if cosx+cosy+cosz=sinx+siny+sinz=0 then prove that sin2x+sin2y+sin2z=cos2x+cos2y+cos2z=0

Question

Nishant Vora · Answer

hello student,,

cosx + cosy = -cosz , cos^2x + cos^2y + 2cosx.cosy = cos^2z , in a similar way : sin^2x+sin^2y+2sinx.siny = sin^2z , add eqns & simplify : 1+1+2(cosx.cosy+sinx.siny) = 1 , 2cos(x-y)=1-2=-1 , cos(x-y)=-1/2

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Prashant Gulia · Answer

then what we have to do

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if cosx+cosy+cosz=sinx+siny+sinz=0 then prove that sin2x+sin2y+sin2z=cos2x+cos2y+cos2z=0

if cosx+cosy+cosz=sinx+siny+sinz=0 then prove that sin2x+sin2y+sin2z=cos2x+cos2y+cos2z=0

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if cosx+cosy+cosz=sinx+siny+sinz=0 then prove that sin2x+sin2y+sin2z=cos2x+cos2y+cos2z=0

if cosx+cosy+cosz=sinx+siny+sinz=0 then prove that sin2x+sin2y+sin2z=cos2x+cos2y+cos2z=0

2 Answers

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