How do I bypass the limit for the tan function on the calculator? So basically, my partner and I are creating problems for a project. She created one where we aren't sure what the correct answer is.
Problem:
$Find \ cos\frac{\theta}{2} \ if \ tan \ \theta = \frac{3}{4}; \pi < \theta<\frac{3\pi}{2}$
I'll show you what we did.
First, we used the pythagorean formula to find $cos \ \theta$ (we need it later). The half-angle formula for cosine is $cos \frac {\theta}{2} = \pm \sqrt{\frac{1 + cos \ \theta}{2}}$.
These are our steps.

*

*$cos \frac {tan^{-1}\frac{3}{4}}{2} = \pm \sqrt{\frac{1 + cos \ (tan^{-1}\frac{3}{4})}{2}}$

*$ = \pm \sqrt{\frac{1 + \frac{4}{5}}{2}}$

*$ = \pm \sqrt{\frac{\frac{9}{5}}{2}}$

*$ = \pm \sqrt{\frac{9}{10}}$

*$ = \pm \frac{3\sqrt{10}}{10}$
To decide whether we have a negative sign or a positive, we have to look at the domain.
$ \pi < \theta<\frac{3\pi}{2} \ changes \ to  \ \frac{\pi}{2} < \frac{\theta}{2} < \frac{3\pi}{4}$
We now know that $cos \frac {\theta}{2}$ is located in the second quadrant, which makes our final answer   $cos \frac {tan^{-1}\frac{3}{4}}{2} = - \frac{3\sqrt{10}}{10}$
But then, I checked the calculator.
The values are equal except for the negative sign
My partner said that the calculator thinks tangent is in the first quadrant, and if it were, cosine would be positive. But since we are in the third quadrant, cosine is in the second quadrant. I know there are domain restrictions, but I don't entirely understand what she meant. Also, how do I get the calculator to display the same answer?

You cannot do the inverse tangent because the range of inverse tangent is $(-\frac{\pi}{2},\frac{\pi}{2})$, and so your answer for step 2 (and also the answer from your calculator) is already incorrect.
I would suggest you do this:
Draw the triangle in the third quadrant (because it is in the range $(\pi, \frac{3\pi}{2})$, with adjacent side (on the x-axis) $x=-4$ and opposite side (on the y-axis) $y=-3$. The hypotenuse is $5$ by Pythagoras. Then, $\cos \theta = -\frac{4}{5}$. Using the cosine half-angle formula, $\cos \frac{\theta}{2}=\pm\sqrt{\frac{1+\cos \theta}{2}}=\pm\sqrt{\frac{1+(-\frac{4}{5})}{2}}=\pm\sqrt{\frac{1}{10}}=\pm \frac{\sqrt{10}}{10}$ Because the angle is in the range $(\pi, \frac{3\pi}{2})$, then half the angle is in the range $(\frac{\pi}{2}, \frac{3\pi}{4})$, so $\cos$ is negative. Therefore, your answer is $\boxed{-\frac{\sqrt{10}}{10}}$.
