Question

sin(a-b) =3/5
cos(a+b)=12/13
sin2a=??

Answer 1

Here ,

From ∆ PQR , Sin ( A - B ) = 3/5 , so Cos ( A - B ) = 4/5

and From ∆ MNO Cos ( A + B ) = 12/13 , so Sin ( A + B ) = 5/13

Sin 2A = Sin [ ( A + B ) + ( A - B ) ]

Sin [ ( A + B ) + ( A - B ) ] = Sin ( A + B ) × Cos ( A - B ) + Cos ( A + B ) × Sin ( A - B )

$= \frac{5}{13} \times \frac{4}{5} + \frac{12}{13} \times \frac{3}{5} \\ \\ = \frac{36}{65} + \frac{20}{65} \\ \\ = \frac{56}{65}$