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Mathematics Question on Trigonometric Identities

In a △\triangle ABC, among the following which one is true?

A

(b + c) cos A2=asin(B+C2)\frac{A}{2} = a sin \bigg( \frac{ B + C }{2} \Bigg)

B

(b+c)cos(B+C2)=asinA2(b + c) cos \bigg( \frac{ B + C }{2} \Bigg) = a sin \frac{A}{2}

C

(b−c)cos(B−C2)=asin(A2)(b - c) cos \bigg( \frac{ B - C }{2} \Bigg) = a sin \bigg(\frac{A}{2}\bigg)

D

(b - c) cos A2=asin(B−C2)\frac{A}{2} = a sin \bigg( \frac{ B - C }{2} \Bigg)

Answer

(b - c) cos A2=asin(B−C2)\frac{A}{2} = a sin \bigg( \frac{ B - C }{2} \Bigg)

Explanation

Solution

Let a,b,c are the sides △\triangle ABC.
Now, b+ca=k (sin B+sin C)k sin A\frac{ b + c}{a} = \frac{ k \, ( sin \, B + sin \, C)}{ k \, sin \, A }
= 2sin(B+C2)cos (B−C2)2 sin A2 cos A2\frac{ 2 sin \bigg( \frac{ B + C }{2} \Bigg) cos \, \bigg( \frac{ B - C }{2} \Bigg)}{ 2 \, sin \, \frac{A}{2} \, cos \, \frac{A}{2} }
⇒b+ca=cos(B−C2)sinA2\Rightarrow \frac{ b + c}{a} = \frac{ cos \bigg( \frac{ B - C }{2} \Bigg) }{ sin \frac{A}{2}}
Also, b−ca=sin(B−C2)sinA2 \frac{ b - c}{a} = \frac{ sin \bigg( \frac{ B - C }{2} \Bigg) }{ sin \frac{A}{2}}