If A, B, C are the angles of a triangle, then find cos A + cos B + cos C.

We will use the angle sum property of the triangle to answer this question.

Answer: cos A + cos B + cos C = 1 + 4 sin(A/2) sin(B/2) sin(C/2)

Let us see how we can find it.

Explanation:

We need to evaluate cos A + cos B + cos C where A, B, and C are angles of triangle.

So, by angle sum property of triangle, A + B + C = 180º

Now, consider cos A + cos B + cos C = ( cos A + cos B ) + cos C

= { 2 · cos[ ( A+B) / 2 ] · cos [ ( A-B) / 2 ] } + cos C

= { 2 · cos [ (π/2) – (C/2) ] · cos [ (A-B) / 2 ] } + cos C

= { 2 · sin (C/2 ) · cos [ (A-B) / 2 ] } + { 1 – 2 · sin² ( C/2 ) }

= 1 + 2 sin (C/2 )· { cos [ (A -B) / 2 ] – sin ( C/2 ) }

= 1 + 2 sin (C/2 )· { cos [ (A-B) / 2 ] – sin [ (π/2) – ( (A + B)/2 ) ] }

= 1 + 2 sin ( C/2 )· { cos [ (A-B) / 2 ] – cos [ (A+B)/ 2 ] }

= 1 + 2 sin ( C/2 )· 2 sin ( A/2 )· sin( B/2 )

= 1 + 4 sin (A/2) sin(B/2) sin(C/2)

Thus, cos A + cos B + cos C = 1 + 4 sin(A/2) sin(B/2) sin(C/2)