## Precalculus (6th Edition) Blitzer

The exact value of $\sin \left( 105{}^\circ \right)$ is $\frac{\sqrt{3}+1}{2\sqrt{2}}$.
Rewrite the expression for sine of $105{}^\circ$ as the sum of two angles as, $\sin 105{}^\circ =\sin \left( 60{}^\circ +45{}^\circ \right)$ Use the sum formula of sines and evaluate the modified expression as, $\sin \left( 60{}^\circ +45{}^\circ \right)=\sin 60{}^\circ \cos 45{}^\circ +\cos 60{}^\circ \sin 45{}^\circ$ Substitute the values $\cos 45{}^\circ =\frac{1}{\sqrt{2}},\text{ }\cos 60{}^\circ =\frac{1}{2},\text{ }\sin 45{}^\circ =\frac{1}{\sqrt{2}},\text{ and }\sin 30{}^\circ =\frac{\sqrt{3}}{2}$. \begin{align} & \sin \left( 60{}^\circ +45{}^\circ \right)=\left( \frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}} \right)+\left( \frac{1}{2}\times \frac{1}{\sqrt{2}} \right) \\ & =\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}} \\ & =\frac{\sqrt{3}+1}{2\sqrt{2}} \end{align} Hence, the exact value of $\sin 105{}^\circ$ is equivalent to $\frac{\sqrt{3}+1}{2\sqrt{2}}$.