8.2 Q-6

Exercise 8.2 — Question 6

Problem

Prove the following identities using double-angle and half-angle formulas:

(i) ${\cos }^{4}x-{\sin }^{4}x=\cos 2x$

(ii) $\frac{1-\cos 2\theta }{\sin 2\theta }=\tan \theta$

(iii) $\sin 2\alpha =\frac{2\tan \alpha }{1+{\tan }^2\alpha }$

(iv) $\cos 2\alpha =\frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha }$

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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:

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

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

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

The sign in half-angle formulas is determined by the quadrant of $\frac{\theta }{2}$, not of $\theta$.

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Worked Solutions

Part (i): Prove ${\cos }^{4}x-{\sin }^{4}x=\cos 2x$

L.H.S. $={\cos }^{4}x-{\sin }^{4}x$

Factor as a difference of squares:

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

Since ${\cos }^{2}x+{\sin }^{2}x=1$:

$=({\cos }^{2}x-{\sin }^{2}x)(1)=\cos 2x=\text{R.H.S.}✓$

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Part (ii): Prove $\frac{1-\cos 2\theta }{\sin 2\theta }=\tan \theta$

L.H.S. $=\frac{1-\cos 2\theta }{\sin 2\theta }$

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

$=\frac{1-(1-2{\sin }^2\theta )}{2\sin \theta \cos \theta }=\frac{2{\sin }^2\theta }{2\sin \theta \cos \theta }=\frac{\sin \theta }{\cos \theta }=\tan \theta =\text{R.H.S.}✓$

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Part (iii): Prove $\sin 2\alpha =\frac{2\tan \alpha }{1+{\tan }^2\alpha }$

R.H.S. $=\frac{2\tan \alpha }{1+{\tan }^2\alpha }$

Substitute $\tan \alpha =\frac{\sin \alpha }{\cos \alpha }$ and use $1+{\tan }^2\alpha ={\sec }^2\alpha$:

$=\frac{2\cdot \frac{\sin \alpha }{\cos \alpha }}{{\sec }^2\alpha }=\frac{\frac{2\sin \alpha }{\cos \alpha }}{\frac{1}{{\cos }^2\alpha }}=2\sin \alpha \cos \alpha =\sin 2\alpha =\text{L.H.S.}✓$

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Part (iv): Prove $\cos 2\alpha =\frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha }$

R.H.S. $=\frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha }$

Substitute $\tan \alpha =\frac{\sin \alpha }{\cos \alpha }$ and $1+{\tan }^2\alpha ={\sec }^2\alpha$:

$=\frac{1-\frac{{\sin }^2\alpha }{{\cos }^2\alpha }}{{\sec }^2\alpha }=\frac{\frac{{\cos }^2\alpha -{\sin }^2\alpha }{{\cos }^2\alpha }}{\frac{1}{{\cos }^2\alpha }}={\cos }^2\alpha -{\sin }^2\alpha =\cos 2\alpha =\text{L.H.S.}✓$

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