Section 7.3

Double-Angle & Half-Angle Formulas

Double-angle formulas show up directly in solving trig equations in Section 7.5 — you'll need $\cos(2x) = 1 - 2\sin^2(x)$ to convert double-angle equations into solvable quadratics. Expect 2–3 problems from this section on Exam 3, including at least one double-angle calculation.

⚡ Quick Summary

The double-angle formulas answer a natural question: if you know the trig values of an angle $\theta$ , can you find the trig values of $2\theta$ without starting over? Yes — and the formulas come straight from the sum formulas in Section 7.2, just with both angles set equal. The sine version is clean: $\sin(2\theta) = 2\sin\theta\cos\theta$ — you need both sine and cosine, so find the missing one first.

Sine Double-Angle

\sin(2\theta) = 2\sin\theta\cos\theta

Cosine Double-Angle (3 forms)

\cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta

Choose the form that best fits what you already know (sin only, cos only, or both).

📋 Before You Start

Tap any item if you need a refresher:

Plain-English version

The cosine double-angle formula comes in three flavors: $\cos^2\theta - \sin^2\theta$ , $2\cos^2\theta - 1$ , or $1 - 2\sin^2\theta$ . They're all the same formula — just rearranged using $\sin^2 + \cos^2 = 1$ . Pick the version that matches what you already know. If you only have sine, use $1 - 2\sin^2\theta$ . If you only have cosine, use $2\cos^2\theta - 1$ .

Half-angle formulas let you go the other direction — finding trig values at "weird" angles like $22.5°$ (which is half of $45°$ ) or $15°$ (half of $30°$ ). The tricky part is a $\pm$ sign: you have to decide positive or negative based on which quadrant the half-angle is in, not the original angle. For instance, $\sin(15°)$ is positive because 15° is in Quadrant I.

Power-reducing formulas turn squared trig functions into non-squared ones: $\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$ and $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$ . Think of them as a way to "downgrade" a squared expression. Quick memory tip: sine gets the minus ("sine is sad"), cosine gets the plus ("cosine is cheerful").

Key Formulas

Sine Double-Angle

\sin(2\theta) = 2\sin\theta\cos\theta

Cosine Double-Angle (3 forms)

\cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta

Choose the form that best fits what you already know (sin only, cos only, or both).

Tangent Double-Angle

\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}

Half-Angle: Sine

\sin\frac{\alpha}{2} = \pm\sqrt{\frac{1 - \cos\alpha}{2}}

The ± depends on the quadrant of α/2.

Half-Angle: Cosine

\cos\frac{\alpha}{2} = \pm\sqrt{\frac{1 + \cos\alpha}{2}}

The ± depends on the quadrant of α/2.

Power-Reducing: Sine

\sin^2\theta = \frac{1 - \cos(2\theta)}{2}

Derived from the cosine double-angle formula; replaces a square with a first-power expression.

Power-Reducing: Cosine

\cos^2\theta = \frac{1 + \cos(2\theta)}{2}

Derived from the cosine double-angle formula; replaces a square with a first-power expression.

Half-Angle: Tangent

\tan\frac{\alpha}{2} = \frac{1 - \cos\alpha}{\sin\alpha} = \frac{\sin\alpha}{1 + \cos\alpha}

Two equivalent forms — pick whichever avoids a zero denominator.

Key Takeaways

✓ $\sin(2\theta) = 2\sin\theta\cos\theta$ — you need both sin and cos, so find the missing one first
✓ $\cos(2\theta)$ has three forms — pick the one matching what you know: sin only → $1 - 2\sin^2\theta$ , cos only → $2\cos^2\theta - 1$
✓Half-angle formulas have a $\pm$ that depends on the quadrant of the half-angle, not the original angle
✓Power-reducing formulas turn squares into first-power expressions: $\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$

⚠️ Common Mistakes

✗Writing $\sin(2\theta) = 2\sin\theta$ instead of $2\sin\theta\cos\theta$ — you need both sine AND cosine
✗Picking the wrong form of $\cos(2\theta)$ — use $1 - 2\sin^2\theta$ when you only know sine, $2\cos^2\theta - 1$ when you only know cosine
✗Forgetting to choose $+$ or $-$ in half-angle formulas — the sign depends on the quadrant of the half-angle $\frac{\alpha}{2}$ , not the full angle $\alpha$
✗Confusing power-reducing formulas: $\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$ (minus for sine) vs. $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$ (plus for cosine)

Next: Section 7.5 Builds on: 7.2

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7.2 Sum & Difference Identities

7.5 Solving Trigonometric Equations