1. ## Question

is arctan(-pi/3) same thing as arctan(2pi/3)
and arctan(-4pi/3) same thing as arctan(5pi/3)?

if not, why is it that one?

hence solve $tanx=-\sqrt{3}$

thanks

2. Originally Posted by BabyMilo

hence solve $tanx=-\sqrt{3}$

thanks
In the case of

$\tan(x) =-\sqrt{3}$

In the first snd and 4th quadrants $x = \pi - \frac{\pi}{3} , 2\pi - \frac{\pi}{3}= \frac{2\pi}{3} , \frac{5\pi}{3}$

As $\tan(x)$ has the period $\pi$ in general you can say

$x = \frac{2\pi}{3} \pm n\pi , n = 0,1,2,\dots$

3. Originally Posted by BabyMilo
is arctan(-pi/3) same thing as arctan(2pi/3)
and arctan(-4pi/3) same thing as arctan(5pi/3)?
if not, why is it that one?
The correct answer is: no they are not the same.
The $\arctan$ is injection (one-to-one function) from $\mathbb{R}$ to $\left( {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right)$.

4. but they do give the same solution on my calculator?

5. Originally Posted by BabyMilo
but they do give the same solution on my calculator?
In that case, I would get a better calculator if I were you.
$\arctan \left( {\frac{{ - \pi }}{3}} \right) = -0.80844879263$
$\arctan \left( {\frac{{2\pi }}{3}} \right) = 1.125338832884$

6. Originally Posted by BabyMilo
is arctan(-pi/3) same thing as arctan(2pi/3)
and arctan(-4pi/3) same thing as arctan(5pi/3)?

These may have been meant to be written using tan instead of arctan,
as the brackets contain common angles.

if not, why is it that one?

hence solve $tanx=-\sqrt{3}$

thanks
$tan\left(-\frac{\pi}{3}\right)=tan\left(\frac{2{\pi}}{3}\rig ht)$

$tan\left(-\frac{4{\pi}}{3}\right)=tan\left(\frac{5{\pi}}{3}\ right)$

$tan\left(\frac{\pi}{3}\right)=\sqrt{3}$

$tan(-A)=-tanA$

$tan\left(-\frac{\pi}{3}\right)=-\sqrt{3}$

$tan\left(-\frac{\pi}{3}\right)=tan\left(\frac{2{\pi}}{3}\rig ht)$

Therefore, if

$tanx=-\sqrt{3}$

then

$x=arctan\left(\frac{2{\pi}}{3}\right)$

7. I just realize what i was asking.
I was supposed to ask tan(2pi/3)
same thing as tan(-pi/3)

thanks