Question

in a triangle ABC prove that cos(A+B/2)=sinc/2?​

Answer 1

In a triangle ABC we have,

$\longrightarrow A+B+C=\pi$

Subtracting C from both sides,

$\longrightarrow A+B=\pi-C$

Dividing both sides by 2,

$\longrightarrow\dfrac{A+B}{2}=\dfrac{\pi-C}{2}$

$\longrightarrow\dfrac{A+B}{2}=\dfrac{\pi}{2}-\dfrac{C}{2}$

Taking cosine function on both sides,

$\longrightarrow\cos\left(\dfrac{A+B}{2}\right)=\cos\left(\dfrac{\pi}{2}-\dfrac{C}{2}\right)\quad\quad\dots(1)$

But we know that,

• $\cos\left(\dfrac{\pi}{2}-x\right)=\sin x$

Then (1) becomes,

$\longrightarrow\underline{\underline{\cos\left(\dfrac{A+B}{2}\right)=\sin\left(\dfrac{C}{2}\right)}}$

Hence Proved!