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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Prove : $\cos(\large\frac{\pi}{4}$$-x)\cos (\large\frac{\pi}{4}$$-y)-\sin(\large\frac{\pi}{4}$$-x)\sin (\large\frac{\pi}{4}$$-y)=\sin (x+y)$

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Toolbox:
  • $\cos (A+B)=\cos A\cos B-\sin A\sin B$
  • $\cos(\large\frac{\pi}{2}$$-\theta)=\sin \theta$
L.H.S
$\cos(\large\frac{\pi}{4}$$-x)\cos (\large\frac{\pi}{4}$$-y)-\sin(\large\frac{\pi}{4}$$-x)\sin (\large\frac{\pi}{4}$$-y)$
Put $\large\frac{\pi}{4}$$-x=A$ and $\large\frac{\pi}{4}$$-y=B$
$\cos A\cos B-\sin A\sin B$
$\Rightarrow \cos(A+B)$
$\Rightarrow \cos[(\large\frac{\pi}{4}$$-x)+\large\frac{\pi}{4}$$-y)]$
$\Rightarrow [\large\frac{\pi}{2}$$-(x+y)]$
$\cos (\large\frac{\pi}{2}$$-\theta)=\sin \theta$
$\Rightarrow \sin(x+y)$=R.H.S
Hence proved.
answered Apr 17, 2014 by sreemathi.v
 

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