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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Prove that $\cos \theta \cos\large\frac{\theta}{2}-$$\cos 3\theta\cos \large\frac{9\theta}{2}=$$\sin 7\theta\sin 8\theta$

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Toolbox:
  • $\cos a\cos b=\large\frac{1}{2}$$[\cos(a+b)-\cos(a-b)]$
  • $\cos a-\cos b=-2\sin (\large\frac{a+b}{2})$$.\sin (\large\frac{a-b}{2})$
  • $\sin (-\theta)=-\sin \theta$
  • $\cos (-\theta)=\cos \theta$
LHS
$\cos \theta.\cos \large\frac{\theta}{2}$$-\cos 3\theta\cos \large\frac{9\theta}{2}$
$\large\frac{1}{2}$$[\cos(\theta+\large\frac{\theta}{2})]-\large\frac{1}{2}$$[\cos(3\theta+\large\frac{9\theta}{2}]$$+\cos(3\theta-\large\frac{9\theta}{2})]$
$\large\frac{1}{2}$$[\cos \large\frac{3\theta}{2}+$$\cos\large\frac{\theta}{2}]-\large\frac{1}{2}$$[\cos \large\frac{15\theta}{2}+$$\cos\large\frac{-3\theta}{2}]$
$\large\frac{1}{2}$$[\cos \large\frac{3\theta}{2}$$+\cos \large\frac{\theta}{2}-$$\cos \large\frac{15\theta}{2}$$-\cos \large\frac{3\theta}{2}]$
$\large\frac{1}{2}$$[\cos \large\frac{\theta}{2}$$-\cos \large\frac{15\theta}{2}]$
$\large\frac{1}{2}$$\times -2\sin (\large\frac{\theta}{2}+\frac{15\theta}{2})$$\sin (\large\frac{\theta}{2}-\frac{15\theta}{2})$
$-1\times \sin \large\frac{16\theta}{2}$$\sin (-\large\frac{14\theta}{2})$
$\sin(-\theta)=-\sin \theta$
$\sin 8\theta.\sin 7\theta$=RHS
Hence proved.
answered May 20, 2014 by sreemathi.v
 

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