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I come across this problem in an advanced maths textbook for grade 11 in my country. And it's marked a star, which means that it's a difficult exercise, and so, no solution for this problem is given.

I can solve problems asking for which conditions do $\sin(\alpha + \beta) = \sin(\alpha) + \sin(\beta)$, and $\tan(\alpha + \beta) = \tan(\alpha) + \tan(\beta)$ hold. They are pretty easy, and straight-forward. But for this problem ($\cos(\alpha + \beta) = \cos(\alpha) + \cos(\beta)$), I have tried using all kinds of formulas, from Sum of Angles, to Sum to Product, and Double Angles, but without any luck.

So, I think there should be some glitch here that I haven't been able to spot it out.

So I hope you guys can give me some hints, or just a little push as a start.

Any help would be greatly appreciated,

Thank you very much,

And have a good day, :D

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    $\begingroup$ The sum formula leads to a quadratic equation $2 x^2 + 2y^2 + 2 x y - 2 x^2 y - 2 x y^2 - 1= 0$ in $x = \cos(\alpha)$ and $y = \cos(\beta)$. $\endgroup$ – Winther Aug 4 '15 at 3:39
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    $\begingroup$ If you graph this, there is in fact a closed curve containing possible solutions within the box $ \ [-1, \ 1] \ \times \ [-1, \ 1 ] \ $ . The only "nice" solutions appear to be $ \ \cos( \pi \ \pm \ \frac{\pi}{3}) \ = \ \cos \pi \ + \ \cos ( \pm \frac{\pi}{3}) \ $ and $ \ \cos( 0) \ = \ \cos ( \frac{\pi}{3}) \ + \ \cos ( - \frac{\pi}{3}) \ $ . $\endgroup$ – colormegone Aug 4 '15 at 3:50
  • $\begingroup$ @RecklessReckoner: Another 'nice' solution I found is $\alpha = -\frac{\pi}{4}$, and $\beta = \frac{\pi}{2}$. $\endgroup$ – user49685 Aug 4 '15 at 3:58
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    $\begingroup$ Sorry, I took a quick look at the $ \ x-$ and $ \ y-$ intercepts and didn't think they were anything simple, but they're at $ \ \pm \frac{\sqrt{2}}{2} \ $ . So there are other such solutions as well. $\endgroup$ – colormegone Aug 4 '15 at 4:04
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The following contour plots hint us why the third equation defies many attempts.

enter image description here

Another possible explanation is that, if we substitute $x = \tan(\alpha/2)$ and $y = \tan(\beta/2)$ then the formulas

$$ \sin \alpha = \frac{2x}{1+x^2}, \quad \cos \alpha = \frac{1-x^2}{1+x^2}, \quad \tan \alpha = \frac{2x}{1-x^2} $$

show that

\begin{align*} \sin(\alpha+\beta) = \sin\alpha+\sin\beta &\quad \Longleftrightarrow \quad xy(x+y) = 0, \\ \tan(\alpha+\beta) = \tan\alpha+\tan\beta &\quad \Longleftrightarrow \quad xy(x+y)(1-xy) = 0, \\ \cos(\alpha+\beta) = \cos\alpha+\cos\beta &\quad \Longleftrightarrow \quad 3x^2y^2 - x^2 - 4xy - y^2 - 1 = 0. \end{align*}

This may be another reason why our equation seems impossible to solve in simple terms. Finally, one interesting observation is that

$$ \cos(\alpha+\beta) + 1 = \cos\alpha + \cos\beta \quad \Longleftrightarrow \quad xy(1-xy) = 0 $$

can be easily solved.

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  • $\begingroup$ I can find the relation between $\cos\alpha$, and $\cos\beta$ as Eric Wofsey, and Winther helped me, and from your way to tackle the problem, we can also find the relation between $\tan\left(\frac{\alpha}{2}\right)$, and $\tan\left(\frac{\beta}{2}\right)$. And from there, of course we can find the relations between $\alpha$, and $\beta$. But they all look pretty nasty. When looking at the plot, I really wonder if there is some straight relation between $\alpha$, and $\beta$? They seem to trace elliptical curves? $\endgroup$ – user49685 Aug 4 '15 at 4:27
  • $\begingroup$ @user49685, My original plan was to find a nice transformation that reveals some simple relationship so that I can parametrize the curve. My second part of the answer is a by-product of this plan, but I am still unsuccessful... $\endgroup$ – Sangchul Lee Aug 4 '15 at 4:36
  • $\begingroup$ By letting $\alpha = x+y$, and $\beta = x-y$, I can get to $\alpha = \pm \arccos\left( \frac{\cos t \pm \sqrt{\cos ^ 2 t + 2}}{2} \right) - t$, and $\beta = \pm \arccos\left( \frac{\cos t \pm \sqrt{\cos ^ 2 t + 2}}{2} \right) + t$. But it's still a mess, do you arrive at any 'nicer' parametric functions? :( $\endgroup$ – user49685 Aug 4 '15 at 4:42
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    $\begingroup$ @user49685, Writing $x = \cos \alpha$ and $y = \cos \beta$, I found that $$ (x, y) = \left( \frac{1}{2}\left( p \pm \sqrt{\frac{(p-1)(p^2 - 2p -2)}{p+1}} \right), \frac{1}{2}\left( p \mp \sqrt{\frac{(p-1)(p^2 - 2p -2)}{p+1}} \right) \right), $$ where $1-\sqrt{3} \leq p \leq 1$. Except for the arc-cosine term that will appear for $\alpha$ and $\beta$, this is now an algebraic expression. This still seems to nasty for me, though... $\endgroup$ – Sangchul Lee Aug 4 '15 at 16:38
  • $\begingroup$ I was wondering if these were ellipsis but they are not, though very close. If you set $a=\cos^{-1}\left(\frac{\sqrt{3}-1}2\right)\cos(t)$ and $b=\frac{2\pi}{3}\sin(t)$ and $\alpha=a+b-\pi, \beta=a-b-\pi$ so as to rotate, center and normalize the ellipsis to a unit circle, then $\cos(2a)+\cos(a+b)+\cos(a-b)\neq 0$ although the maximum is $\approx 0.03$, so nearly but not exactly an ellipsis. desmos.com/calculator/syo8kus6qc $\endgroup$ – zwim Mar 21 at 13:02
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Write $x=\cos\alpha$, $y=\cos\beta$, and now write $\cos(\alpha+\beta)$ in terms of just $x$ and $y$. Rearrange the terms of the equation $\cos(\alpha+\beta)=\cos\alpha+\cos\beta$ and square both sides. You should now get an equation you can use to solve for $y$ in terms of $x$. It doesn't look like the solution is going to be very pretty though!

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  • $\begingroup$ Thank you very much Eric, I solved it using your suggestion (and Winther's comment). Thank you very much for your help, and I really hope you don't mind if I leave this topic open for one, or two more days before accepting your answer, in case someone else wants to post their answer. :D $\endgroup$ – user49685 Aug 4 '15 at 4:14
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There may be a better way using the aforementioned transforms with $x$ and $y$, but just looking at the trig as is you can find the "nice" solutions as $\alpha = 2 \pi n_1+pi, \hspace{.1cm} \beta = \frac{1}{3} (6 \pi n_2 \pm \pi)$ and $\alpha = \frac{1}{3} (6 \pi n_2 \pm \pi), \hspace{.1cm} \beta = 2 \pi n_1+\pi$ for some integers $n_1$ and $n_2$.

I found those just from checking numbers and without much sophistication. Examining the result with a CAS I found there is a rather general and very nasty result involving tangent for a range of $\alpha$'s.

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