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}

Collected in the board: Trigonometry

Steven Zheng posted 1 month ago


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

Steven Zheng posted 1 month ago

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