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VasheelAhmed Pathan Grade: 11
        

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

6 years ago

Answers : (1)

Rhinithaa PT
19 Points
										

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

6 years ago
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