## Precalculus (6th Edition) Blitzer

The required value is $\frac{\sqrt{3}}{2}$
We have the double angle formulas, ${{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta$. Therefore, applying the formula in the given expression where $\theta ={{15}^{\circ }}$, we get: \begin{align} & {{\cos }^{2}}15-{{\sin }^{2}}15=\cos \left( {{2.15}^{\circ }} \right) \\ & =\cos \left( {{30}^{\circ }} \right) \\ & =\frac{\sqrt{3}}{2} \end{align} Hence, the required value is $\frac{\sqrt{3}}{2}$