# Prove the following: cos^{2}2x - cos^{2}6x = sin 4x sin 8x

**Solution:**

LHS = cos^{2}2x - cos^{2}6x

= (cos 2x + cos 6x) (cos 2x - cos 6x)

[Using a² - b² formula]

= [2cos {(2x + 6x) / 2} cos {(2x - 6x) / 2}] × [ -2 sin {(2x + 6x) / 2} sin {(2x - 6x) / 2}]

{Since cos A + cos B = 2cos [(A + B) / 2] cos [(A - B) / 2] and cos A - cos B = -2sin [(A + B) / 2] sin [(A - B) / 2]}

= [2cos 4x cos (-2x)] × [-2sin 4x sin (-2x)]

= [2cos 4x cos 2x] × [-2sin 4x (-sin 2x)]

[By trigonometric formulas, cos (-A) = cos A and sin (-A) = -sin A]

= [2cos 4x cos 2x] × [2sin 4x sin 2x]

= [2cos 4x sin 4x] × [2cos 2x sin 2x]

= [sin (4x + 4x) - sin (4x - 4x)] × [sin (2x + 2x) - sin (2x - 2x)]

[Since 2cos A sin B = sin (A + B) - sin (A - B)]

= [sin 8x + sin 0] × [sin 4x + sin 0]

= sin 8x × sin 4x

[by trigonometric table sin 0 = 0]

= sin 4x sin 8x

= RHS

NCERT Solutions Class 11 Maths Chapter 3 Exercise 3.3 Question 13

## Prove the following: cos^{2}2x - cos^{2}6x = sin 4x sin 8x

**Summary:**

We got, cos^{2}2x - cos^{2}6x = sin 4x sin 8x. Hence Proved

visual curriculum