If sin A + sin B = frac{1}{2} and cos A + cos | vedclass

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(B) Given $\sin A + \sin B = \frac{1}{2}$ and $\cos A + \cos B = 1$.Squaring and adding both equations:$(\sin A + \sin B)^2 + (\cos A + \cos B)^2 = (\

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$\pm \frac{\sqrt{11}}{4}$

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(B) Given $\sin A + \sin B = \frac{1}{2}$ and $\cos A + \cos B = 1$.Squaring and adding both equations:$(\sin A + \sin B)^2 + (\cos A + \cos B)^2 = (\frac{1}{2})^2 + (1)^2$$\sin^2 A + \sin^2 B + 2 \sin A \sin B + \cos^2 A + \cos^2 B + 2 \cos A \cos B = \frac{1}{4} + 1$$(\sin^2 A + \cos^2 A) + (\sin^2 B + \cos^2 B) + 2(\cos A \cos B + \sin A \sin B) = \frac{5}{4}$$1 + 1 + 2 \cos(A-B) = \frac{5}{4}$$2 + 2 \cos(A-B) = \frac{5}{4}$$2 \cos(A-B) = \frac{5}{4} - 2 = -\frac{3}{4}$$\cos(A-B) = -\frac{3}{8}$Using the identity $\cos 	heta = 1 - 2 \sin^2(\frac{	heta}{2})$:$1 - 2 \sin^2(\frac{A-B}{2}) = -\frac{3}{8}$$2 \sin^2(\frac{A-B}{2}) = 1 + \frac{3}{8} = \frac{11}{8}$$\sin^2(\frac{A-B}{2}) = \frac{11}{16}$$\sin(\frac{A-B}{2}) = \pm \frac{\sqrt{11}}{4}$

If $\sin A + \sin B = \frac{1}{2}$ and $\cos A + \cos B = 1$,then $\sin \left(\frac{A-B}{2}\right)$ equals

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