Question

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

Answer 1

all about expansion

Answer 2

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