Question
In \triangle ABC, prove that
\cos A+\cos B+ \cos C= 1+4\sin\dfrac{A}{2} \sin\dfrac{B}{2} \sin\dfrac{C}{2}
Answer 1
The sum of internal angles of a triangle is equal to \pi
\because A+B+C = \pi
\therefore \cos C = -\cos(A+B)
\cos(\dfrac{\pi}{2}-\dfrac{C}{2})=\cos \dfrac{A+B}{2}
Applying Cofunction Identity
\sin \dfrac{C}{2} =\cos \dfrac{A+B}{2}
LHS = \cos A+\cos B - \cos(A+B)
Using sum identity and double angle identity,
RHS = 1+4\sin\dfrac{A}{2} \sin\dfrac{B}{2} \cos \dfrac{A+B}{2}
=1+4\sin\dfrac{A}{2} \sin\dfrac{B}{2}(\cos\dfrac{A}{2}\cos\dfrac{B}{2}-\sin\dfrac{A}{2}\sin\dfrac{B}{2} )
= 1+\dfrac{1}{4}\sin A \sin B - \sin^2\dfrac{A}{2}\sin^2\dfrac{B}{2}
=1+[\sin A \sin B - (1-\cos A)(1-\cos B)]
=1+[\sin A \sin B - (1-\cos A-\cos B+\cos A\cos B)]
=\cos A+\cos B - \cos (A+B)
=LHS
Answer 2
The sum of internal angles of a triangle is equal to \pi
\because A+B+C = \pi
\therefore \cos C = -\cos(A+B)
\cos(\dfrac{\pi}{2}-\dfrac{C}{2})=\cos \dfrac{A+B}{2}
Applying Cofunction Identity
\sin \dfrac{C}{2} =\cos \dfrac{A+B}{2}
Using Sum to Product identity and double angle identity
LHS = 2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2}-\cos(A+B)
= 2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2} - (2\cos^2\dfrac{A+B}{2}-1 )
=2\cos\dfrac{A+B} {2}(\cos\dfrac{A-B}{2}-\cos\dfrac{A+B}{2} )+1
=4\sin \dfrac{C}{2} \sin \dfrac{B}{2} \sin \dfrac{A}{2} +1
=RHS