Question

If cos a + cos b + cos c = 0 = sin a + sin b + sin c

cos²a + cos²b + cos²c = ?

Answer 1

Sina + Sinb = -Sinc -----------------(1)
Cosa + Cosb = -Cosc ----------------(2)

Sq and add (1),(2)
2+2Cos(a-b) = 1 -----------------------(3)

Sq and Sub (2),(1)
Cos2a + Cos2b + 2Cos(a+b) = Cos2c
2Cos(a+b)Cos(a-b) + 2Cos(a+b) = Cos2c
or Cos(a+b){2Cos(a-b) + 2} = Cos2c
or Cos(a+b) = Cos2c ----------- [Using (3)]

Now Sina+Sinb+Sinc = 0
Squaring, Sin²A+Sin²B+Sin²C = -2∑SinASinB ----------------------------(4)

Now Cos(a+b) = Cos2c
Similarly Cos(b+c) = Cos2a
and Cos(c+a) = Cos2b
So adding these, ∑CosaCosb - ∑SinaSinb = ∑Cos2a

Also Cos(a-b) = -1/2 [from (3)]
similarly Cos(b-c) = -1/2
and Cos(c-a) = -1/2
Adding ∑Cosa Cosb + ∑SinaSinb = -3/2

So -2∑Sina Sinb = ∑Cos2a + 3/2

Hence from (4) Sin²a+Sin²b+Sin²c = -2∑SinaSinb= ∑Cos2b + 3/2

Now Cos2a+Cos2b+Cos2c = 2Cos(a+b)Cos(a-b) + Cos2c = -Cos(a+b) + Cos2c = 0 ------ [because Cos(a+b) = Cos2c; and Cos(a-b) = -1/2 which was proved previously]
Hence ∑Cos2a = 0.

Cos²a+Cos²b+Cos²c=3/2