Express cos4x as powers of cosx. ?

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Answer 1 · 2018-05-18T12:33:44

#cos4x=cos2(2x)=color(red)[2cos^2(2x)-1#

#cos2(2x)=cos^2(2x)-sin^2(2x)#

#=cos^2(2x)-1+cos^2(2x)=color(red)[2cos^2(2x)-1]#

#=2[cos2x*cos2x]-1=2[(cos^2x-sin^2x)*(cos^2x-sin^2x)]-1#

#=2[cos^4x-sin^2x*cos^2x-sin^2x*cos^2x+sin^4x]-1#

#=[2cos^4x-4sin^2x*cos^2x+2sin^4x]-1#

Answer 2 · 2018-05-18T15:55:43

#rarrcos4x=8cos^4x-8cos^2x+1#

#rarrcos4x#

#=cos2*(2x)#

#=cos^2(2x)-sin^2(2x)#

#=[cos^2x-sin^2x]^2-[2sinx*cosx]^2#

#=cos^4x-2cos^2x*sin^2x+(sin^2x)^2-4sin^2x*cos^2x#

#=cos^4x-2cos^2x(1-cos^2x)+(1-cos^2x)^2-4(1-cos^2x)*cos^2x#

#=cos^4x-2cos^2x+2cos^4x+1-2cos^2x+cos^4x-4cos^2x+4cos^4x#

#=8cos^4x-8cos^2x+1#