# cosA+cosB+cosC+cos(A+B+C)=4 cos(A+B/2)cos(B+C/2)+cos(C+A/2) Considering the angles A,B,C as the interior angles of a triangle, hence, the sum A+B+C is of `180^o` .

`A+B+C = pi => A+B = pi - C => (A+B)/2 = pi/2 - C /2`

`cos ((A+B)/2) = cos(pi/2 - C /2) = sin (C/2)`

Hence, reasoning by analogy, yields:

`cos ((B + C)/2) = cos(pi/2 -A /2) = sin (A/2)`

`cos (C + A)/2 = cos(pi/2 - B/2) = sin (B/2)`

Hence, converting the sum of cosines into products yields:

`cos A + cos B = 2 cos ((A+B)/2) cos ((A-B)/2)`

`cos C + cos (A+B+C) = cos C + cos pi = cos C - 1`

Notice that you may consider C as double of half angle such that:

`cos C = cos 2(C/2) = 2 cos^2(C/2) - 1`

`cos C - 1 = 2 cos^2(C/2) - 2 = 2(cos^2 (C/2) - 1) = -2 sin^2 (C/2)`

You should write the left side using the converted form such that:

`2 cos ((A+B)/2) cos ((A-B)/2) - 2 sin^2 (C/2) = 4cos ((A+B)/2)cos ((B + C)/2)cos ((C + A)/2)`

You may substitute `sin (C/2)`  for `cos ((A+B)/2)`  such that:

`2sin (C/2) cos ((A-B)/2) - 2 sin^2 (C/2) = 4cos ((A+B)/2)cos ((B + C)/2)cos ((C + A)/2)`

You should factor out `2sin (C/2)`  such that:

`2 sin (C/2) (cos ((A-B)/2) -sin (C/2)) = 4sin (C/2)cos ((B + C)/2)cos ((C + A)/2)`

Reducing by `2sin (C/2)`  both sides yields:

`(cos ((A-B)/2) - cos ((A+B)/2)) = 2cos ((B + C)/2)cos ((C + A)/2)`

You need to convert the difference of cosines into a product such that:

`2sin (A/2 - B/2 + A/2 + B/2)/2 sin (A/2 + B/2 - A/2 + B/2)/2 = 2cos ((B + C)/2)cos ((C + A)/2)`

`2 sin (A/2)*sin (B/2) = 2cos ((B + C)/2)cos ((C + A)/2)`

But `sin (A/2) = cos ((B + C)/2)`  and `sin(B/2) =cos ((C + A)/2)` , hence, substituting  `sin (A/2)`  for `cos ((B + C)/2)`  and `sin(B/2)`  for `cos ((C + A)/2)`  yields:

`2 sin (A/2)*sin (B/2) = 2 sin (A/2)*sin (B/2) `

Hence, considering `A+B+C = pi`  and using trigonometric identities yields `cos A + cos B + cosC + cos(A+B+C) = 4cos ((A+B)/2)cos ((B + C)/2)cos ((C + A)/2).`

Approved by eNotes Editorial Team LHS

cosA+cosB+cosC+Cos (A+B+C)

remember A+B+C= pi

Then Cos(A+B+C) = Cos (pi) = -1

So LHS;

=( cos A + cos B ) + cos C-1

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

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

= { 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 ) }

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

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

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

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

=RHS

Approved by eNotes Editorial Team 