Question

if cos (A-B)=5/13 and sin (A+B)=4/5, then find sin 2B.​

Answer 1

Answer:

Step-by-step explanation:

let A-B=P,A+B=Q

Q-P=2B

Both side take sin

And use formula get ans

Answer 2

Given :

$\cos (A-B)= \frac{5}{13}\\\\\sin (A+B)=\frac{4}{5}$

To find : sin 2B

solution :

$\cos (A-B)= \frac{5}{13}\\\\\cos A \cos B +\sin A \sin B = \frac{5}{13}..(1)\\\\\sin (A+B)= \frac{4}{5}\\\\\sin A \cos B +\cos A \sin B =\frac{4}{5}\\\\\cos (A+B)=\frac{3}{5}\\\\\cos A \cos B +\sin A \sin B =\frac{3}{-5}..(2)\\\\\text{subtracting 1 from 2 }\\\\2\sin A \sin B=\frac{5}{13}-\frac{3}{5}\\\\2\sin A \sin B=\frac{25-39}{65}\\\\2\sin A \sin B=\frac{-14}{65}\\\\\sin 2B = \sin (B+B)\\\\=\sin B \cos B+\sin B \cos B\\\\2\sin B \cos B\\\\2\sin B \sin B= \frac{-14}{65}$

hence ,

$\sin 2B= \frac{-14}{65}$

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