Question

If sin ( A-B) =3/5and sin (A+B )=4/5,then sin 2a is equal to?

Answer 1

Step-by-step explanation:

The value of \sin 2A=1sin2A=1

Step-by-step explanation:

We have,

\sin (A-B)=\dfrac{3}{5}sin(A−B)=53 and \sin(A+B)=\dfrac{4}{5}sin(A+B)=54

To find, the value of \sin 2A=?sin2A=?

∴ \sin (A-B)=\dfrac{3}{5}sin(A−B)=53

⇒ A-B=\sin^{-1}(\dfrac{3}{5})A−B=sin−1(53)       .....(1)

and \sin(A+B)=\dfrac{4}{5}sin(A+B)=54

⇒ A+B=\sin^{-1}(\dfrac{4}{5})A+B=sin−1(54)       .....(2)

Addind (1) and (2), we get

2A=\sin^{-1}(\dfrac{3}{5})+\sin^{-1}(\dfrac{4}{5})2A=sin−1(53)+sin−1(54)

⇒ 2A=\sin^[{\dfrac{3}{5} \sqrt{1-\dfrac{16}{25}}+\dfrac{4}{5} \sqrt{1-\dfrac{9}{25}}}]2A=sin[531−2516+541−259]

⇒ 2A=\sin^[{\dfrac{3}{5} \sqrt{\dfrac{25-16}{25}}+\dfrac{4}{5} \sqrt{\dfrac{25-9}{25}}}]2A=sin[532525−16+542525−9]

⇒ 

⇒ 

⇒ \sin 2A=1sin2A=1

Thus, the value of \sin 2A=1sin2A=1