# Domain/range finding of composite trig functions

I am stuck on this problem in my book for finding the domain and range of composite functions

For problems 8a-e I used a developed method to solve for the implied domain of these functions which produced correct results

Steps were:

1. Use the domain of the first function and the range of the other trig function contained inside it
2. Find the intersection of these two values
3. substitute the contained trig function between the intersection value to find the domain
4. substitute the domain values into the function to find the range

However, once I got to "8f" I found that the method I was using for the previous questions didn't carry over into arctan problems.

I see now that I probably have a fundamental misunderstanding of the process needed to solve these forms of problems. I'm guessing that question 10 is also related to this type of question

Thanks

• Part of the problem is that te inverse trig functions are multivalued. For example a. for $(-\pi/2,\pi/2)$ then $0\le cos(x)\le 1$ so $sin^{-1}$ will have two ranges $(0,\pi/2)$ and $(\pi/2.\pi)$. For $(0,\pi)$ we have $-1\le cos(x)\le 1$, so the range of $y$ has two pairs of pieces covering either $(0,2\pi)$ or $-\pi,\pi)$. – herb steinberg Jan 19 at 4:22
• I think there may be a typo in 10, it should have $\cos^{-1}(\sin^{-1}(-0.5))$ and so on. For instance, you can definitely evaluate $\cos(\sin^{-1}(-0.5))$, given that $\sin^{-1}(-0.5)=\frac{-\pi}{6}$ the cosine is $\frac{\sqrt{3}}{2}$ at that point. – Nico Terry Jan 19 at 5:35
• @herbsteinberg could you please lay out the process you would use to solve one of these questions? Like the above steps I listed – CT-27-3555 Jan 20 at 4:25
• @NicoTerry okay but how would you then solve the amended problem for question 10? Having cos−1(-π/6) – CT-27-3555 Jan 20 at 4:34
• @herbsteinberg so the range in your above example would be (0,π) \ {π/2}. Meaning that you basically combined the two domains of the tri functions. And you are saying that the composite trig function has two ranges $"(0,2π) or (−π,π)."$. I'm still not following – CT-27-3555 Jan 20 at 4:40

It's not your fault that at least some of the questions are clearly incorrectly framed. In particular, 10 seems to be missing the inverse function on each of its outside $$\sin$$ or $$\cos$$ terms.
On 8f, you would want to note that $$\cos x\in[-1,1]$$ for all real numbers, and then identify the portion of the range of $$\tan^{-1}$$ corresponding to those values, i.e. find $$\tan^{-1}([-1,1])$$. Let me know if that gets you where you need to go!
• As originally written, I would argue that the description portion of the question is missing $^{-1}$ on its $\sin$ and $\cos$ as well. The domains of $\sin,\cos$ are restricted in that way when defining their inverses, but the domains of $\sin,\cos$ are all real numbers more generally. On further thought, though, I think your observation may suffice to solve all of those problems as written. Unfortunately, they seem to be very poorly written and do not further your understanding of the material. – Nico Terry Jan 31 at 14:11