Question

write interms of first powers of Cosine
sin4 (3x)​

Answer 1

Answer:

sin4(2x) = sin2(2x) * sin2(2x)

An alternate version of the half-angle formula is that

sin2x = (1/2) [1 - cos(2x)] or sin2(2x) = (1/2) [ 1 - cos(4x) ]

So we replace both sin2(2x) with the half-angle version:

sin2(2x) * sin2(2x) = (1/2) [1-cos(4x)] (1/2) [1 - cos(4x)]

this simplifies to

(1/4)[1-cos(4x)]2

Then we "FOIL" the squared expression.

(1/4) [ 1 - 2cos(4x) + cos2(4x)]

Distribute:

(1/4) - (1/2) cos(4x) + (1/4) cos2(4x)

Another alternate form of the half-angle formula is

cos2x = (1/2)[ 1 + cos(2x)] or cos2(4x) = (1/2)[ 1 + cos(8x) ]

Replacing the remaining squared term with it's half-angle equivalent

(1/4) - (1/2)cos(4x) + (1/4)[(1/2)(1+cos(8x))]

Simplify:

(1/4) - (1/2)cos(4x) + (1/8)[ 1 + cos(8x) ]

Distribute:

(1/4) - (1/2)cos(4x) + (1/8) + (1/8)cos(8x)

Last step: combine both constants and get

(3/8) - (1/2)cos(4x) + (1/8)cos(8x)