cos x + cos2 x = 1, then sin8 x + 2 sin6 x + sin4 x =?

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Answer 1 · 2018-07-05T14:40:11

#color(red)"LONG PROOF AHEAD"#

#color(blue)(cosx+cos^2x=1,#
then #color(green)(cosx=1-cos^2x=sin^2x.#

#sin^12x+3sin^10x+3 sin^8x+sin^6x+2 sin^4x+2sin^2x-2 #

#= cos^6x+3cos^5x+3 cos^4x+cos^3x+2 cos^2x+2cosx-2 #

#=cos^6x+3cos^5x+3 cos^4x+cos^3x #

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

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

#= (1-cosx)^2 (4-cosx)+cosx((1-cosx)(4-3cosx) #

#= (1-2cosx+1-cosx)(4-cosx)+cosx(4-7cosx+3-cos x #

#= (2-3cosx)(4-cosx)+ cosx(7-10cosx) #

#= 8-14cosx+3cos^2x+7cosx-10cos^2x #

#= 8-7cosx-7cos^2x=8-7(1) #

#= 1#

Answer 2 · 2018-07-09T18:35:36

See a Proof in Explanation.

Given that, #cosx+cos^2x=1#.

#:. cosx=1-cos^2x=sin^2x...................(star)#.

Now, #sin^8x+2sin^6x+sin^4x#,

#=sin^4x(sin^4x+2sin^2x+1)#,

#=sin^4x(sin^2x+1)^2#,

#={sin^2x(sin^2x+1)}^2#,

#={cosx(cosx+1)}^2.............[because, (star)]#,

#=(cos^2x+cosx)^2#,

#=1^2...............[because," Given]"#,

#=1#, as Reyan Roberth has already derived!