Question
In \triangle ABC, show that \sin A+\sin B + \sin C = 4\cos\dfrac{A}{2}\cos \dfrac{B}{2}\cos\dfrac{C}{2}
Answer
\because A+B+C = \pi
\therefore \sin C = \sin(A+B)
\sin(\dfrac{\pi}{2}-\dfrac{C}{2})=\sin \dfrac{A+B}{2}
\cos \dfrac{C}{2} =\sin \dfrac{A+B}{2}
LHS = \sin A+\sin B +\sin(A+B)
RHS =4 \sin \dfrac{A+B}{2} \cos\dfrac{A}{2}\cos \dfrac{B}{2}
=4 \cos\dfrac{A}{2}\cos \dfrac{B}{2}(\sin\dfrac{A}{2}\cos\dfrac{B}{2}+\cos\dfrac{A}{2}\sin\dfrac{B}{2} )
=4\sin\dfrac{A}{2}\cos\dfrac{A}{2}\cos^2 \dfrac{B}{2}+4\sin\dfrac{B}{2}\cos\dfrac{B}{2}\cos^2\dfrac{A}{2}
=\sin A\cdotp 2\cos^2 \dfrac{B}{2}+\sin B\cdotp 2\cos^2 \dfrac{A}{2}
=\sin A(1+\cos B)+\sin B(1+\cos A)
=\sin A+\sin B+\sin (A+B)
=LHS