Question

Question #7cfc8

Answer 1

Proof below

Explanation:

First we will find the expansion of `sin(3x)` separately (this will use the expansion of trig functions formulae):
`sin(3x)=sin(2x+x)`
`=sin2xcosx+cos2xsinx`
`=2sinxcosx*cosx+(cos^2x-sin^2x)sinx`
`=2sinxcos^2x+sinxcos^2x-sin^3x`
`=3sinxcos^2x-sin^3x`
`=3sinx(1-sin^2x)-sin^3x`
`=3sinx-3sin^3x-sin^3x`
`=3sinx-4sin^3x`

Now to solve the original question:
`(sin3x)/(sinx)=(3sinx-4sin^3x)/sinx`
`=3-4sin^2x`
`=3-4(1-cos^2x)`
`=3-4+4cos^2x`
`=4cos^2x-1`
`=4cos^2x-2+1`
`=2(2cos^2x-1)+1`
`=2(cos2x)+1`