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I had a question in my Math mcq test.

If $\cos x + \cos y = 2$ find the value of $\cos(x-y)$.

I couldn't get a way to calculate the value. So I just substituted $x = y = 0$. (It seemed obvious to me) So I got $\cos(x-y) = \cos 0 = 1$.

But can we actually solve it? I am looking for ways that can be done in short time ($2-3$) minutes. As longer solutions may not be feasible in a time bound examination...

Michael Albanese
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Sudhanshu
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    Think about the maximum value of $\cos$ to find all solutions. – Git Gud Aug 08 '14 at 12:22
  • I see there are good answers below, but this question raises a different question to me. This is a kind of test question that always makes me uncomfortable, from a problem-writing perspective. As a test question it always is the case that the student has the meta-information for each problem that it is meant to be solvable, and therefore (in this specific problem) the latter value is truly determined by the former. Thus in a sense simply testing $x=0$ is a reasonable solution! – Jason Knapp Aug 08 '14 at 12:43

3 Answers3

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Since $\cos(x)\in[-1,1]$ you are solving $s+t=2$ with $(s,t)\in[-1,1]\times[-1,1]$. Now the line $s=-t+2$ only touches the corner of the square $[-1,1]\times[-1,1]$, namely at $s=t=1$.

Note that here $s=\cos(x)$ and $t=\cos(y)$.

enter image description here

String
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  • I like your representation of the question. Makes it really easy to understand!! – Sudhanshu Aug 08 '14 at 13:17
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    @Sudhanshu: Yes, I felt that since you already figured out that $x=y=0$ would work, the problem was to be certain that no essentially different solutions existed. Dmoreno is quite correct that it should be stated that multiples of $2\pi$ does not affect the result, though! – String Aug 08 '14 at 13:25
  • Thanks for mentioning! Very nice answer by the way. – Dmoreno Aug 08 '14 at 13:33
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Since the maximum value of $\cos{\theta}$ is $1$, this means that $x$ and $y$ must be those angles that reach $\cos{x} = \cos{y} = 1$ and hence $\cos{x} + \cos{y} = 2$. So, when is $\cos{\theta} = 1$? When $\theta = 2 \pi k$ for any integer $k$ so the difference between $x$ and $y$ is given by:

$$x - y = 2 \pi m - 2 \pi n = 2 \pi (m - n) = 2 \pi k,$$ being $m$ and $n$ integers as well. Therefore, $\cos{(x-y)} = \cos{2 \pi k } = \ldots$

Hope this helps.

Dmoreno
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$\cos x+\cos y=2$ implies $\cos x=\cos y=1$, and ....

Hamou
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