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# Prove that: (cos x - cos y)² + (sin x - sin y)² = 4sin²[(x - y) / 2]

**Solution:**

LHS = (cos x - cos y)^{2} + (sin x - sin y)^{2}

= cos^{2}x + cos^{2}y - 2cos x cos y + sin^{2}x + sin^{2}y - 2sin x sin y [Because (a - b)² = a² + b² - 2ab ]

= 1 + 1 - 2cos (x - y) [Because cos^{2}A + sin^{2}A = 1 and cos (A + B) = cos A cos B + sin A sin B]

= 2 - 2cos (x - y)

= 2[1 - cos {2(x - y) / 2} ]

= 2[1 - (1 - 2sin^{2}{(x - y) / 2})] [By double angle formulas, cos 2A = 1 - 2sin^{2}A]

= 2[1 - 1 + 2sin^{2}{(x - y) / 2}]

= 4sin^{2}[(x - y) / 2]

= RHS

NCERT Solutions Class 11 Maths Chapter 3 Exercise ME Question 4

## Prove that: (cos x - cos y)² + (sin x - sin y)² = 4sin²[(x + y) / 2]

**Summary:**

We got, (cos x - cos y)^{2} + (sin x - sin y)^{2} = 4sin^{2}[(x + y) / 2]. Hence Proved.

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