Answer
The value of $\sin \frac{\pi }{3}\cos \frac{\pi }{4}-\tan \frac{\pi }{4}$ is $\frac{\sqrt{6}-4}{4}$.
Work Step by Step
Consider the given expression, $\sin \frac{\pi }{3}\cos \frac{\pi }{4}-\tan \frac{\pi }{4}$.
Consider the provided right angled triangle.
Now, to find $\sin \left( \frac{\pi }{3} \right)$ apply the definition of $\sin \theta $ to the above triangle,
$\sin \theta =\frac{\text{Length of side opposite of }\theta }{\text{hypotenuse}}$.
But, here, $\theta =\frac{\pi }{3}$ , and $\sin \left( \frac{\pi }{3} \right)=\sin 60{}^\circ $.
Therefore,
$\begin{align}
& \sin \left( \frac{\pi }{3} \right)=\frac{\text{Length of side opposite of 60}{}^\circ }{\text{Length of hypotenuse}} \\
& =\frac{\sqrt{3}}{2}
\end{align}$
To find $\cos \left( \frac{\pi }{4} \right)$ apply the definition of $\cos \theta $ to the provided triangle,
$\cos \theta =\frac{\text{Length of side adjacent of }\theta }{\text{hypotenuse}}$.
But, here, $\theta =\frac{\pi }{4}$ , and $\cos \left( \frac{\pi }{4} \right)=\cos 45{}^\circ $.
Therefore,
$\begin{align}
& \cos \left( \frac{\pi }{4} \right)=\frac{\text{Length of side adjacent of 45}{}^\circ }{\text{Length of hypotenuse}} \\
& =\frac{1}{\sqrt{2}}
\end{align}$
Rationalize the denominator,
$\begin{align}
& \cos \left( \frac{\pi }{4} \right)=\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}} \\
& =\frac{\sqrt{2}}{2}
\end{align}$
To find $\tan \left( \frac{\pi }{4} \right)$ apply the definition of $\tan \theta $ to the provided triangle,
$\tan \theta =\frac{\text{Length of side opposite of }\theta }{\text{Length of side adjacent of }\theta }$.
But, here, $\theta =\frac{\pi }{4}$ and $\tan \left( \frac{\pi }{4} \right)=\tan 45{}^\circ $.
Therefore,
$\begin{align}
& \tan \left( \frac{\pi }{4} \right)=\frac{\text{Length of side opposite of 45}{}^\circ }{\text{Length of side adjacent of 45}{}^\circ } \\
& =\frac{1}{1} \\
& =1
\end{align}$
Now, substitute the value of $\sin \left( \frac{\pi }{3} \right),\text{ cos}\left( \frac{\pi }{4} \right),\text{ and tan}\left( \frac{\pi }{4} \right)$ in the provided expression $\sin \frac{\pi }{3}\cos \frac{\pi }{4}-\tan \frac{\pi }{4}$.
Therefore,
$\begin{align}
& \sin \frac{\pi }{3}\cos \frac{\pi }{4}-\tan \frac{\pi }{4}=\left( \frac{\sqrt{3}}{2} \right)\left( \frac{\sqrt{2}}{2} \right)-1 \\
& =\frac{\sqrt{6}}{4}-1 \\
& =\frac{\sqrt{6}-4}{4}
\end{align}$
Hence the value of $\sin \frac{\pi }{3}\cos \frac{\pi }{4}-\tan \frac{\pi }{4}$ is $\frac{\sqrt{6}-4}{4}$.
