Expand #sin^6 x#?

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Øko Share
Mar 9, 2018

Answer:

#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)#

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