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Question Number 116493 by bemath last updated on 04/Oct/20
Givensin2x+sinx=1,findthevalueofcos8x+cos6x+cos4x=?
Answered by bobhans last updated on 04/Oct/20
⇔sin2x+sinx=1;sinx=1−52⇔1−cos2x+sinx=1;cos2x=sinx⇒cos8x+cos6x+cos4x=⇒cos4x(cos4x+cos2x+1)=⇒cos4x(cos4x+sinx+1)=⇒cos4x(sin2x+sinx+1)=⇒cos4x(1+1)=2sin2x=2(1−52)2=2(6−254)=3−5
Answered by som(math1967) last updated on 04/Oct/20
sin2x+sinx=1sinx=1−sin2xsinx=cos2xcos8x+cos6x+cos4x=cos4x(cos4x+cos2x)+cos4x=cos4x(sin2x+cos2x)+cos4x=cos4x+cos4x=2cos4x
Answered by Dwaipayan Shikari last updated on 04/Oct/20
cos2x=sinxAndsinx=5−12andsin2x=1−sinxcos4x(cos4x+cos2x+1)sin2x(sin2x+cos2x+1)=2sin2x=2(1−sinx)=3−5
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