Answer
The solution is $0$.
Work Step by Step
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 $0$.
