Prove this immidiately
asked Sep 28, 2013 in Trigonometry Answers by bharat (120 points)

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We will take RHS and prove LHS, so it is given that, cosx+cosy+cosz=0......(1)sinx+siny+sinz=0.......(2)Now multiplying equation 1 by cosy , cos z, cosx one by one we will get threeequation as, cosycosx+ cos 2 y+coszcosy=0......(3) cosxcosz+cosycosz+ cos 2 z=0......(4) cos 2 x+cosycosx+coszcosx=0......(5) ANd now multiply equation 2 by sinx, siny , sinz then we will get three equation as, sin 2 x+sinysinx+sinzsinx=0.......(6) sinxsiny+ sin 2 y+sinzsiny=0.......(7) sinxsinz+sinysinz+ sin 2 z=0.......(8) Add all the six equation we get,2cosxcosy+2sinxsiny+2cosycosz+2sinysinz+2coszcosx+2sinzsinx+ + sin 2 x+ cos 2 x+ cos 2 y+ sin 2 y+ sin 2 z + cos 2 z=0 2cos(x−y)+2cos(y−z)+2cos(z−x)+3=02cos(x−y)+2cos(y−z)+2cos(z−x)=−3cos(x−y)+cos(y−z)+cos(z−x)=− 3 2 =LHS
answered Jun 11 by R.Hari Charan.

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