Question

If sina + sinB + siny = 0 = cosa + cosB + cosy, then show that cos(a - b) + cos(B - y) + cos(y - a) = -3/2

Answer 1

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Answer 2

Given,

Sin a + Sin B + Sin y = 0 = Cos a + Cos B + Cos y

To prove,

Cos (a - B) + Cos (B - y) + Cos (y - a) = -3/2

Solution,

We can simply solve this mathematical problem using the following process:

Mathematically,

For any two angles x and y;

(a) Cos (x-y) = Cos x.Cos y + Sin x.Sin y

(b) Sin^2 a + Cos^2 a = 1

{Statement-1}

For three variables a, b, and c, there exists an algebraic identity as follows;(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

{Statement-2}

Now, according to the question;

L.H.S.

= Cos (a - B) + Cos (B - y) + Cos (y - a)

= Cos a.Cos B + Sin a.Sin B + Cos B.Cos y + Sin B.Sin y + Cos a.Cos y + Sin a.Sin y

{according to statement-1 (a)}

= 1/2 × 2{Cos a.Cos B + Sin a.Sin B + Cos B.Cos y + Sin B.Sin y + Cos a.Cos y + Sin a.Sin y}

= 1/2 × {2Cos a.Cos B + 2Sin a.Sin B + 2Cos B.Cos y + 2Sin B.Sin y + 2Cos a.Cos y + 2Sin a.Sin y}

= 1/2 × {(2Cos a.Cos B + 2Cos B.Cos y + 2Cos a.Cos y) + (2Sin a.Sin B + 2Sin B.Sin y + 2Sin a.Sin y)}

= 1/2 × [{(Cos^2 a + Cos^2 B + Cos^2 y) + (2Cos a.Cos B + 2Cos B.Cos y + 2Cos a.Cos y)} + {(Sin^2 a + Sin^2 B + Sin^2 y) + (2Sin a.Sin B + 2Sin B.Sin y + 2Sin a.Sin y)} - {Cos^2 a + Cos^2 B + Cos^2 y + Sin^2 a + Sin^2 B + Sin^2 y}]

= 1/2 × [{(Cos^2 a + Cos^2 B + Cos^2 y) + (2Cos a.Cos B + 2Cos B.Cos y + 2Cos a.Cos y)} + {(Sin^2 a + Sin^2 B + Sin^2 y) + (2Sin a.Sin B + 2Sin B.Sin y + 2Sin a.Sin y)} - {(Sin^2 a + Cos^2 a) + (Sin^2 B+ Cos^2 B) + (Sin^2 y + Cos^2 y)}]

= 1/2 × [(Cos a + Cos B + Cos y)^2 + (Sin a + Sin B + Sin y)^2 - {(Sin^2 a + Cos^2 a) + (Sin^2 B+ Cos^2 B) + (Sin^2 y + Cos^2 y)}]

{according to statement-2}

= 1/2 × [(0)^2 + (0)^2 - {1+1+1}]

{according to statement-1 (b)}

= 1/2 × [0 + 0 - 3]

= (-3)/2 = R.H.S.

Hence, it is proved that Cos (a - B) + Cos (B - y) + Cos (y - a) = -3/2.