Question

i need help, i am able to reduce cosine but get stuck with sin, any help is appreciated! with steps please

Answer 1

It's d. 1/16( 1 - cos 2t - cos 4t + cos4t cos 2t).

Explanation:

cos 2t = 2 cos^2 t - 1

cos^2 t = 1/2( 1 + cos 2t) -----------(1)

cos 2t = 1 -2 sin^2 t

sin^2 t = 1/2 ( 1 - cos 2t)

sin^4 t = ( sin^2 t)^2 = 1/4(1 - cos 2t)^2

= 1/4( 1 - 2 cos 2t + cos^2 2t)

Now cos 4t = 2 cos^2 2t - 1

cos^2 2t = 1/2(1 + cos 4t)

So sin^4 t = 1/4(1 - 2 cos 2t + 1/2 + 1/2 cos 4t)

= 1/4( 3/2 - 2 cos 2t + 1/2 cos 4t)

= 3/8 - 1/2 cos 2t + 1/8 cos 4t. .......... (2).

Multiplying equations (1) and (2):

sin^2 t cos^2 t = 1/2( 1 + cos 2t)(3/8 - 1/2 cos 2t + 1/8 cos 4t)

= 1/2(1 + cos 2t)1/8(3 - 4cos 2t + cos 4t)

= 1/16( 3 - 4 cos 2t + cos 4t + 3 cos 2t - 4 cos^2 2t + cos 2t cos 4t)

Now cos^2 2t = 1/2 + 1/2cos 4t so we have

1/16( 3 - cos 2t + cos 4t - 4(1/2 + 1/2 cos4t ) + cos 2t cos 4t)

= 1/16( 3 - cos 2t + cos 4t - 2 - 2 cos 4t + cos 2t cos 4t)

= 1/16( 1 - cos 2t - cos 4t + cos4t cos 2t) Answer.