8.2 Q-4

Exercise 8.2 — Question 4

Topic: Double-Angle and Half-Angle Identities

Relevant SLO: Apply fundamental law and its deductions to derive trigonometric ratios of allied angles, double angle, half angle identities (M-11-B-23).

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Key Identities Used

Double-Angle Identities:

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

$\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta =2{\cos }^2\theta -1=1-2{\sin }^2\theta$

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

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

LHS: ${\cos }^{4}x-{\sin }^{4}x$

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

Apply the Pythagorean identity ${\cos }^{2}x+{\sin }^{2}x=1$: $=({\cos }^{2}x-{\sin }^{2}x)(1)$

Apply the double-angle identity $\cos 2x={\cos }^{2}x-{\sin }^{2}x$: $=\cos 2x=\text{RHS}✓$

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Worked Example 2 — Express ${\cos }^2\theta$ in terms of $\cos 2\theta$

Starting from: $\cos 2\theta =2{\cos }^2\theta -1$

Solve for ${\cos }^2\theta$: ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$

Similarly, from $\cos 2\theta =1-2{\sin }^2\theta$: ${\sin }^2\theta =\frac{1-\cos 2\theta }{2}$

These are the power-reduction formulas, directly derived from the double-angle identities.

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Worked Example 3 — Find $\sin 15°$ using a half-angle formula

Note that $15°=\frac{30°}{2}$, so $\theta =30°$:

$\sin 15°=\sin \frac{30°}{2}=\sqrt{\frac{1-\cos 30°}{2}}$

(Positive because $15°$ is in the first quadrant.)

$=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}=\sqrt{\frac{2-\sqrt{3}}{4}}=\frac{\sqrt{2-\sqrt{3}}}{2}$

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