Question

Question #d42f6

Answer 1

Expand this function, using trig identities:
`sin^2 x = 1 - cos^2 x`
`sin 2x = 2sin x.cos x`
We get:
`sin^4 x = sin^2 x.sin^2 x = sin^2 x(1 - cos^2 x) =`
`= sin^2 x - sin^2 x.cos^2 x = sin^2 x - (sin^2 (2x))/4 =`
`= (sin x - (sin 2x)/2)(sin x + (sin 2x)/2)`

Answer 2

`rarrsin^4x`

`=(sin^2x)^2`

`=((1-cos2x)/2)^2`

`=(1-2cos2x+cos^2(2x))/4`

`=1/4[1-2cos2x+(1+cos4x)/2)]`

`=1/4[(2-4cos2x+1+cos4x)/2]`

`=1/8[3-4cos2x+cos4x]`