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# Prove the following: cos 6x = 32cos⁶x - 48cos⁴x + 18cos²x - 1

**Solution:**

LHS = cos 6x

= cos 3(2x)

= 4cos^{3}2x - 3cos2x [As cos 3A = 4cos^{3}A - 3cosA]

= 4(2cos^{2}x - 1)^{3} - 3(2cos^{2}x - 1) [By double angle formula, cos 2A = 2cos^{2}A - 1]

= 4(8cos^{6}x - 1 - 12cos^{4}x + 6cos^{2}x) - 6cos^{2}x + 3 [Because (a - b)³ = a³ - b³ - 3a²b+ 3ab²]

= 32cos^{6}x - 4 - 48cos^{4}x + 24cos^{2}x - 6cos^{2}x + 3

= 32cos^{6}x - 48cos^{4}x + 18cos^{2}x - 1

= RHS

NCERT Solutions Class 11 Maths Chapter 3 Exercise 3.3 Question 25

## Prove the following: cos 6x = 32cos⁶x - 48cos⁴x + 18cos²x - 1

**Summary:**

We got, cos 6x = 32cos^{6}x - 48cos^{4}x + 18cos^{2}x - 1. Hence Proved

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