## Precalculus (6th Edition) Blitzer

The solution is $0$.
Convert $\frac{\pi }{4}$ into degrees. \begin{align} & \frac{\pi }{4}=\frac{\pi }{4}\left( \frac{180\text{ }\!\!{}^\circ\!\!\text{ }}{\pi } \right) \\ & =\left( \frac{\pi }{\pi } \right)\left( \frac{180\text{ }\!\!{}^\circ\!\!\text{ }}{4} \right) \\ & =45\text{ }\!\!{}^\circ\!\!\text{ } \end{align} The expressions for $\sin \theta$ and $\cos \theta$ are $\sin \theta =\frac{a}{c}$ …… (I) $\cos \theta =\frac{b}{c}$ …… (II) Here, $a$ is the length of the side opposite to $\theta$ , $b$ is the length of the side adjacent to $\theta$ and $c$ is the hypotenuse. In triangle 1, For $\theta =45\text{ }\!\!{}^\circ\!\!\text{ }$ , $a=1$ , $b=1$ and $c=\sqrt{2}$. Substitute $\frac{\pi }{4}$ for $\theta$ , $1$ for $a$ and $\sqrt{2}$ for $c$ in equation (I). $\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$ …… (III) Substitute $\frac{\pi }{4}$ for $\theta$ , $1$ for $b$ and $\sqrt{2}$ for $c$ in equation (II). $\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$ …… (IV) Subtract equation (IV) from equation (III). \begin{align} & \sin \frac{\pi }{4}-\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \\ & =0 \end{align} Therefore, the solution is $0$.