Question

prove that cos 3 X is equals to cos 4 cube x minus 3 cos x​

Answer 1

Answer:

$4 { \cos }^{3} x - 3 \cos(x) \\ \\ = cosx \: (4 \: { \cos }^{2} x - 3) \\ \\ = cosx ({ \cos }^{2} x + 3( { \cos }^{2} x - 1) \\ \\ \\ = \cos(x) ( { \cos }^{2}x \: - 3 { \sin }^{2} x) \\ \\ = \cos(x)( { \cos }^{2} x - { \sin }^{2} x - 2 { \sin }^{2} x) \\ \\ = cos(x) ( \cos(2x) - 2 + 2 { \cos }^{2} x) \\ \\ = \cos(x) \cos(2x) - 2 \cos(x) + 2 { \cos }^{3} x \\ \\ = \frac{1}{2} \times ( \cos(3x) + \cos(x) ) - 2 \cos(x) + 2 { \cos }^{3} x \\ \\ = \frac{ \cos(3x) }{2} + \frac{ \cos(x) }{2} - 2 \cos(x) + 2 { \cos }^{3} x \\ \\ = \frac{ \cos(3x) }{2} \: - \frac{3 \cos(x) }{2} + 2 { \cos }^{3} x \\ \\ = \frac{ \cos3x }{2} - \frac{3 \cos(x) + 4 { \cos }^{3}x }{2} \\ \\ = \frac{ \cos(3x) }{2} + \frac{ \cos(3x) }{2} \\ \\ = \cos(3x) \: as \: proved$