Expand sin^6 xsin6x?

1 Answer
Øko
Mar 9, 2018

sin^6(x)=-1/32(cos(6x)-6cos(4x)+15cos(2x)-10)sin6(x)=−132(cos(6x)−6cos(4x)+15cos(2x)−10)

Explanation:

We want to expand

sin^6(x)sin6(x)

One way is to use these identities repeatedly

  • sin^2(x)=1/2(1-cos(2x))sin2(x)=12(1−cos(2x))
  • cos^2(x)=1/2(1+cos(2x))cos2(x)=12(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))sin(x)=eix−e−ix2i and color(red)(cos(x)=(e^(ix)+e^(-ix))/2)cos(x)=eix+e−ix2

Thus

sin^6(x)=((e^(ix)-e^(-ix))/(2i))^6sin6(x)=(eix−e−ix2i)6

=(e^(ix)-e^(-ix))^6/(-64)=(eix−e−ix)6−64

=(e^(6ix)-6e^(4ix)+15e^(2ix)-20+15e^(-2ix)-6e^(-4ix)+e^(-6ix))/(-64)=e6ix−6e4ix+15e2ix−20+15e−2ix−6e−4ix+e−6ix−64

=-1/32(e^(6ix)+e^(-6ix)-6e^(4ix)-6e^(-4ix)+15e^(2ix)+15e^(-2ix)-20)/2=−132e6ix+e−6ix−6e4ix−6e−4ix+15e2ix+15e−2ix−202

=-1/32(cos(6x)-6cos(4x)+15cos(2x)-10)=−132(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)2cos(nx)=zn+1zn and color(blue)(2isin(nx)=z^n-1/z^n2isin(nx)=zn−1zn

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

Thus

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

=>sin^6(x)=-1/64(z-1/z)^6⇒sin6(x)=−164(z−1z)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)BIN=(z6+1z6−6z4−6z4+15z2+15z2−20)

color(white)(RHS)=(z^6+1/z^6-6*(z^4+1/z^4)+15*(z+1/z)+20)RHS=(z6+1z6−6⋅(z4+1z4)+15⋅(z+1z)+20)

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

Thus

sin^6(x)=-1/64(2cos(6x)-12cos(4x)+30cos(2x)-20)sin6(x)=−164(2cos(6x)−12cos(4x)+30cos(2x)−20)

sin^6(x)=-1/32(cos(6x)-6cos(4x)+15cos(2x)-10)sin6(x)=−132(cos(6x)−6cos(4x)+15cos(2x)−10)

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