## Precalculus (6th Edition) Blitzer

The value of $\sin \frac{\pi }{3}\cos \frac{\pi }{4}-\tan \frac{\pi }{4}$ is $\frac{\sqrt{6}-4}{4}$.
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 $\frac{\sqrt{6}-4}{4}$.