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NCERT Class XI Mathematics - Trigonometric Functions - Solutions

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Question : 55 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
= [−2sin⁡(x+y2)sin⁡(x−y2)]2\left[-2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)\right]^2 + [2cos⁡(x+y2)sin⁡(x−y2)]2\left[2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)\right]^2
= 4sin⁡2(x+y2)sin⁡2(x−y2)4\sin^2\left(\frac{x+y}{2}\right)\sin^2\left(\frac{x-y}{2}\right) + 4cos⁡2(x+y2)sin⁡2(x−y2)4\cos^2\left(\frac{x+y}{2}\right)\sin^2\left(\frac{x-y}{2}\right)
= 4sin⁡2(x−y2)4\sin^2\left(\frac{x-y}{2}\right) [sin⁡2(x+y2)+cos⁡2(x+y2)]\left[\sin^2\left(\frac{x+y}{2}\right)+\cos^2\left(\frac{x+y}{2}\right)\right]
= 4sin⁡2(x−y2)4\sin^2\left(\frac{x-y}{2}\right) = R.H.S.
Since sin⁡2x+cos⁡2x\sin^2 x + \cos^2 x = 1
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