Given sinθ =3/5 and (π/2<θ<π), how do you find sine (2π)?

Given ( \sin\theta = \frac{3}{5} ) and ( \frac{\pi}{2} < \theta < \pi ), we can find ( \sin(2\theta) ) using the double-angle formula for sine.

The double-angle formula for sine is: [ \sin(2\theta) = 2\sin\theta\cos\theta ]

First, we need to find the value of ( \cos\theta ) using the given information about ( \sin\theta ). Since ( \sin\theta = \frac{3}{5} ), we can use the Pythagorean identity to find ( \cos\theta ): [ \cos^2\theta = 1 - \sin^2\theta ] [ \cos^2\theta = 1 - \left(\frac{3}{5}\right)^2 ] [ \cos^2\theta = 1 - \frac{9}{25} ] [ \cos^2\theta = \frac{16}{25} ] [ \cos\theta = \pm\frac{4}{5} ]

Since ( \frac{\pi}{2} < \theta < \pi ), we know that ( \cos\theta ) is negative in this quadrant, so we take ( \cos\theta = -\frac{4}{5} ).

Now we can use the double-angle formula for sine: [ \sin(2\theta) = 2\sin\theta\cos\theta ] [ \sin(2\theta) = 2\left(\frac{3}{5}\right)\left(-\frac{4}{5}\right) ] [ \sin(2\theta) = -\frac{24}{25} ]

Therefore, ( \sin(2\theta) = -\frac{24}{25} ).