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Dear Student,
We have,sinx+sin2x=1We can write it as;sinx=1-sin2xsinx=cos2x---1Squaring both sides we get;sin2x=cos4x---2Againg squaring both sides we get;sin4x=cos8x---3Now, multiplying 1 & 2 and 1 & 3 and 2 & 3 we get;sin3x=cos6x---4sin5x=cos10x ---5and,sin6x=cos12x ---6And we have the expression;cos12x+3cos10x+3cos8x+cos6xsubstituting from 4, 5 and 6 we get;sin6x+3sin5x+3sin4x+sin3xWe can write it as,sin2x3+3×sin2x2×sinx+3×sin2x×sinx2+sinx3Using identity: a+b3=a3+b3+3a2b+3ab2we get;sinx+sin2x3=13=1

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