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# Prove that: (cos x + cos y)² + (sin x - sin y)² = 4cos²[(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 and (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 + 2cos^{2}{(x + y) / 2} - 1] [By double angle formulas, cos 2A = 2cos^{2}A - 1]

= 4cos^{2}[(x + y) / 2]

= RHS

NCERT Solutions Class 11 Maths Chapter 3 Exercise ME Question 3

## Prove that: (cos x + cos y)² + (sin x - sin y)² = 4cos²[(x + y) / 2]

**Summary:**

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

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