Simplify Inverse Trigonometric Expression The problem asks to simplify the expression
$\arccos (\frac 3 5 \cos x + \frac 4 5 \sin x)$
where $x \in \; [\frac {-3\pi} 4 , \frac \pi 4]$.
Here's my approach.
Let $\frac 3 5 = r \cos y$ and $\frac 4 5 = r \sin y$.
Therefore, $r^2 = 1 \implies r = \pm 1\\
y = \arctan \frac 4 3$
Replacing $\frac 3 5$ and $\frac 4 5$ with their supposed values in the given expression we get,
$\arccos \; [r(\cos x \cos y + \sin x \sin y)]$
Now, 2 cases arise.
Case-I $(r = 1)$
The given expression becomes,
$\arccos \; [\cos(x - y)]\\
= x - y\\
= x - \arctan \frac 4 3$
Another equivalent answer is $\arctan \frac 4 3 - x$. This is the only answer according to my book.
Case-II $(r = -1)$
The given expression becomes,
$\arccos \; [-\cos(x - y)]\\
= \pi - (x - y)\\
= \pi - x + \arctan \frac 4 3$
Now this answer is what created the problem. My book doesn't give this answer. Is this answer wrong? Does it have to do something with the interval in which $x$ lies? Or is there something wrong with my approach. I mean if I suppose $\frac 3 5$ and $\frac 4 5$ as only $\cos y$ and $\sin y$ respectively, I wouldn't face this problem. Any help would be appreciated.
Edit
As I can suppose $\frac 3 5$ and $\frac 4 5$ to be $\cos y$ and $\sin y$ respectively, I believe there is nothing wrong is assuming them as $-\cos y$ and $-\sin y$ either. Or is there something wrong with this assumption? If there's nothing wrong with it then what about the answers we get with this assumption? Aren't they correct too? So shouldn't I suppose $\frac 3 5$ and $\frac 4 5$ as $\pm \cos y$ and $\pm \sin y$ respectively?
 A: First of all, you don't need to consider two cases. Since $y$ is not part of the original problem, but a new quantity that you're introducing (defining, choosing) yourself, you're free to define it to be $y=\arctan\frac{4}{3}$, i.e. you can choose $r=1$.
Now, to actual mistakes. The claim that "$x-\arctan\frac{4}{3}$ is equivalent to $\arctan\frac{4}{3}-x$" is plain wrong — they are different values, negatives of each other. Instead, you need to understand why your answer isn't correct but the one in the textbook is (or is it?). And the problem is that you can NOT  deduce that $\color{magenta}{\arccos(\cos(\alpha))=\alpha}$ in general — it is only true when $\color{red}{\alpha\in[0,\pi]}$.
In this problem, by the definition of the arctangent function, $y=\arctan\frac{4}{3}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. Moreover, since $\frac{4}{3}>1=\tan\frac{\pi}{4}$, we can actually conclude that $y\in\left(\frac{\pi}{4},\frac{\pi}{2}\right)$. Combining that with the given range for $x$, we see that
$$x\le\frac{\pi}{4} \quad \text{and} \quad y>\frac{\pi}{4} \quad \Longrightarrow \quad x-y<0,$$
so $\color{red}{x-y\notin[0,\pi]}$, and therefore $\color{magenta}{\arccos(\cos(x-y))\neq x-y}$. (I'm not going to go over the second case in detail, unless you'd like me to, but you made a similar mistake there.)
And then, here's the funniest part. The answer in the textbook isn't correct either, at least not with the given range for $x$.
$$x\in\left[-\frac{3\pi}{4},\frac{\pi}{4}\right] \quad \text{and} \quad y\in\left(\frac{\pi}{4},\frac{\pi}{2}\right) \quad \Longrightarrow \quad y-x\in\left(0,\frac{5\pi}{4}\right),$$
which can't guarantee that $x$ is withing $\color{red}{[0,\pi]}$. So for some values of $x$ this is going to work, but for some it doesn't. For example, if $x=-\frac{3\pi}{4}$ , then $\arccos\left(\frac{3}{5}\cos x+\frac{4}{5}\sin x\right)\approx3$, while $\arctan\frac{4}{3}-x\approx3.3$.
A: Let $cos \theta=3/5$
=$arccos(cos\theta * cos x+sin\theta * sin x)$
=$arccos( cos (\theta-x))$
=$\theta-x$
=$arc cos 3/5-x$
