Rewrite #2sin^6(x)# in terms of an expression containing only cosines to the power of one?

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Answer 1 · 2018-03-08T09:51:32

#2sin^6x=(10-cos(6x)+6cos(4x)-15cos(2x))/16#

We are given #2sin^6x#

Using De Moivre's Theorem we know that:
#(2isin(x))^n=(z-1/z)^n# where #z=cosx+isinx#

#(2isin(x))^6=-64sin^6x=z^6-6z^4+15z^2-20+15/z^2-6/z^4+1/z^6#

First we arrange everything together to get:
#-20+(z+1/z)^6-6(z+1/z)^4+15(z+1/z)^2#

Also, we know that #(z+1/z)^n=2cos(nx)#

#-64sin^6x=-20+(2cos(6x))-6(2cos(4x))+15(2cos(2x))#

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

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

#2sin^6x=2*(-20+2cos(6x)-12cos(4x)+30cos(2x))/-64=(-20+2cos(6x)-12cos(4x)+30cos(2x))/-32=(10-cos(6x)+6cos(4x)-15cos(2x))/16#

Answer 2 · 2018-03-08T14:50:44

#rarr2sin^6x=1/16[10-15cos2x+6cos4x-cos6x]#

#rarr2sin^6x#

#=1/4[(2sin^2x)^3]#

#=1/4[(1-cos2x)^3]#

#=1/4[1-3cos2x+3cos^2(2x)-cos^3(2x)]#

#=4/(4*4)[1-3cos2x+3cos^2(2x)-cos^3(2x)]#

#=1/16[4-12cos2x+3*2*{2cos^2(2x)}-4cos^3(2x)]#

#=1/16[4-12cos2x+3*2*{1+cos4x}-cos6x-3cos2x]#

#=1/16[4-15cos2x+6+6cos4x-cos6x]#

#=1/16[10-15cos2x+6cos4x-cos6x]#