# 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.

1. $$cos \frac {tan^{-1}\frac{3}{4}}{2} = \pm \sqrt{\frac{1 + cos \ (tan^{-1}\frac{3}{4})}{2}}$$
2. $$= \pm \sqrt{\frac{1 + \frac{4}{5}}{2}}$$
3. $$= \pm \sqrt{\frac{\frac{9}{5}}{2}}$$
4. $$= \pm \sqrt{\frac{9}{10}}$$
5. $$= \pm \frac{3\sqrt{10}}{10}$$

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 $$\boxed{-\frac{\sqrt{10}}{10}}$$.

• so is there a specific reason you used cosine? Feb 25, 2021 at 21:39
• @Spongebob I used cosine because you did. Feb 25, 2021 at 21:41
• oh so you took my $\theta = tan^-1 \frac{3}{4}$ and changed it to $\theta = cos^-1 \frac{4}{5}$, but the range of cosine is between 0 and $\pi$, so how does that work Feb 25, 2021 at 21:46
• Range of cosine is between $0$ and $\pi$?? It's between $[-1, 1]$ Feb 25, 2021 at 21:54
• oh i meant inverse cosine Feb 25, 2021 at 22:28

The "limit" you're trying to bypass is the fact that $$\tan^{-1}$$ only produces output values between $$-\frac\pi2$$ and $$\frac\pi2.$$ It will never directly give you an angle in the third quadrant.

The way you "get around" this is you remember that $$\tan\theta = \tan(\theta + \pi) = \tan(\theta + 2\pi) = \tan(\theta - \pi),$$ that is, no matter what whole multiple of $$\pi$$ you add or subtract from $$\theta,$$ the tangent of the resulting angle is always the same as $$\tan\theta.$$

Since you know $$\pi < \theta<\frac{3\pi}{2}$$, you know that if you subtract $$\pi$$ from $$\theta$$ you will get an angle $$\alpha = \theta - \pi$$ where $$0 < \alpha < \frac\pi2.$$ Since $$\tan\alpha = \tan\theta = \frac34,$$ and $$\alpha$$ is an angle that you can get as an answer from $$\tan^{-1},$$ you know that $$\alpha = \tan^{-1}\frac34.$$ Therefore $$\theta = \alpha + \pi = \pi + \tan^{-1}\frac34.$$

To continue along the lines you were trying to follow, you will have to compute $$\cos\frac{\pi + \tan^{-1}(3/4)}{2}.$$ One way to continue is to use the cosine half-angle formula, just as you attempted, but with the correct angle, so you get $$\cos\frac\theta2 = \cos\frac{\pi + \tan^{-1}\frac34}{2} = \pm\sqrt{\frac{1 + \cos\left(\pi + \tan^{-1}\frac34\right)}{2}}.$$ Now use the fact that $$\cos(\pi + \beta) = -\cos(\beta).$$ So now you have $$\cos\frac\theta2 = \pm\sqrt{\frac{1 - \cos\left(\tan^{-1}\frac34\right)}{2}}.$$ You can get to the correct answer from there using similar methods to what you already used in your attempt for this question, but you get a different final answer since you have a difference inside the square root rather than a sum.

Alternatively, write $$\cos\frac{\pi + \tan^{-1}(3/4)}{2} = \cos\left(\frac\pi2 + \frac{\tan^{-1}(3/4)}{2}\right),$$ and use the fact that $$\cos\left(\frac\pi2 + \phi\right) = -\sin\phi.$$ Therefore $$\cos\frac\theta2 = -\sin\left(\frac{\tan^{-1}\frac34}{2}\right) = \pm\sqrt{\frac{1 - \cos\left(\tan^{-1}\frac34\right)}{2}}.$$ The same result follows.

Having said all that, I think the easier way to solve this is not to "get around" the $$\tan^{-1}$$ function at all, but rather draw a triangle as recommended in the other answer. That answer uses straightforward geometric intuition that is much less likely to lead you astray.