Question
In a $\triangle\text{ABC},$ Prove that: $\cos\Big(\frac{\text{A+B}}{2}\Big)=\sin\frac{\text{C}}{2}$
✓
Answer
We have $\text{A + B + C} = \pi$ $\big(\because$ sum of 3 angle of a triangle is $\pi=180^\circ\big)$ $\Rightarrow \text{A+B}={\pi}-\text{C}$ $$$\Rightarrow \frac{\text{A+B}}{2}=\frac{\pi-\text{C}}{2}$ $\Rightarrow \frac{\text{A+B}}{2}=\frac{\pi}{2}-\frac{\text{C}}{2}$ $\Rightarrow \cos=\Big(\frac{\text{A+B}}{2}\Big)=\cos\Big(\frac{\pi}{2}-\frac{C}{2}\Big)$ $\Rightarrow = \sin\frac{\text{C}}{2}$ $\Big(\because\cos\Big(\frac{\pi}{2}-\theta\Big)=\sin\theta\Big)$ Hence, $\cos\Big(\frac{\text{A+B}}{2}\Big)=\sin\frac{\text{C}}{2}$ $\text{Proved}$
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