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. If sin x + sin2x + sin3x= 1, then cos6x-4cos4x+8cos2x = ?

VasheelAhmed Pathan , 14 Years ago
Grade 11
anser 1 Answers
Rhinithaa PT

Last Activity: 14 Years ago

sinx+ (1-cos2x)+sinx(1-cos2x)=1

sinx(1+1-cos2x)+1-cos2x=1

sinx(1+1-cos2x)-cos2x=1-1=0

sinx(2-cos2x)-cos2x=0

2-cos2x = cos2x / sinx       [sinx =√1-cos2x]

2-cos2x = cos2x /√1-cos2x

squaring on both sides

(2-cos2x)2 = (cos2x)2 /(√1-cos2x)2

(4+cos4x-4cos2x)(1-cos2x) = cos4x

4+cos4x-4cos2x-4cos2x-cos6x+4cos4x= cos4x

4-8cos2x-cos6x+4cos4x=0

on rearranging

8cos2x+cos6x-4cos4x=4

hence solved

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