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.