Exercise 8.2 — Double Angle and Half Angle Identities

Key Identities

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

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

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

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

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Worked Examples

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

LHS: ${\cos }^{4}x-{\sin }^{4}x=({\cos }^{2}x-{\sin }^{2}x)({\cos }^{2}x+{\sin }^{2}x)$

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

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Example 2 — Verify: $\frac{1-\cos 2\theta }{\sin 2\theta }=\tan \theta$

LHS: $\frac{1-\cos 2\theta }{\sin 2\theta }=\frac{1-(1-2{\sin }^2\theta )}{2\sin \theta \cos \theta }=\frac{2{\sin }^2\theta }{2\sin \theta \cos \theta }=\frac{\sin \theta }{\cos \theta }=\tan \theta =\text{RHS}✓$

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Example 3 — Verify: $\sin 3\theta =3\sin \theta -4{\sin }^3\theta$

LHS: $\sin 3\theta =\sin (2\theta +\theta )=\sin 2\theta \cos \theta +\cos 2\theta \sin \theta$ $=2\sin \theta {\cos }^2\theta +(1-2{\sin }^2\theta )\sin \theta$ $=2\sin \theta (1-{\sin }^2\theta )+\sin \theta -2{\sin }^3\theta$ $=2\sin \theta -2{\sin }^3\theta +\sin \theta -2{\sin }^3\theta$ $=3\sin \theta -4{\sin }^3\theta =\text{RHS}✓$

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Example 4 — Verify: $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$

LHS: $\cos 3\theta =\cos (2\theta +\theta )=\cos 2\theta \cos \theta -\sin 2\theta \sin \theta$ $=(2{\cos }^2\theta -1)\cos \theta -2{\sin }^2\theta \cos \theta$ $=2{\cos }^3\theta -\cos \theta -2(1-{\cos }^2\theta )\cos \theta$ $=2{\cos }^3\theta -\cos \theta -2\cos \theta +2{\cos }^3\theta$ $=4{\cos }^3\theta -3\cos \theta =\text{RHS}✓$

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Flashcards

MCQs