(a) It is given that sin ( A - B )= sin A cos B - cos A sin B. Use it to find the value of sin 15^°.

\[\begin{aligned} \sin 15^{\circ} & =\sin \left(45^{\circ}-30^{\circ}\right) \\ & =\sin 45^{\circ} \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ} \\ & =\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2} \\ & =\frac{\sqrt{3}-1}{2 \sqrt{2}} \text { or } \frac{\sqrt{6}-\sqrt{2}}{4} \end{aligned} \]

\(\displaystyle \cos \mathrm{A}=\sqrt{1-\sin ^{2} \mathrm{~A}}=\sqrt{1-\mathrm{y}^{2}}\)

\(\displaystyle \tan \mathrm{A}=\frac{\sin \mathrm{A}}{\cos \mathrm{A}}=\frac{\mathrm{y}}{\sqrt{1-\mathrm{y}^{2}}}\)