Question

In a \[\Delta ABC\], prove that: \[\cos \left( \dfrac{A+B}{2} \right)=\sin \dfrac{C}{2}\]

Answer 1

Hint: We know that the sum of the angles of the triangle is 180 and hence we will use this to solve this question. First we will isolate angle C to the right hand side of the equation after dividing the whole equation by 2 then we will apply cos to both sides and finally we will use the trigonometry identity to prove.

Complete step-by-step answer:

We know that the sum of all the angles of the triangle is 180 degrees. So using this information we get,

\[A+B+C={{180}^{\circ }}.......(2)\]

So now dividing both sides of the equation (2) by 2 we get,

\[\Rightarrow \dfrac{A+B+C}{2}=\dfrac{{{180}^{\circ }}}{2}.......(3)\]

Now rearranging and simplifying equation (3) we get,

\[\Rightarrow \dfrac{A}{2}+\dfrac{B}{2}+\dfrac{C}{2}={{90}^{\circ }}.......(4)\]

Now sending the third term in the right hand side of the equation (4) we get,

\[\Rightarrow \dfrac{A}{2}+\dfrac{B}{2}={{90}^{\circ }}-\dfrac{C}{2}.......(5)\]

Now applying cos to both sides of equation (5) we get,

\[\Rightarrow \cos \left( \dfrac{A}{2}+\dfrac{B}{2} \right)=\cos \left( {{90}^{\circ

}}-\dfrac{C}{2} \right).......(6)\]

Now we know that cos(90-x) is equal to sin x. So applying this identity in equation (6) we get,


\[\Rightarrow \cos \left( \dfrac{A+B}{2} \right)=\sin \dfrac{C}{2}\]

Hence we have proved.

Note: In trigonometry remembering the formulas and the identities is very important because then it becomes easy. We in a hurry can make a mistake in applying the identities as we can write cos(90-x) equal to –sinx in place of sinx. Also if we don’t remember the formula then we can get confused about how to proceed further after equation (6). Also we have to keep in mind that the sum of the angles of the triangle is always 180.