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sina+simb+sinc=cosa+cosb+cosc=0 than prove that cos2a+cos2b+cos2c=0 and sin2a+sin2b+sin2c+0

sina+simb+sinc=cosa+cosb+cosc=0
than prove that cos2a+cos2b+cos2c=0 and sin2a+sin2b+sin2c+0
 

Grade:12th pass

1 Answers

Y RAJYALAKSHMI
45 Points
6 years ago
Let cosa + i sina = x ; cosb + i sinb = y; cosc + i sinc = z
then 1/x = cosa –  i sina ; 1/y = cosb –  i sinb ; 1/z = cosc –  i sinc 
x + y + z = cosa + i sina + cosb + i sinb + cosc + i sinc  = (cosa + cosb + cosc) + i (sina + sinb + sinc) = 0  (Since cosa + cosb + cosc = sina + sinb + sinc = 0)
(x + y + z)2  = 0 
=> x2 + y2 + z2 = – 2(xy + yz + zx)
= – 2xyz ( 1/x + 1/y + 1/z)
= –2xyz (  cosa –  i sina + cosb –  i sinb +  cosc –  i sinc)
= – 2xyz [( cosa + cosb +  cosc) –  i(sina + sinb + sinc) ] = 0  (Since cosa + cosb + cosc = sina + sinb + sinc = 0)
So, x2 + y2 + z2 = 0
=> (cosa –  i sina)2 + (cosb –  i sinb)2 +  (cosc –  i sinc)2 = 0  
=> cos2a – sin2a – 2i cosa sina   + cos2b – sin2b – 2i cosb sinb + cos2c – sin2c – 2i cosc sinc = 0 
=> (cos 2a + cos 2b + cos 2c) – i (sin 2a + sin 2b + sin 2c) = 0
=> (cos 2a + cos 2b + cos 2c) =  (sin 2a + sin 2b + sin 2c) = 0
 
 
 
 
 
 
 
 

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