Exercise 8.2 — Question 1: Double-Angle and Half-Angle Identities

Double-Angle Identities

The double-angle identities are derived from the sum formulas by setting $\alpha =\beta =\theta$.

Sine Double-Angle

$\sin 2\theta =2\sin \theta \cos \theta$

Cosine Double-Angle

The cosine double-angle identity has three equivalent forms: $\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$ $\cos 2\theta =2{\cos }^2\theta -1$ $\cos 2\theta =1-2{\sin }^2\theta$

The second and third forms are obtained by substituting ${\sin }^2\theta =1-{\cos }^2\theta$ or ${\cos }^2\theta =1-{\sin }^2\theta$ respectively.

Tangent Double-Angle

$\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$

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Half-Angle Identities

Rearranging the cosine double-angle formulas gives the half-angle identities:

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

$\cos \frac{\theta }{2}=\pm \sqrt{\frac{1+\cos \theta }{2}}$

$\tan \frac{\theta }{2}=\pm \sqrt{\frac{1-\cos \theta }{1+\cos \theta }}$

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Worked Examples

Example 1: Verify ${\cos }^{4}x-{\sin }^{4}x=\cos 2x$

Solution:

Factor the left-hand side as a difference of squares: ${\cos }^{4}x-{\sin }^{4}x=({\cos }^{2}x-{\sin }^{2}x)({\cos }^{2}x+{\sin }^{2}x)$

Since ${\cos }^{2}x+{\sin }^{2}x=1$: $=({\cos }^{2}x-{\sin }^{2}x)(1)={\cos }^{2}x-{\sin }^{2}x=\cos 2x✓$

Example 2: Find $\sin 2\theta$ if $\sin \theta =\frac{3}{5}$ and $\theta$ is in Quadrant I

Solution:

First find $\cos \theta$: Since ${\cos }^2\theta =1-{\sin }^2\theta =1-\frac{9}{25}=\frac{16}{25}$, so $\cos \theta =\frac{4}{5}$ (positive in Q I).

$\sin 2\theta =2\sin \theta \cos \theta =2\cdot \frac{3}{5}\cdot \frac{4}{5}=\frac{24}{25}$

Example 3: Find $\cos \frac{\theta }{2}$ if $\cos \theta =\frac{1}{2}$ and $\frac{\theta }{2}$ is in Quadrant I

Solution:

$\cos \frac{\theta }{2}=+\sqrt{\frac{1+\cos \theta }{2}}=\sqrt{\frac{1+\frac{1}{2}}{2}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$