Question

cos 4x = 1 - 8 cos^2x + 8 cos^4x
Prove LHS= RHS​

Answer 1

To prove :

$\sf \: \cos(4x) = 1 - 8 \cos {}^{2} (x) + 8 { \cos }^{4} (x)$

Consider LHS,

$\sf \: LHS = \cos(4x) \\ \\ \implies \sf \: LHS = 2 { \cos }^{2} (2x) - 1 \\ \\ \implies \sf \: LHS = 2 \bigg \{ { 2 { \cos}^{2}(x) }^{} -1 \bigg \} {}^{2} - 1 \\ \\ \implies \sf \: LHS = 2 \bigg \{ 4 { \cos {}^{4} (x) }^{} - 4 \cos {}^{2} (x) + 1 \bigg \}- 1 \\ \\ \implies \sf \: LHS = 8 { \cos {}^{4} (x) }^{} - 8 \cos {}^{2} (x) + 2- 1 \\ \\ \implies \sf \: LHS = 1 - 8 \cos {}^{2} (x) + 8 { \cos }^{4} (x) \\ \\ \implies \sf \: LHS = RHS$