Question

sin(A-B)=3/5 and sin(A+B)=4/5.Find value of sin 2A?

Answer 1

A+B=Sin^-1 (4/5)
A-B=Sin^-1 (3/5)
adding
2A=Sin^-1 (4/5)+sin^-1 ( 3/5)
simplify
sin2A=1

Answer 2

The value of $\sin 2A=1$

Step-by-step explanation:

We have,

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

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

∴ $\sin (A-B)=\dfrac{3}{5}$

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

and $\sin(A+B)=\dfrac{4}{5}$

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

Addind (1) and (2), we get

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

⇒ $2A=\sin^[{\dfrac{3}{5} \sqrt{1-\dfrac{16}{25}}+\dfrac{4}{5} \sqrt{1-\dfrac{9}{25}}}]$

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

⇒ $2A=\sin^[{\dfrac{3}{5}.\dfrac{3}{5} }+\dfrac{4}{5}.\dfrac{4}{5}}]$

⇒ $\sin 2A=\dfrac{9}{255} }+\dfrac{16}{5}}=\dfrac{25}{25}$

⇒ $\sin 2A=1$

Thus, the value of $\sin 2A=1$