Question

Expand `sin^6 x`?

Answer 1

`sin^6(x)=-1/32(cos(6x)-6cos(4x)+15cos(2x)-10)`

Explanation:

We want to expand

`sin^6(x)`

One way is to use these identities repeatedly

• `sin^2(x)=1/2(1-cos(2x))`
• `cos^2(x)=1/2(1+cos(2x))`

This often gets quite long, which sometimes leads to mistake

Another way is to use the complex numbers (and Euler's formula)

We can express sine and cosine as

`color(red)(sin(x)=(e^(ix)-e^(-ix))/(2i))` and `color(red)(cos(x)=(e^(ix)+e^(-ix))/2)`

Thus

`sin^6(x)=((e^(ix)-e^(-ix))/(2i))^6`

`=(e^(ix)-e^(-ix))^6/(-64)`

`=(e^(6ix)-6e^(4ix)+15e^(2ix)-20+15e^(-2ix)-6e^(-4ix)+e^(-6ix))/(-64)`

`=-1/32(e^(6ix)+e^(-6ix)-6e^(4ix)-6e^(-4ix)+15e^(2ix)+15e^(-2ix)-20)/2`

`=-1/32(cos(6x)-6cos(4x)+15cos(2x)-10)`

The third way is using De Moivre's theorem, we can express

`color(blue)(2cos(nx)=z^n+1/z^n)` and `color(blue)(2isin(nx)=z^n-1/z^n`

where `z^n=(cos(x)+isin(x))^n=cos(nx)+isin(nx)`

Thus

`(2isin(x))^6=(z-1/z)^6`

`=>sin^6(x)=-1/64(z-1/z)^6`

Expand the binomial on the right hand side

`BIN=(z^6+1/z^6-6z^4-6/z^4+15z^2+15/z^2-20)`

`color(white)(RHS)=(z^6+1/z^6-6*(z^4+1/z^4)+15*(z+1/z)+20)`

`color(white)(RHS)=(2cos(6x)-12cos(4x)+30cos(2x)-20)`

Thus

`sin^6(x)=-1/64(2cos(6x)-12cos(4x)+30cos(2x)-20)`

`sin^6(x)=-1/32(cos(6x)-6cos(4x)+15cos(2x)-10)`