Question

Prove that: cos6x= 32cos^6x - 48 cos^4x +18cos^2x -1

Answer 1

LHS :

cos 6x

=> cos2(3x)

=> 2 cos² (3x) - 1

=> 2 (cos 3x)² - 1

=> 2(4 cos³x - 3 cos x)² - 1

=> 2(16 cos⁶ x + 9 cos² x - 24 cos⁴ x) - 1

=> 32 cos⁶x + 18 cos² x - 48 cos⁴ x - 1

RHS

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Answer 2

$\large \: cos6x \\ \\ = cos3.2x \\ \\ = 4 {cos}^{3} 2x - 3cos2x \\ \\ = 4(2 {cos}^{2} x - 1) ^{3} - 3(2 {cos}^{2} x - 1) \\ \\ = 4(8 {cos}^{6} x - 12 {cos}^{4} x + 6 {cos}^{2} x - 1) - (6 {cos}^{2} x - 3) \\ \\ = 32 {cos}^{6} x - 48 {cos}^{4} x + 24 {cos}^{2} x - 4 - 6 {cos}^{2} x + 3 \\ \\ = 32 {cos}^{6} x - 48 {cos}^{4} x + 18 {cos}^{2} x - 1$