Question

Show using complex numbers that 8sin^4x = cos4x - 4cosx + 3?

Please help. I don't know how to solve this using complex numbers

Answer 1

Please see the explanation below.

Explanation:

With Euler's Identity

`e^(ix)=cosx+isinx`

`e^(-ix)=cosx-isinx`

From those equations , we get

`sinx=(e^(ix)-e^(-ix))/(2i)`

`cosx=(e^(ix)+e^(-ix))/2`

`i^2=-1`

`i^4=1`

Therefore, by the binomial theorem

`(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4`

`8sin^4x=8*(e^(ix)-e^(-ix))^4/(2i)^4`

`=1/2(e^(i4x)+e^(-i4x)-4e^(i3x)*e^(-ix)-4e^(ix)*e^(-3ix)+6e^(2ix)*e^(-2ix))`

`=(e^(i4x)+e^(-i4x))/2-4(e^(i2x)+e^(-i2x))/2+6/2`

`=cos4x-2cos2x+3`

I think that the middle term is `2cos2x` instead of `4cosx`