Question

in a triangle ABC prove that sin(A+B)/2=cosC/2

Answer 1

Answer on Question #57330 – Math – Trigonometry

Question

In a triangle ABC prove that $\sin(A+B)/2 = \cos C/2$

Solution

It is known that in a triangle, the three interior angles always add to $180{}^{\circ}$:

$$A + B + C = 180{}^{\circ}$$ $$A + B = 180{}^{\circ} - C$$ $$\frac{A + B}{2} = 90{}^{\circ} - \frac{C}{2} \quad \Rightarrow \quad 90{}^{\circ} - \frac{A + B}{2} = \frac{C}{2}$$

The first identity in the next system holds true due to the well-known identity expressing the sine function in terms of its complement. The second identity in the next system holds true due to identity (1).

$$\left\{ \begin{array}{l} \sin \frac{A + B}{2} = \cos \left(90{}^{\circ} - \frac{A + B}{2}\right) \\ \cos \left(90{}^{\circ} - \frac{A + B}{2}\right) = \cos \left(\frac{C}{2}\right) \end{array} \right. \Rightarrow \sin \frac{A + B}{2} = \cos \left(\frac{C}{2}\right)$$

Q.E.D.

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