If \sin x+\sin^2x+\sin^3x=1, then find \cos^6x-4\cos^4x+8\cos^2x

mairie0708zl

mairie0708zl

Answered question

2022-01-27

If sinx+sin2x+sin3x=1, then find cos6x4cos4x+8cos2x

Answer & Explanation

stamptsk

stamptsk

Beginner2022-01-28Added 23 answers

Let us denote y=sinx. The relation we have for y is then y+y2+y3=1 or also if we multiply by y1, we get y4=2y1. The idea is simply to write the expression in cos2x given in terms of y, and use the relation to simplify it. We have
cos6x4cos4x+8cos2x=(1y2)34(1y2)2+8(1y2)
=(1y2)[(12y2+y4)+(4y24)+8]
=(1y2)[5+2y2+y4]
=(1y2)[5+2y2+(2y1)]
=2(1y2)[2+y+y2]
=2[2+y+y22y2y3y4]
=2[2+y+(y)(y+y2+y3)]
=2[2+yy]
=4
Jacob Trujillo

Jacob Trujillo

Beginner2022-01-29Added 13 answers

Given
sinx+sin2x+sin3x=1
sinx+sin3x=1sin2x
(sinx+sin3x)2=(1sin2x)2
sin2x+sin6x+2sin4x=cos4x
1cos2x+(1cos2x)3+2(1cos2x)2=cos4x
1cos2x+13cos2x+3cos4xcos6x+14cos2x+2cos4x=cos4x
cos6x4cos4x+8cos2x=4

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