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Question: If \(\sin A + \sin B = C,\cos A + \cos B = D,\) then the value of \(\sin(A + B) =\)...

If sin⁡A+sin⁡B=C,cos⁡A+cos⁡B=D,\sin A + \sin B = C,\cos A + \cos B = D, then the value of sin⁡(A+B)=\sin(A + B) =

A

CDCD

B

CDC2+D2\frac{CD}{C^{2} + D^{2}}

C

C2+D22CD\frac{C^{2} + D^{2}}{2CD}

D

2CDC2+D2\frac{2CD}{C^{2} + D^{2}}

Answer

2CDC2+D2\frac{2CD}{C^{2} + D^{2}}

Explanation

Solution

As given sin⁡A+sin⁡Bcos⁡A+cos⁡B=CD\frac{\sin A + \sin B}{\cos A + \cos B} = \frac{C}{D}

⇒2sin⁡A+B2.cos⁡A−B22cos⁡A+B2.cos⁡A−B2=CD⇒tan⁡A+B2=CD\Rightarrow \frac{2\sin\frac{A + B}{2}.\cos\frac{A - B}{2}}{2\cos\frac{A + B}{2}.\cos\frac{A - B}{2}} = \frac{C}{D} \Rightarrow \tan\frac{A + B}{2} = \frac{C}{D}

Thus, sin⁡(A+B)=2tan⁡A+B21+tan⁡2A+B2=2CD1+C2D2=2CD(C2+D2)\sin(A + B) = \frac{2\tan\frac{A + B}{2}}{1 + \tan^{2}\frac{A + B}{2}} = \frac{2\frac{C}{D}}{1 + \frac{C^{2}}{D^{2}}} = \frac{2CD}{(C^{2} + D^{2})}.