# Solve this: 1) sin(7 pi)(6)=? 2) cos(4 pi)(3)=? 3) tan(3 pi)(2)=? 4) cot(11 pi)(4)=?

Solve this:
$1\right)\mathrm{sin}\left(\frac{7\pi }{6}\right)=?\phantom{\rule{0ex}{0ex}}2\right)\mathrm{cos}\left(-\frac{4\pi }{3}\right)=?\phantom{\rule{0ex}{0ex}}3\right)\mathrm{tan}\left(\frac{3\pi }{2}\right)=?\phantom{\rule{0ex}{0ex}}4\right)\mathrm{cot}\left(\frac{11\pi }{4}\right)=?$
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ambivalentnoe1
1) $\mathrm{sin}\left(\frac{7\pi }{6}\right)=\mathrm{sin}\left(\pi +\frac{\pi }{6}\right)\phantom{\rule{0ex}{0ex}}\mathrm{sin}\left(A+B\right)=\mathrm{sin}A\mathrm{cos}B+\mathrm{cos}A\mathrm{sin}B\phantom{\rule{0ex}{0ex}}=\mathrm{sin}\pi \mathrm{cos}\frac{\pi }{6}+\mathrm{cos}\pi \mathrm{sin}\left(\frac{\pi }{6}\right)\phantom{\rule{0ex}{0ex}}=0\cdot \left(\frac{\sqrt{3}}{2}\right)+\left(-1\right)\left(\frac{1}{2}\right)=-\frac{1}{2}$
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rivasguss9
2) $\mathrm{cos}\left(-4\frac{\pi }{3}\right)\phantom{\rule{0ex}{0ex}}\mathrm{cos}\left(-x\right)=\mathrm{cos}\left(x\right)\mathrm{cos}\left(4\frac{\pi }{3}\right)=\mathrm{cos}\left(\pi +\frac{\pi }{3}\right)\phantom{\rule{0ex}{0ex}}\mathrm{cos}\left(A+B\right)=\mathrm{cos}A\mathrm{cos}B-\mathrm{sin}A\mathrm{sin}B\phantom{\rule{0ex}{0ex}}=\mathrm{cos}\pi \mathrm{cos}\frac{\pi }{3}-\mathrm{sin}\pi \mathrm{sin}\frac{\pi }{3}\phantom{\rule{0ex}{0ex}}=\left(-1\right)\left(\frac{1}{2}\right)-0\cdot \frac{\sqrt{3}}{2}=-\frac{1}{2}$
3)$\mathrm{tan}\left(\frac{3\pi }{2}\right)=\mathrm{tan}\left(\frac{2+1}{2}\pi \right)=\mathrm{tan}\left(\pi +\frac{\pi }{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{tan}\left(x+\pi \cdot k\right)=\mathrm{tan}\left(x\right)\phantom{\rule{0ex}{0ex}}\mathrm{tan}\left(\pi +\frac{\pi }{2}\right)=\mathrm{tan}\left(\frac{\pi }{2}\right)=\mathrm{\infty }$
4)$\mathrm{cot}\left(\frac{11\pi }{4}\right)=\frac{1}{\mathrm{tan}\left(\frac{11\pi }{4}\right)}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\left(\frac{11\pi }{4}\right)=\mathrm{tan}\left(\frac{8+3}{4}\pi \right)=\mathrm{tan}\left(2\pi +\frac{3}{4}\pi \right)\phantom{\rule{0ex}{0ex}}\mathrm{tan}\left(x+\pi \cdot k\right)=\mathrm{tan}\left(x\right)\phantom{\rule{0ex}{0ex}}\mathrm{tan}\left(2\pi +\frac{3}{4}\pi \right)=\mathrm{tan}\left(\frac{3}{4}\pi \right)\phantom{\rule{0ex}{0ex}}\mathrm{tan}\left(\frac{3\pi }{4}\right)=\mathrm{tan}\left(\frac{\pi }{4}+\frac{\pi }{2}\right)=\mathrm{tan}\left(\frac{\pi }{4}\right)=-1$