If sin x + sin2 x = 1, then write the value of cos12 x + 3 cos10 x + 3 cos8 x + cos6 x.

Given: sin x + sin2x = 1


To find the value of cos12 x + 3 cos10 x + 3 cos8 x + cos6 x.


⇒ sin x = 1 – sin2x


⇒ sin x = cos2x


⇒ cos12x = sin6x, cos10x = sin5x, cos8x = sin4x, cos6x = sin3x.


Substituting above values in given equation we get


⇒ sin6x + 3sin5x + 3sin8x + sin3x [(a+b)3 = a3+3a2b+3ab2+b3]


⇒ (sin x + sin2 x)3 = (1)3


⇒ 1.


1