Question

if cos(a-b)+cos(b-c)+cos(c-a)=-1/2 ,then prove that cos a+cos b+cos c =0

Answer 1

Answer:

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}

2

−3

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 = 1Sin

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} aSin

2

a+Cos

2

a +Sin^{2}b + Cos^{2} bSin

2

b+Cos

2

b + Sin^{2}c + Cos^{2} cSin

2

c+Cos

2

c = 0 ........ (2)

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

2

+y

2

+z

2

+2xy+2yz+2yz=0

Using above identity to equation (2),

⇒ (Cos a + Cos b + Cos c)^{2}(Cosa+Cosb+Cosc)

2

+ (Sin a + Sin b + Sin c)^{2}(Sina+Sinb+Sinc)

2

= 0

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

Hence proved.

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

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