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

= cos²x + cos²y + 2cos x cos y + sin²x + sin²y - 2sin x sin y [Because (a + b)² = a² + b² + 2ab and (a - b)² = a² + b² - 2ab ]

= 1 + 1 + 2cos (x + y) [Because cos²A + sin²A = 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 + 2cos²{(x + y) / 2} - 1] [By double angle formulas, cos 2A = 2cos²A - 1]

= 4cos²[(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 = 4cos²[(x + y) / 2]. Hence Proved.