If 0

Question

(C) Given $0 < A < B < \frac{\pi}{4}$,$\cos (A+B) = \frac{11}{61}$ and $\sin (A-B) = \frac{24}{25}$.Since $0 < A+B < \frac{\pi}{2}$,$\sin (A+B) = \sqr

Vedclass · Accepted Answer

$\frac{168}{305}$

Vedclass · Answer

(C) Given $0 Since $0 Since $0 Using $\cos (A-B) = \sqrt{1 - (\frac{24}{25})^2} = \frac{7}{25}$.We know $\sin 2A = \sin ((A+B) + (A-B)) = \sin (A+B) \cos (A-B) + \cos (A+B) \sin (A-B) = (\frac{60}{61} 	imes \frac{7}{25}) + (\frac{11}{61} 	imes \frac{24}{25}) = \frac{420 + 264}{1525} = \frac{684}{1525}$.We know $\sin 2B = \sin ((A+B) - (A-B)) = \sin (A+B) \cos (A-B) - \cos (A+B) \sin (A-B) = (\frac{60}{61} 	imes \frac{7}{25}) - (\frac{11}{61} 	imes \frac{24}{25}) = \frac{420 - 264}{1525} = \frac{156}{1525}$.Therefore,$\sin 2A + \sin 2B = \frac{684}{1525} + \frac{156}{1525} = \frac{840}{1525} = \frac{168}{305}$.

If $0 < A < B < \frac{\pi}{4}$,$\cos (A+B) = \frac{11}{61}$ and $\sin (A-B) = \frac{24}{25}$,then $\sin 2A + \sin 2B = $

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