Question

if cos(a-b)+cos(b-c)+cos(c-a)=-3/2 then prove that cos a +cos b+ cos c = sin a+ sin b+sin c=0

Answer 1

its a bit confusing so if you face any problem..feel free to ask me....

Answer 2

Cos a + Cos b + Cos c = Sin a + Sin b + Sin c = 0

Step-by-step explanation:

From question,

Cos(a-b) = Cos a Cos b +Sin a SIn b

Using above identity in the given question and taking $\frac{-3}{2}$ on other side, the result is:

⇒ 2(Cos a Cos b + Sin a Sin b + Cos c Cos b + Sin c Sin b + Cos a Cos c + Sin a Sin c) + 3 = 0

We can write 3 as 1 + 1 + 1,     ...........    (1)

So, we know  $Sin^{2}a + Cos^{2} a = 1$

applying above equation in equation (1),

⇒ 2(Cos a Cos b + Sin a Sin b + Cos c Cos b + Sin c Sin b + Cos a Cos c + Sin a Sin c) + $Sin^{2}a + Cos^{2} a$ +$Sin^{2}b + Cos^{2} b$ + $Sin^{2}c + Cos^{2} c$  = 0     ........   (2)

We know, $x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2yz = 0$

Using above identity to equation (2),

⇒ $(Cos a + Cos b + Cos c)^{2}$ + $(Sin a + Sin b + Sin c)^{2}$ = 0

Hence, sum of squares of above terms are zero. So, these terms are also 0.

Hence proved.