 = 1 - \cos^2\theta \).

Simplify: \( 2\cos^2\theta + 2\cos\theta = 0 \).

Factor: \( 2\cos\theta(\cos\theta + 1) = 0 \) → \( \cos\theta = 0 \) or \( \cos\theta = -1 \).

Candidates: \( \theta = \dfrac{\pi}{2}, \dfrac{3\pi}{2}, \pi \).

Check for extraneous roots (squaring can introduce them): \( \dfrac{\pi}{2} \) ✓, \( \pi \) ✓, \( \dfrac{3\pi}{2} \) ✗. Solutions: \( \theta = \dfrac{\pi}{2} \) and \( \theta = \pi \). > 📝 ADMIN NOTE: Any time you square both sides of an equation, you must check every candidate in the original equation. Squaring is not reversible and routinely manufactures fake solutions — Example 3's \( 3\pi/2 \) is exactly such a case.

EXAMPLE 4

Find the values of \( x \) where \( f(x) = \sin x \) and \( g(x) = \cos x \) intersect.

Solution

Equations with Functions of Multiples of Angles

When the unknown appears inside a multiple (like \( 3\theta \)) or a fraction (like \( \tfrac{\theta}{2} \)), first solve for that whole quantity, then divide or multiply to isolate \( \theta \). Remember to generate enough \( k \)-values to capture every solution in the requested interval.

EXAMPLE 5

For \( 2\sin 3\theta - 1 = 0 \): (a) find all solutions; (b) list those in \( [0, 2\pi) \).

Solution

EXAMPLE 6

For \( \sqrt{3}\tan\dfrac{\theta}{2} - 1 = 0 \): (a) find all solutions; (b) list those in \( [0, 4\pi) \).

Solution

Modeling a Wave Shape

A point on a vibrating string undergoes simple harmonic motion \( y = A\sin\omega t \). Photographing the whole string at a fixed instant gives a sine curve in position \( x \): \( y = A\sin kx \), where \( y \) is the string's height at point \( x \).

Traveling Waves

A wave moving right with velocity \( v \) shifts the shape by \( vt \) after time \( t \).

— Traveling Wave

\[ y(x, t) = A\sin k(x - vt) \] Fixing \( x \) gives the motion of one point over time; fixing \( t \) gives the string's shape at that instant. The velocity is read off by writing the function in this standard form.

EXAMPLE 7

A wave is \( y(x, t) = 3\sin\!\left(2x - \dfrac{\pi}{2}t\right) \), \( x \ge 0 \). Find the motion of the point \( x = \dfrac{\pi}{6} \), and the wave velocity.

Solution

Substitute \( x = \dfrac{\pi}{6} \): \( y\!\left(\dfrac{\pi}{6}, t\right) = 3\sin\!\left(\dfrac{\pi}{3} - \dfrac{\pi}{2}t\right) \) — simple harmonic motion, amplitude 3, period \( \dfrac{2\pi}{\pi/2} = 4 \).

Factor to standard form: \( y(x, t) = 3\sin 2\!\left(x - \dfrac{\pi}{4}t\right) \), so velocity \( v = \dfrac{\pi}{4} \).

Standing Waves

When a wave and its reflection travel in opposite directions along the same string, they superpose. Adding them and applying a Sum-to-Product formula yields a wave that vibrates in place rather than traveling.

— Standing Wave

\[ y(x, t) = A\sin k(x - vt) + A\sin k(x + vt) = 2A\sin kx\,\cos kvt \] **Nodes** are points where \( kx \) is a multiple of \( \pi \); there \( y = 0 \) for all \( t \), so they never move.

EXAMPLE 8

In a 30-ft tank the waves \( y = 1.5\sin\!\left(\dfrac{\pi}{5}x - 3t\right) \) and \( y = 1.5\sin\!\left(\dfrac{\pi}{5}x + 3t\right) \) combine. Find the combined wave and its nodes.

Solution

Add and apply Sum-to-Product: \( y = 3\sin\dfrac{\pi}{5}x\,\cos 3t \).

Nodes where \( \sin\dfrac{\pi}{5}x = 0 \), i.e. \( \dfrac{\pi}{5}x = k\pi \) → \( x = 5k \).

Within the tank: \( x = 0, 5, 10, 15, 20, 25, 30 \). > 📝 ADMIN NOTE: This is the mathematical backbone of resonance and vibration analysis in mechanical engineering — natural frequencies of beams, shafts, and strings all trace back to the node structure of standing waves. The Sum-to-Product step from 7.3 is what makes the "vibrates in place" behavior fall out cleanly. Figure 7.M.1 — A traveling wave shifting right at successive instants \( t \). (See the source course, Focus on Modeling.) Figure 7.M.2 — A standing wave: fixed nodes with the string vibrating up and down between them. (See the source course, Focus on Modeling.) --- End of Chapter 7 — Analytic Trigonometry. Exercises, review questions, and chapter-review problems are intentionally omitted per the MechForge content rules.