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Question: In a triangle ABC, the value of \(\sin A + \sin B + \sin C\) is...

In a triangle ABC, the value of sin⁡A+sin⁡B+sin⁡C\sin A + \sin B + \sin C is

A

4sin⁡A2sin⁡B2sin⁡C24\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}

B

4cos⁡A2cos⁡B2cos⁡C24\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}

C

4cos⁡A2sin⁡B2sin⁡C24\cos\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}

D

4cos⁡A2sin⁡B2cos⁡C24\cos\frac{A}{2}\sin\frac{B}{2}\cos\frac{C}{2}

Answer

4cos⁡A2cos⁡B2cos⁡C24\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}

Explanation

Solution

In ΔAB⥂C,A+B+C=180∘⇒sin⁡A+sin⁡B+sin⁡C=\Delta AB ⥂ C,A + B + C = 180{^\circ} \Rightarrow \sin A + \sin B + \sin C =

2sin⁡A+B2cos⁡A−B2+2sin⁡C2cos⁡C22\sin\frac{A + B}{2}\cos\frac{A - B}{2} + 2\sin\frac{C}{2}\cos\frac{C}{2}

=2sin⁡(π2−C2)cos⁡A−B2+2cos⁡C2sin⁡(π2−A+B‾2)= 2\sin\left( \frac{\pi}{2} - \frac{C}{2} \right)\cos\frac{A - B}{2} + 2\cos\frac{C}{2}\sin\left( \frac{\pi}{2} - \frac{\overline{A + B}}{2} \right)

=2cos⁡C2cos⁡A−B2+2cos⁡C2cos⁡A+B2= 2\cos\frac{C}{2}\cos\frac{A - B}{2} + 2\cos\frac{C}{2}\cos\frac{A + B}{2}

=2cos⁡C2[cos⁡A−B2+cos⁡A+B2]= 2\cos\frac{C}{2}\left\lbrack \cos\frac{A - B}{2} + \cos\frac{A + B}{2} \right\rbrack

=2cos⁡C2(2cos⁡A2cos⁡B2)=4cos⁡A2cos⁡B2cos⁡C2= 2\cos\frac{C}{2}\left( 2\cos\frac{A}{2}\cos\frac{B}{2} \right) = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} .