8.2 Double, Triple and Half Angle Identities

Double Angle Identities

These identities express trigonometric functions of $2\theta$ in terms of functions of $\theta$. They are derived from the sum formulas by setting $\alpha =\beta =\theta$.

Sine Double Angle

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

Alternate form in terms of $\tan \theta$: $\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }$

Cosine Double Angle

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

Alternate form in terms of $\tan \theta$: $\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$

Tangent Double Angle

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

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Triple Angle Identities

These are derived by writing $3\theta =2\theta +\theta$ and applying sum and double angle formulas.

$\sin 3\theta =3\sin \theta -4{\sin }^3\theta$ $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$ $\tan 3\theta =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$

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Half Angle Identities

Rearranging the cosine double angle formulas gives the power-reduction and half-angle identities.

Power-Reduction Identities

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

Half-Angle Formulas

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

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Power Reduction Strategy

To reduce higher powers of trig functions (e.g., ${\sin }^4\alpha$, ${\cos }^4\alpha$) to first-power cosines:

1. Apply ${\sin }^2\theta =\frac{1-\cos 2\theta }{2}$ and ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$.
2. Square the result if needed, then apply the identities again.

Example: Reduce ${\sin }^4\alpha$ ${\sin }^4\alpha =({\sin }^2\alpha {)}^2={\left(\frac{1-\cos 2\alpha }{2}\right)}^2=\frac{1-2\cos 2\alpha +{\cos }^{2}2\alpha }{4}$ $=\frac{1-2\cos 2\alpha +\frac{1+\cos 4\alpha }{2}}{4}=\frac{3-4\cos 2\alpha +\cos 4\alpha }{8}$

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Key Identity Verification Technique

To verify ${\cos }^{4}x-{\sin }^{4}x=\cos 2x$: ${\cos }^{4}x-{\sin }^{4}x=({\cos }^{2}x-{\sin }^{2}x)({\cos }^{2}x+{\sin }^{2}x)=({\cos }^{2}x-{\sin }^{2}x)(1)=\cos 2x✓$