## Precalculus (6th Edition) Blitzer

The required value is $\frac{\sqrt{2}-\sqrt{6}}{4}$
We know that the term $\sin \left( {{195}^{\circ }} \right)$ can be expressed as $\sin \left( {{135}^{\circ }}+{{60}^{\circ }} \right)$. Now, apply the sum formula for sin. The identity $\sin \left( C+D \right)$ is equal to the product of sine of the first angle and cosine of the second angle plus the product of the cosine of the first angle and the sine of the second angle. $\sin \left( C+D \right)=\sin C\text{cos }D+\cos C\sin D$ Now, use the sum formula in the given expression and apply the values of the trigonometric functions. \begin{align} & sin\left( {{195}^{\circ }} \right)=\sin \left( {{135}^{\circ }}+{{60}^{\circ }} \right) \\ & =\sin {{135}^{\circ }}\cos {{60}^{\circ }}+\cos {{135}^{\circ }}\sin {{60}^{\circ }} \\ & =\frac{\sqrt{2}}{2}.\frac{1}{2}+\left( -\frac{\sqrt{2}}{2} \right).\frac{\sqrt{3}}{2} \\ & =\frac{\sqrt{2}}{4}-\frac{\sqrt{6}}{4} \\ & =\frac{\sqrt{2}-\sqrt{6}}{4} \end{align}