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NCERT Class XI Mathematics - Trigonometric Functions - Solutions

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Question : 54 of 61
Marks: +1, -0
(cos⁡x+cos⁡y)2(\cos x + \cos y)^2 + (sin⁡x−sin⁡y)2(\sin x - \sin y)^2 = 4cos⁡2x+y24\cos^2 \frac{x+y}{2}
Solution:  
We have,
L.H.S. = (cos⁡x+cos⁡y)2(\cos x + \cos y)^2 + (sin⁡x−sin⁡y)2(\sin x - \sin y)^2 = 4cos⁡2x+y24\cos^2 \frac{x+y}{2}
= (2cos⁡x+y2cos⁡x−y2)2\left(2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}\right)^2 + (2cos⁡x+y2sin⁡x−y2)2\left(2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}\right)^2
= 4cos⁡2x+y24 \cos^2 \frac{x+y}{2} [cos⁡2x−y2+sin⁡2x−y2]\left[ \cos^2 \frac{x-y}{2} + \sin^2 \frac{x-y}{2} \right]
= 4cos⁡2x+y24 \cos^2 \frac{x+y}{2} = R.H.S.
Since sin⁡2x+cos⁡2x\sin^2 x + \cos^2 x = 1
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