Question #16a52

1 Answer
P dilip_k
Jan 26, 2018

Given #sinx+sin^2x=1 #

#=>sinx=1-sin^2x #

#=>sinx=cos^2x #

Now

#cos^12x+3cos^10x+3cos^8x+cos^6x-1#

#=cos^6x(cos^6x+3cos^4x+3cos^2x+1)-1#

#=cos^6x(cos^2x+1)^3-1#

#={cos^2x(cos^2x+1)}^3-1#

#=(cos^4x+cos^2x)^3-1#

#=((cos^2x)^2+cos^2x)^3-1#

Putting #cos^2x=sinx #

#=(sin^2x+cos^2x)^3-1#

#=1^3-1=0#

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