## Precalculus (10th Edition)

$3+\sqrt {2}$
RECALL: $\tan{\theta}=\dfrac{\sin{\theta}}{\cos{\theta}}$ With $\sin{\frac{\pi}{4}}=\frac{\sqrt2}{2}$ and $\cos{\frac{\pi}{4}}=\frac{\sqrt2}{2}$, then $2\sin \dfrac {\pi }{4}+3\tan \dfrac {\pi }{4} \\=2\times \left(\dfrac{\sqrt {2}}{2}\right)+3\times \left(\dfrac {\sin \dfrac {\pi }{4}}{\cos \dfrac {\pi }{4}}\right) \\=\sqrt {2}+3\times \left(\dfrac {\dfrac {\sqrt {2}}{2}}{\dfrac {\sqrt {2}}{2}}\right) \\=\sqrt {2}+(3\times 1) \\=\sqrt2+3$