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I'm trying to solve the following system:

$$ \sin(x) + \cos(y) = 0.6\\ \cos(x) - \sin(y) = 0.2\\ $$

Solving for y in terms of x:

$$ y=\arccos(0.6-\sin(x))=\arcsin(\cos(x) -0.2) $$

Therefore: $$ \frac{\pi}{2} - \arcsin(0.6-\sin(x)) = \arcsin(\cos(x) - 0.2) $$ Forgot to add: $$ \frac{-\pi}{2}\le x,y \le \frac{\pi}{2} $$ Not too sure how to proceed from here, any hints?

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The assumption $$-\frac{\pi}{2} \leq x,y \leq \frac{\pi}{2}$$ is beautiful. Why? Because then we can use the substitution $$\left\{\begin{align} x &= \arctan\alpha \\ y &= \arctan\beta \end{align}\right.$$

Why do we want to use this substitution? Because of the identities $$\begin{align}\sin(\arctan\alpha) &= \frac{\alpha}{\sqrt{1+\alpha^2}} \\ \cos(\arctan\beta) &= \frac{1}{\sqrt{1+\beta^2}}\end{align}$$ Now write the new equations in $\alpha,\beta$ $$\begin{align}\frac{\alpha}{\sqrt{1+\alpha^2}} + \frac{1}{\sqrt{1+\beta^2}} &= 0.6 \\ \frac{1}{\sqrt{1+\beta^2}} - \frac{\alpha}{\sqrt{1+\alpha^2}} &= 0.4\end{align}$$ Some rearranging gets us $$\begin{align}\frac{(\alpha-0.6)\sqrt{1+\beta^2}+1}{\sqrt{(1+\alpha^2)(1+\beta^2)}} & = 0 \\[2ex] \frac{1-(\alpha+0.4)\sqrt{1+\beta^2}}{\sqrt{(1+\alpha^2)(1+\beta^2)}} &= 0\end{align}$$ The only parts that will make the expressions null are the numerators. Put the $1$ on the RHS in both equations, and square both sides. After rearranging again, you'll get $$\begin{align}\alpha^2 -1.2\alpha + 0.36 &= 1/(1+\beta^2) \\ \alpha^2 + 0.8\alpha + 0.16 & = 1/(1+\beta^2) \end{align}$$ One possible value for $\alpha$ comes easily! Just subtract the last two equations: $$2\alpha + 0.52 = 0 \implies \fbox{$\alpha = -0.26$}$$ We can unravel the rest now. Divide $P(\alpha) = (\alpha^2 - 1.2\alpha+0.36-1/(1+\beta^2))$ by $(\alpha +0.26)$ (as polynomials). You should do the calculations yourself to double check, but mine give a quotient $q$ and remainder $r$ of $$\begin{align}q(\alpha) &= \alpha - 1.46 \\ r(\alpha) &= 0.7396 - \frac{1}{1+\beta^2}\end{align}$$

The remainder must be zero, since we found $\alpha = -0.26$ to be a solution. This gives two pairs of (possible) solutions $(\alpha,\beta) \approx (-0.26,\pm 0.59337)$. Now, equating the quotient to zero gives a possible value $\alpha = 1.46$. Note that the remainder can also be obtained by substituting in $P(\alpha)$. We could have done this before but we also wanted the quotient which is why we carried out the entire division. Now we don't care about the quotient (it's $(\alpha + 0.26)$), so just substitute and equate to zero to get $$P(1.46) = 0.6244 -\frac{1}{1+\beta^2} = 0$$

This gets you two more pairs of solutions. Now you need to do this entire process over again with the second equation $$\alpha^2 + 0.8\alpha + 0.16 = 1/(1+\beta^2)$$

again starting with $\alpha = -0.26$.

In the end you'll have multiple candidates (I keep saying "possible solutions" or candidates because we have possibly introduced, when we squared the equations, values that don't solve the original equations). Check which ones work with the first $\alpha,\beta$ equations. Since the exercise obviously is to find numerical solutions, going back to $x,y$ is then a simple matter of using the tan^(-1) key on your calculator (in fact, I already had to use the square root key).

Oh, and $\arctan$ doesn't ever reach $-\pi/2$ or $\pi/2$, so let's check real quick if those are solutions: $$\quad1\neq 0.6 \\ -1 \neq 0.2$$ Nope, they aren't. We've covered every possibility now.

Edit: I've noticed a mistake in subtracting the two equations, the first value for alpha (in the box) should be $0.1$. This changes the other results of course, but I'll edit those tommorow (all in all it's just solving quadratics).

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Let's rewrite $\sin t=-i\frac{e^{it}-e^{-it}}{2}, \cos t=\frac{e^{it}+e^{-it}}{2}$: $$\begin{cases}-i(e^{ix}-e^{-ix})+e^{iy}+e^{-iy}=1.2&|\cdot i\\ e^{ix}+e^{-ix}+i(e^{iy}-e^{-iy})=0.4\end{cases}$$ $$\begin{cases}e^{ix}-e^{-ix}-ie^{iy}-ie^{-iy}=1.2i\\ e^{ix}+e^{-ix}+i(e^{iy}-e^{-iy})=0.4\end{cases}$$ Add/subtract them: $$\begin{cases}2e^{-ix}+2ie^{iy}=0.4-1.2i\\ 2e^{ix}-2ie^{-iy}=0.4+1.2i\end{cases}$$ Now we see it's only one equation, the other is the complex conjugate. $$e^{-ix}+ie^{iy}=0.2-0.6i$$ How do we extract all the information from it? Yeah, $|\,\cdot\,|$ and $\arg$.

$$(e^{-ix}+ie^{iy})(e^{ix}-ie^{-iy})=0.04+0.36$$ $$1-ie^{-i(x+y)}+ie^{i(x+y)}+1=0.04+0.36$$ $$\sin(x+y)=0.8$$ $$\arg (e^{-ix}+ie^{iy})=-x+\arg (1+e^{i(y+x+\frac{\pi}{2})})=-x+\frac{y+x+\frac{\pi}{2}}{2}=\arg (0.2-0.6i)=-\arctan 3+2\pi k$$ $$\begin{cases}\cos\left(\frac{\pi}{2}-(x+y)\right)=0.8\\ y-x+\frac{\pi}{2}=-2\arctan 3+4\pi k\end{cases}$$ $$\begin{cases}\frac{\pi}{2}-(x+y)=\pm\arccos 0.8+2\pi k\\ y-x+\frac{\pi}{2}=-2\arctan 3+4\pi k\end{cases}$$ A bit dirty, though.

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here is one way to do this. $$\begin{align}\sin x + \cos y &= 0.6\\ \sin (x) + \sin(\pi/2- y) &= 0.6 \\ \sin(\pi/4+x/2-y/2)\cos(x/2+y/2-\pi/4) &= 0.3\tag 1\end{align}$$

in the same way $$\begin{align}\cos x - \sin y &= 0.2\\ \cos(x) - \cos(\pi/2- y) &= 0.2 \\ \sin(\pi/4+x/2-y/2)\sin(x/2+y/2-\pi/4) &= -0.1\tag 2\end{align}$$ dividing $(1)$ by $2$, we have $$\tan(x/2+y/2-\pi/4) = -\frac13\tag 3 $$ which gives you $$x+y=\pi/2 -2\tan^{-1}(1/3)+ 2k\pi \tag 4 $$

using $(3)$ in $(2)$ we find that $$\sin(\pi/4+x/2-y/2)\sin(x/2+y/2-\pi/4) = \sin(\pi/4+x/2-y/2)\left(\mp\frac1{\sqrt{10}}\right)=-0.1 $$ so that $$x-y = -\pi/2 \pm 2\sin^{-1}(1/\sqrt{10}), -\pi/2 \mp 2\sin^{-1}(1/\sqrt{10}) \tag 5$$

solving $(4)$ and $5$ you get four pairs of solutions. i will do one pair.

$$x = 0, y = \pi/2-2\tan^{-1}(1/3) $$ you can verify that $$\sin x + \cos y = 0.6, \cos x - \sin y = 0.2 $$

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Square the two equations, add, and simplify, giving $$\sin(x-y) = -0.74.$$ Then subtract the two squared equations and simplify, giving $$\left(\cos\left(\frac{x-y}{2}\right) + \sin\left(\frac{x-y}{2}\right)\right)^2 \sin(x+y) = 0.1.$$ The first equation allows you to determine $x-y$; from that information, using the half-angle formulas and the second equation, you can determine $x+y$.

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  • $\begingroup$ Yeah, to re-write to $\frac{x+y}{2}$ and $\frac{x-y}{2}$ seems a good way to me. $\endgroup$ – Alexey Burdin May 11 '15 at 18:57
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here is still another, more geometric way, to do this problem. we will use complex numbers to make the computations easier. let the point $b$ be given by $$b = 0.2+0.6i = e^{it}, t = \tan^{-1}(3).$$ we have $$\sin x + \cos y = 0.6, \quad \cos x - \sin y = 0.2 $$ multiplying the first one by $i$ and adding it to the second one, we get $$e^{ix} + e^{i(y+\pi/2)} = b\tag 1$$ we interpret $(1)$ as looking for two complex numbers of unit magnitude and their sum is $b.$ that is we are looking for a rhombus $$0,a = 1, b, c= -0.8+0.6j = e^{2it}$$ whose centre is $b/2 = 0.1+0.3j.$

there are two solutions: $$x = 2t, y+\pi/2 = 0 $$ corresponding to the identification $e^{ix} = e^{2it}, e^{i(y+\pi/2)} = 1,$ the other one is $$x = 0, y+\pi/2 = 2t $$ corresponding to the identification $e^{ix} = 1, e^{i(y+\pi/2)} = e^{2it}.$

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