c) First, we need to convert 7π/6 radians to degrees. π radians are equivalent to 180°, then:
[tex]\frac{7\pi}{6}radians=\frac{7\pi}{6}radians\cdot\frac{180\text{ \degree}}{\pi\text{ radians}}=210\text{ \degree}[/tex]From the table:
[tex]\sin (\frac{7\pi}{6})=\sin (210^o)=-\frac{1}{2}[/tex]8π/3 can be expressed as follows:
[tex]\frac{8}{3}\pi=2\pi+\frac{2}{3}\pi[/tex]The function tan(x) is periodic, with a period of π. This means that evaluating:
[tex]\tan (\frac{8}{3}\pi)[/tex]is the same as evaluating:
[tex]\tan (\frac{2}{3}\pi)[/tex]In this case, x (the input in the function) is translated 2π units to the left. From the periodicity of the function, the values are the same.
2π/3 radians is converted to degrees as follows:
[tex]\frac{2\pi}{3}radians=\frac{2\pi}{3}radians\cdot\frac{180\text{ \degree}}{\pi\text{ radians}}=120\text{ \degree}[/tex]From the table:
[tex]\tan (\frac{2}{3}\pi)=\tan (120^o)=-\sqrt[]{3}[/tex]Substituting these values into the original expression:
[tex]\begin{gathered} \sin (\frac{7\pi}{6})\cdot\tan (\frac{8}{3}\pi)= \\ =\sin (\frac{7\pi}{6})\cdot\tan (\frac{2}{3}\pi)= \\ =(-\frac{1}{2})\cdot(-\sqrt[]{3})= \\ =\frac{\sqrt[]{3}}{2} \end{gathered}[/tex]