If #sinx+sin^(2)x+sin^3x=1#, then what is the correct option for #cos^6x-4cos^4x+8cos^2x#?

#a)1#
#b)2#
#c)3#
#d)4#

2 Answers
Abhishek K.
Jan 14, 2018

The correct option is #(d)#

Explanation:

Provided that
#rarrsinx+sin^2x+sin^3x=1#

#rarrsinx+sin^3x=1-sin^2x#

#rarrsinx(1+sin^2x)=cos^2x#

#rarr[sinx(1+sin^2x)]^2=[cos^2x]^2#

#rarrsin^2x(2-cos^2x)^2=cos^4x#

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

#rarr4-4cos^2x+cancel(cos^4x)-cos^2x(4-4cos^2x+cos^4x)=cancel(cos^4x)#

#rarr4-4cos^2x-4cos^2x+4cos^4x-cos^6x=0#

#rarr4-8cos^2x+4cos^4x-cos^6x=0#

#rarrcos^6x-4cos^4x+8cos^2x=4#

So the correct option is #4#.

P dilip_k
Jan 14, 2018

#sinx+sin^(2)x+sin^3x=1#

#=>sinx+sin^3x=1-sin^2x#

#=>sinx+sin^3x=cos^2x#

#=>(sinx+sin^3x)^2=cos^4x#

#=>(sin^2x+2sin^4x+sin^6x=cos^4x#

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

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

#=>1-8cos^2x+2+5cos^4x+1-cos^6x=cos^4x#

#=>cos^6x-4cos^4x+8cos^2x=4#

Then the correct option for #cos^6x-4cos^4x+8cos^2x# is (d)

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