Question

What is the value of tan (-2pi/3)?

Answer 1

This revolves around exact trig values - no easy way to say this, you just need to memorise them. They are there for sin cos and tan, but I will give you the main tan ones below - note this is RADIANS (always work in them when you can, everything is better):

tan0: 0
tanpi/6: 1/sqrt(3) 
tanpi/4: 1
tanpi/3: sqrt(3)
tanpi/2: undefined

Now we just need to equate -2pi/3 to something we understand. 2pi/3 is 1/3 of the way round a circle, so -2pi/3 is 1/3 of the way round the circle going backwards (anticlockwise), so on a diagram we already know it's in the third quadrant of the circle (somewhere between pi and 3pi/2 rads).
We also know it is pi/3 away from pi, so we are looking at sqrt(3) or -sqrt(3) because of those exact values.

Now we just need to work out if it's positive or negative. You can look up a graph of tan and it'll show that the graph intercepts y at (0,0) and has a period of pi rads. Therefore between pi and 3pi/2 rads, the values of tan are positive. Therefore, this gives us our answer of sqrt(3).

Answer 2

Answer:

$tan(\frac{-2\pi}{3} ) =\sqrt{3} \approx1.732$

Step-by-step explanation:

First of all let's define the tangent function:

$tan(\theta )=\frac{Opposite}{Adjacent} =\frac{sin(\theta)}{cos(\theta)}$

Now, let's define the standard angles, standard angles are those that have values that appear very often in everyday life. These angles are 30°=π/6, 45°=π/4, and 60°=π/3, and the angles 0°, 90°=π/2, 120°= 2π/3, 180°=π, 270°=3π/2, and 360°=2π. The latter, although not defined as 'standard', are also very common. Here are the values:

$cos(0)=1\hspace{25}sin(0)=0\\cos(\frac{\pi}{6} )=\frac{\sqrt{3}}{2} \hspace{15}sin(\frac{\pi}{6})=\frac{1}{2} \\cos(\frac{\pi}{4} )=\frac{\sqrt{2}}{2} \hspace{15}sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$

$cos(\frac{\pi}{3})=\frac{1}{2}\hspace{28} sin(\frac{\pi}{3} )=\frac{\sqrt{3}}{2} \\cos(\frac{\pi}{2})=0\hspace{28} sin(\frac{\pi}{2} )=1}\\cos(\frac{2\pi}{3})=-\frac{1}{2}\hspace{15} sin(\frac{2\pi}{3} )=\frac{\sqrt{3}}{2}$

$cos(\pi)=-1\hspace{25}sin(\pi)=0\\cos(\frac{3\pi}{2} )=0 \hspace{28}sin(\frac{3\pi}{2})=-1 \\cos(2\pi )=1 \hspace{27}sin(2\pi)=0$

Also you need to keep in mind that cosine function is an even function, and sine function is an odd function, that is:

$cos(-\theta)=cos(\theta)\\\\sin(-\theta)=-sin(\theta)$

Using these definitions you are able to solve the problem:

$tan(\frac{-2\pi}{3} ) =\frac{sin(\frac{-2\pi}{3} )}{cos(\frac{-2\pi}{3} )} = \frac{\frac{-\sqrt{3} }{2} }{\frac{-1}{2} } = \sqrt{3} \approx1.732$