How would you use the formulas for lowering powers to rewrite the expression in terms of the first power of cosine? sin^6(x)

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Answer 1 · 2018-02-28T06:59:43

#rarrsin^6x=1/32[cos6x+6cos4x-9cos2x+10]#

#rarrsin^6x#

#=(sin^2x)^3#

#=[(1-cos2x)/2]^3#

#=1/8[1-3cos2x+3cos^2(2x)+cos^3(2x)]#

#=1/16[2-6cos2x+3*(2cos^2(2x))+2cos^2(2x)*cos(2x)]#

#=1/16[2-6cos2x+3*(1+cos4x)+(1+cos4x)cos2x]#

#=1/16[2-6cos2x+3+3cos4x+cos2x+cos4x*cos2x]#

#=1/16[3cos4x-5cos2x+cos4x*cos2x+5]#

#=1/32[6cos4x-10cos2x+2cos4x*cos2x+10]#

#=1/32[6cos4x-10cos2x+cos6x+cos2x+10]#

#=1/32[cos6x+6cos4x-9cos2x+10]#