Expand #sin^6 x#?

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Answer 1 · 2018-03-09T08:39:44

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

We want to expand

#sin^6(x)#

One way is to use these identities repeatedly

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)#