Question

how to find value of sin 15

Answer 1

$\sin 45 = \cos 45 = \frac{1}{\sqrt{2}} \\ \\ \sin 30 = \frac{1}{2} \\ \\ \cos 30 = \frac{\sqrt{3}}{2} \\ \\ \\ \sin 15 = \sin\ (45 - 30)$

$\\ \\ \\ We know that, \\ \\ \sin\ (x - y) = \sin\ x \cdot \cos\ y - \cos\ x \cdot \sin\ y \\ \\ \\ Let \ x = 45\ \ \ \&\ \ \ y = 30$

$\sin\ (45 - 30) \\ \\ = \sin 45\ \cdot\ \cos 30\ \ - \ \ \cos 45\ \cdot\ \sin 30 \\ \\ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \\ \\ = \frac{1}{\sqrt{2}}(\frac{\sqrt{3} - 1}{2}) \\ \\ = \frac{\sqrt{3} - 1}{2\sqrt{2}} \\ \\ \\ \therefore\ \sin 15 = \frac{\sqrt{3} - 1}{2\sqrt{2}} \approx 0.259$