Precalculus Honors · Study guide

2.3 · Trigonometric identities and equations

Choose identities from expression structure, retain domain restrictions, and list every solution in the requested interval.

Precalculus Honors · Reading: Demana 10th ed. §§5.1–5.4; Stewart 8th ed. §§7.1–7.5

Learning objectives

Simplify with Pythagorean and angle formulas.
Verify identities without assuming the conclusion.
Solve trigonometric equations over a stated interval.

An identity is true on a common domain

The Pythagorean identity $\displaystyle \sin^2x+\cos^2x=1$ supports many rewrites. For $\displaystyle (1-\cos^2x)/\sin x$ , replace the numerator by $\displaystyle \sin^2x$ , then cancel only where the original denominator is nonzero.

To verify an identity, transform one side or transform both toward an independent expression. Domain exclusions remain in force after cancellation.

Addition and double-angle formulas create exact values

The formulas $\displaystyle \sin(a+b)=\sin a\cos b+\cos a\sin b$ and $\displaystyle \cos(a+b)=\cos a\cos b-\sin a\sin b$ let you evaluate angles built from known ones.

The double-angle identities include $\displaystyle \sin2x=2\sin x\cos x$ and $\displaystyle \cos2x=1-2\sin^2x=2\cos^2x-1$ . Select a form matching the equation you need to solve.

Find all angles, not just the inverse output

For $\displaystyle 2\sin x=1$ on $\displaystyle [0,2\pi)$ , the principal reference angle is $\displaystyle \pi/6$ but Quadrants I and II both have positive sine. The solutions are $\displaystyle \pi/6$ and $\displaystyle 5\pi/6$ .

Factoring a trigonometric equation is often safer than dividing by a trig factor, which can discard zero-factor solutions. Check endpoints against whether the interval is closed or half-open.

Worked examples

Simplify with a restriction

Simplify $\displaystyle (1-\cos^2x)/\sin x$ .

The numerator becomes $\displaystyle \sin^2x$ .
For $\displaystyle \sin x\ne0$ , cancellation leaves $\displaystyle \sin x$ .
The original excluded inputs remain excluded.

A double-angle equation

Solve $\displaystyle \cos2x=0$ on $\displaystyle [0,2\pi)$ .

The angle $\displaystyle 2x$ runs over $\displaystyle [0,4\pi)$ .
Its solutions are $\displaystyle \pi/2,3\pi/2,5\pi/2,7\pi/2$ .
Dividing by $\displaystyle 2$ gives $\displaystyle \pi/4,3\pi/4,5\pi/4,7\pi/4$ .

Common mistakes

Dividing by a trig factor that could be zero.
Reporting only the inverse-function output.
Dropping an original denominator restriction.

Try it yourself

No calculator required.

1. Solve $\displaystyle 2\sin x=1$ on $\displaystyle [0,2\pi)$ .

The sine graph intersects the horizontal level one half twice on [0, 2 pi). — Find both intersections in the stated interval.

Show worked solution

The horizontal level $\displaystyle 1/2$ meets sine in Quadrants I and II: $\displaystyle x=\pi/6$ and $\displaystyle x=5\pi/6$ .

The sine graph intersects the horizontal level one half twice on [0, 2 pi). The intersections occur at pi over 6 and 5 pi over 6. — The intersections occur at pi over 6 and 5 pi over 6.

2. Find $\displaystyle \cos(\pi/4+\pi/6)$ exactly.

Show worked solution

Use the addition formula to get $\displaystyle (\sqrt{2}/2)(\sqrt{3}/2)-(\sqrt{2}/2)(1/2)=(\sqrt{6}-\sqrt{2})/4$ .

3. Solve $\displaystyle \sin x\cos x=0$ on $\displaystyle [0,2\pi)$ .

Show worked solution

One factor must be zero. The solutions are $\displaystyle 0,\pi/2,\pi,3\pi/2$ . Dividing by either factor would lose some solutions.

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