Answer:
[tex]tan(\frac{11\pi}{6}) =-\frac{\sqrt{3} }{3}[/tex]
Step-by-step explanation:
Notice that [tex]\frac{11\pi}{6}[/tex] is an angle in the fourth quadrant (where the tangent is negative), and the angle is in fact equivalent to [tex]-\frac{\pi}{6}[/tex]. This is one of the special angles for which the sine and cosine functions, as well as the tangent function have well know values:
Recall that the tangent is defined as
[tex]tan(\theta)=\frac{sin(\theta)}{cos(\theta)}[/tex]
and for this angle ( [tex]\frac{11\pi}{6}[/tex] ) the value of the sine and cosine functions are well known:
[tex]sin (\frac{11\pi}{6}) =-\frac{1}{2} \\cos( \frac{11\pi}{6}) =\frac{\sqrt{3} }{2}[/tex]
Then, the tangent would be:
[tex]tan(\theta)=\frac{sin(\theta)}{cos(\theta)}\\tan(\frac{11\pi}{6}) = \frac{-\frac{1}{2} }{\frac{\sqrt{3} }{2} } \\tan(\frac{11\pi}{6}) =-\frac{1}{\sqrt{3} } \\tan(\frac{11\pi}{6}) =-\frac{\sqrt{3} }{3}[/tex]
