Simultaneous equations with a parameter

mathlover2358

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Feb 6, 2020
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Hello. I've been trying for some time but I cannot figure out how to solve this :/
The question is what value of a gives a unique solution?

sinx + siny = 0.25
cosx + cosy = a
 

Romsek

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Nov 16, 2013
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We're not supposed to just give you answers but I found this problem interesting so you lucked out.

\(\displaystyle \sin(x) + \sin(y) = \dfrac 1 4\\
\sin^2(x) + \sin^2(y) + 2\sin(x)\sin(y) = \dfrac{1}{16}\\~\\

\cos(x) + \cos(y) = a\\
\cos^2(x) + \cos^2(y) + 2\cos(x)\cos(y) = a^2\\~\\
\text{now add them}\\
1 + 1 + 2(\sin(x)\sin(y) + \cos(x)\cos(y) = a^2 + \dfrac{1}{16}\\

2(1 + cos(x + y)) = a^2 + \dfrac{1}{16}\\

\cos(x+y) = \dfrac{a^2 + \frac{1}{16}}{2}-1\\

y = -x + \cos^{-1}\left(\dfrac{a^2 + \frac{1}{16}}{2}-1\right),~-x - \cos^{-1}\left(\dfrac{a^2 + \frac{1}{16}}{2}-1\right)\\

\text{In order for the solution to be unique we must have both arccosine terms evaluate to zero}\\
\text{This means that the argument of both must be 1}\\
\dfrac{a^2 + \frac{1}{16}}{2}-1 = 1\\

a = \pm \dfrac{3\sqrt{7}}{4}
\)
 

mathlover2358

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Feb 6, 2020
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Thank You so much
 

apple2357

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Mar 9, 2018
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I wondering if there is a geometric way to tackle this. I will have a think!

Can't see an easy way but tried sketching them implicitly

1581516840428.png


1581516879672.png
 
Last edited:

apple2357

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Zooming in

1581516936069.png
 
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