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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?

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2 Answers 2

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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 $\arctan\frac{4}{3}-x\approx3.3$.

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  • $\begingroup$ Thanks! You certainly did correct me for that mistake in Case-I. I didn't keep in mind that $x \in [\frac {-3\pi} 4 , \frac \pi 4]$ and treated $x$ as a real number. I have edited my question and elaborated what I really want to ask about. Do check it out up there. $\endgroup$
    – Shinsekai no Kami
    Commented Feb 3, 2017 at 11:28
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    $\begingroup$ @SamInuyashaANMF: Regarding the question in your latest edit. No, nothing stops us from making those other choices.Remember that $y$ is NOT part of the original problem -- $3/5$ and $4/5$ are. We choose $y$ as a tool to solve the problem. We can make those other choices as well, if we want to. Sometimes we have good reasons to choose one over another -- because it would lead to a more efficient solution. But sometimes it literally doesn't matter, and we go with whatever we like better. In the end it only changes the form of the answer, but not the answer. $\endgroup$
    – zipirovich
    Commented Feb 4, 2017 at 2:40
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    $\begingroup$ @SamInuyashaANMF: Just like you pointed out in your comment to another answer: we can end up with $\arctan(4/3)$ or $\arccos(3/5)$ in our answer, but it makes no difference because it's the same thing. So all those other choices are just as fine. For each choice, you'd have to make sure to determine the respective $y$ correctly, though. For example, if you set $-\sin y=\frac{3}{5}$ and $-\cos y=\frac{4}{5}$, then your $y$ will be in the third quadrant, so you can say $y=\arcsin\frac{3}{5}+\pi$, but NOT "$y=\arcsin\left(-\frac{3}{5}\right)$". If you do that correctly, you'll be fine. $\endgroup$
    – zipirovich
    Commented Feb 4, 2017 at 2:48
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    $\begingroup$ You already proved that the answer in my book is incorrect as all values of $x$ don't satisfy both, the expression, and the solution. But I found out that this was the best answer of all. The contradiction in the values of the two expressions was because all values of $\arctan ⁡\frac 4 3 − x$ don't lie in the interval $[\frac {3 \pi} {4} , \frac \pi 4]$. Thus, for the values of $\arctan \frac 4 3 − x$ which lie outside this interval, the two expressions are unequal. $\endgroup$
    – Shinsekai no Kami
    Commented Feb 5, 2017 at 14:11
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    $\begingroup$ As I tried replacing $\frac 3 5$ and $\frac 4 5$ with different combinations of $\cos y$ , $\sin y$ , $-\cos y$ and $-\sin y$, I found that I get a similar answer when $y$ is taken to be in the second quadrant, i.e., $\frac 3 5 = -\cos y$ and $\frac 4 5 = \sin y$ (this one isn't the best since it requires unnecessary working and thinking). When I took $y$ to be in the fourth quadrant, I got an answer similar to the one in which I had made the mistake, the $x - \arctan \frac 4 3$ one. So yeah, now I have clearly understood this. Thanks again! $\endgroup$
    – Shinsekai no Kami
    Commented Feb 5, 2017 at 14:12
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Let $cos \theta=3/5$

=$arccos(cos\theta * cos x+sin\theta * sin x)$

=$arccos( cos (\theta-x))$

=$\theta-x$

=$arc cos 3/5-x$

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  • $\begingroup$ My friend, this does not make a lot of difference, as $\arctan \frac 4 3$ is the same as $\arccos \frac 3 5$. In your context, I wish to know that why don't we consider $\cos \theta$ as $\pm \frac 3 5$ instead of just $\frac 3 5$. $\endgroup$
    – Shinsekai no Kami
    Commented Feb 3, 2017 at 10:43

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