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Prove that $(\cos x-\cos y)^2+(\sin x-\sin y)^2=4\sin^2\large\frac{x-y}{2}$

1 Answer

Toolbox:
  • $\cos A-\cos B=-2\sin \large\frac{A+B}{2}$$\sin \large\frac{A-B}{2}$
  • $\sin A-\sin B=-2\cos\large\frac{A+B}{2}$$\sin \large\frac{A-B}{2}$
  • $\cos^2A+\sin^2A=1$
L.H.S
$(\cos x-\cos y)^2+(\sin x-\sin y)^2$
$\Rightarrow (-2\sin\large\frac{x+y}{2}$$\sin \large\frac{x-y}{2})^2$$+(2\cos\large\frac{x+y}{2}$$\sin \large\frac{x-y}{2})$
$\Rightarrow 4\sin^2\large\frac{x+y}{2}$$\sin^2\large\frac{x-y}{2}$$+4\cos^2\large\frac{x+y}{2}$$\sin^2\large\frac{x-y}{2}$
$\Rightarrow 4\sin^2\large\frac{x-y}{2}$$(\sin^2\large\frac{x+y}{2}+$$\cos^2\large\frac{x+y}{2})$
$\Rightarrow 4\sin^2\large\frac{x-y}{2}$=R.H.S
Hence proved.
answered Apr 18, 2014 by sreemathi.v
 

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