If $$\sin \theta = \frac{4}{5}$$, find the value of $$\sin 3 \theta$$

Given,  $$\sin \theta = \frac{4}{5}$$

$$=$$>  $$\cos\theta\ =\sqrt{1-\sin^2\theta\ }$$

$$=$$>  $$\cos\theta\ =\sqrt{1-\left(\frac{4}{5}\right)^2\ }$$

$$=$$>  $$\cos\theta\ =\sqrt{1-\frac{16}{25}}$$

$$=$$>  $$\cos\theta\ =\sqrt{\frac{9}{25}}$$

$$=$$>  $$\cos\theta\ =\frac{3}{5}$$

$$\therefore\ $$ $$\sin3\theta\ =\sin\left(2\theta\ +\theta\ \right)$$

$$=\sin2\theta\ \cos\theta\ +\cos2\theta\ \sin\theta\ $$

$$=\left(2\sin\theta\ \cos\theta\ \right)\cos\theta\ +\left(\cos^2\theta\ -\sin^2\theta\ \right)\sin\theta\ $$

$$=\left(2.\frac{4}{5}.\frac{3}{5}\right)\frac{3}{5}\ +\left(\left(\frac{3}{5}\right)^2\ -\left(\frac{4}{5}\right)^2\ \right)\frac{4}{5}$$

$$=\frac{72}{125}+\left(\frac{9}{25}\ -\frac{16}{25}\ \right)\frac{4}{5}$$

$$=\frac{72}{125}+\left(-\frac{7}{25}\ \right)\frac{4}{5}$$

$$=\frac{72}{125}-\frac{28}{125}$$

$$=\frac{44}{125}$$

Hence, the correct answer is Option B