Verified answer

Remember the formula

use the formula cos(A - B) + cos(A+B) = 2cosA cosB --------------1

LHS

=cos x+cos 3x+cos 5x+cos 7x

=cos x + cos(7x) + cos(3x) + cos(5x)

=cos(4x - 3x) + cos(4x + 3x) + cos(4x - x) + cos(4x + x)

= 2 cos(4x)cos(3x) + 2cos(4x)cos(x) (Using the formula 1)

= 2 cos(4x) [cos(3x) + cosx]

= 2 cos(4x)[cos(2x + x) + cos(2x - x)]

= 2cos(4x)[2cos(2x)cosx] (Using the formula 1)

=4 cos(4x) cos(2x) cos(x) = 4cosxcos2xcos4x = RHS

Cos 3x Formula

cos x+cos 3x+cos 5x+cos 7x

rearrange

=cosx + cos(7x) + cos(3x) + cos(5x)

write x = 4x - 3x : 7x = 4x + 3x : 3x = 4x - x and 5x = 4x + x

=cos(4x - 3x) + cos(4x + 3x) + cos(4x - x) + cos(4x + x)

use the formula cos(A - B) + cos(A+B) = 2cosA cosB

= 2cos(4x)cos(3x) + 2cos(4x)cos(x)

take out cos(4x) common

=2cos(4x)(cos(3x) + cosx)

again write 3x = 2x + x and x = 2x - x

=2 cos(4x)[cos(2x + x) + cos(2x - x)]

=2cos(4x)(2cos(2x)cosx)

=8 cos(4x) cos(2x) cos(x)

Yes, this is an identity. Is that the question?