Finding solutions of tan - confused where i went wrong

[wduk]

Hello

I am trying to find all solutions for tan theta =-3 [-pi,pi]

So i calculated the inverse and drew a graph and this is what i calculated:

Solution one:
pi - (-1.249) = 4.39 radians

Solution two:
0 - 1.249 = -1.25 radians (got this one correct)

According to my book the first solution is wrong, and that the answer is 1.89 but i don't see how. From my graph i am subtracting the negative value of the inverse function solution from pi, but the book is adding the negative value... i don't understand why i am suppose to add from pi?

 

[ksdhart2]

Well, if you draw out the graph, you can see that -1.2489 and 1.8926 are the only two solutions within the interval [-pi, pi]. 4.3906 isn't a solution to the equation at all (because $\displaystyle tan(4.3906) \approx 3$), although -4.3906 is. Intuitively, we can see that this makes sense by looking at the principal value, $\displaystyle tan^{-1}(-3) \approx -1.2489$, and keeping in mind that the period of tangent is pi (i.e. the graph repeats every pi radians or 180 degrees). That means that the full set of solutions is $\displaystyle \theta=n \pi + tan^{-1}(-3) = n \pi - tan^{-1}(3) \: \: \: n \in \mathbb{Z}$. Hopefully that clears up your confusion about whether to add or subtract.

 

[wduk]

I kinda understand it but the logic throws me when tan^-1(-3) is a negative value, yet on the graph in order to get the second solution between 0 and pi we have to use the positive version of it to have pi - (+1.24). But what i don't get is why this would be correct, since tan^-1(-3) is never positive. Isn't that a bit like saying f^-1(x) = -3 then suddenly just deciding to use +3 when we know f^1(x) is not +3 it's just mathematically incorrect..

Visually on my graph is makes complete sense how ever It must be a positive value to subtract from pi, so some where between the visual on my graph and number logic is where i don't get why its okay to use the positive value.

 

[ksdhart2]

I kinda understand it but the logic throws me when tan^-1(-3) is a negative value, yet on the graph in order to get the second solution between 0 and pi we have to use the positive version of it to have pi - (+1.24). But what i don't get is why this would be correct, since tan^-1(-3) is never positive. Isn't that a bit like saying f^-1(x) = -3 then suddenly just deciding to use +3 when we know f^1(x) is not +3 it's just mathematically incorrect..

Visually on my graph is makes complete sense how ever It must be a positive value to subtract from pi, so some where between the visual on my graph and number logic is where i don't get why its okay to use the positive value.

It's not so much that we're just "deciding" arbitrarily to use either positive or negative 3, in order to make the answers fit. Rather, we're using a known property of tan(x): tan(x) is an odd function, so that means that, for any x, tan(-x)=-tan(x). It then follows that the inverse tangent is also odd and obeys the same relationship. From that, we can see that we're always adding something to n pi, but we can choose what to add - either $\displaystyle tan^{-1}(-3) \approx -1.2489$ or $\displaystyle -tan^{-1}(3) \approx -1.2489$. But if we choose to add the negative "version" of the inverse tangent of 3, that means the sign will "appear" to change because we're adding a negative (i.e. subtracting). Does that make any sense? I fear I might be accidentally muddling the concept even further.