Question

In a triangle $ABC,$the value of $\sin \left(A\right)+\sin \left(B\right)+\sin \left(C\right)$ is :

• A. $4\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)\sin \left(\frac{C}{2}\right)$
• B. $4\cos \left(\frac{A}{2}\right)\cos \left(\frac{B}{2}\right)\cos \left(\frac{C}{2}\right)$
• C. $4\cos \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)\sin \left(\frac{C}{2}\right)$
• D. $4\cos \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)\cos \left(\frac{C}{2}\right)$

Answer 1

The correct option is B

$4\cos \left(\frac{A}{2}\right)\cos \left(\frac{B}{2}\right)\cos \left(\frac{C}{2}\right)$

Finding the value of $\sin \left(A\right)+\sin \left(B\right)+\sin \left(C\right)$ in a triangle.

In any triangle, the sum of all the interior angles is always $180°$.

Therefore, In the triangle $ABC,$ $A+B+C=180°$

Now, solving for $\sin \left(A\right)+\sin \left(B\right)+\sin \left(C\right)$:

$$\begin{gathered} \sin \left(A\right)+\sin \left(B\right)+\sin \left(C\right)=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+\sin \left(C\right) \\ =2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+2\sin \left(\frac{C}{2}\right)\cos \left(\frac{C}{2}\right) \\ =2\sin \left(\frac{180°-C}{2}\right)\cos \left(\frac{A-B}{2}\right)+2\sin \left(\frac{C}{2}\right)\cos \left(\frac{C}{2}\right) \\ =2\cos \left(\frac{C}{2}\right)\cos \left(\frac{A-B}{2}\right)+2\sin \left(\frac{C}{2}\right)\cos \left(\frac{C}{2}\right) \\ =2\cos \left(\frac{C}{2}\right)\left(\cos \left(\frac{A-B}{2}\right)+\sin \left(\frac{C}{2}\right)\right) \\ =2\cos \left(\frac{C}{2}\right)\left(\cos \left(\frac{A-B}{2}\right)+\sin \left(\frac{180°-(A+B)}{2}\right)\right) \\ =2\cos \left(\frac{C}{2}\right)\left(\cos \left(\frac{A-B}{2}\right)+\cos \left(\frac{A+B}{2}\right)\right) \\ =4\cos \left(\frac{C}{2}\right)\cos \left(\frac{B}{2}\right)\cos \left(\frac{A}{2}\right) \\ \sin \left(A\right)+\sin \left(B\right)+\sin \left(C\right)=4\cos \left(\frac{A}{2}\right)\cos \left(\frac{B}{2}\right)\cos \left(\frac{C}{2}\right) \end{gathered}$$

Hence, Option (B) is correct.