1. ## A trigonometry question

Hello,
I am doing a trigonometric derivative problem, but I think I have forgotten the right step of when to add a $\pi n$ or when to add a 2 $\pi n$. For example in my book, one of the problem says cosx = $
\frac{{ - 1}}{2}
$
$
\Rightarrow
$
x = $
2\frac{\pi }{3} + 2\pi n
$
or $
4\frac{\pi }{3} + 2\pi n
$
. Now at other problem I see it says tanx = 1 $
\Rightarrow
$
x = $
\frac{\pi }{4} + \pi n
$
(n is an integer). Can you give me an explanation? Thank you for the help.

2. Originally Posted by Mathlv
Hello,
I am doing a trigonometric derivative problem, but I think I have forgotten the right step of when to add a $\pi n$ or when to add a 2 $\pi n$. For example in my book, one of the problem says Cosx = $
\frac{{ - 1}}{2}
$
$
\Rightarrow
$
x = $
2\frac{\pi }{3} + 2\pi n
$
or $
4\frac{\pi }{3} + 2\pi n
$
. Now at other problem I see it says tanx = 1 $
\Rightarrow
$
x = $
\frac{\pi }{4} + \pi n
$
(n is an integer). Thank you for the help.
The period of the sine and cosine functions is $2\pi$. So you would find all solutions from the unit circle then add $2\pi n$.

The period of the tangent function is $\pi$. So you would find a solution then add $\pi n$.

3. Originally Posted by Mathlv
Hello,
I am doing a trigonometric derivative problem, but I think I have forgotten the right step of when to add a $\pi n$ or when to add a 2 $\pi n$. For example in my book, one of the problem says cosx = $
\frac{{ - 1}}{2}
$
$
\Rightarrow
$
x = $
2\frac{\pi }{3} + 2\pi n
$
or $
4\frac{\pi }{3} + 2\pi n
$
. Now at other problem I see it says tanx = 1 $
\Rightarrow
$
x = $
\frac{\pi }{4} + \pi n
$
(n is an integer). Can you give me an explanation? Thank you for the help.

linking trig functions with the unit circle is the best way to get a feel for them in the early going. they are defined by the ratio of the sides one can inscribe therein, where the sides are those of a right triangle. if you study the geometry for a little while, it'll be like riding a bike thereafter.

4. ## A trigonometric question

Thank you very much for the response and the help Prove It and Vince. I get it now.