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. If sin x + sin 2 x + sin 3 x= 1, then cos 6 x-4cos 4 x+8cos 2 x = ?

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

Grade:11

1 Answers

Rhinithaa PT
19 Points
10 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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