0

I have this equation:

$$\frac{\arcsin(\sin(\cos(\Delta)\cdot\frac{2\pi\cdot r}{T+\phi}))-\phi}{\frac{2\pi}{T}}=\cos(\Delta)\cdot r$$

And I want to find an effective way to solve for the Delta variable, but I´m not pretty sure if this is possible, because if I move the terms from one side to the other, I can never mix the two Delta variables together, since they always get stuck inside the cos/acos functions.

Example 1:

(for the first member)

$$\Delta=\arccos\left(\frac{\arcsin(\sin(\cos(\Delta)\cdot\frac{2\pi\cdot r}{T}+\phi))}{\frac{2\pi\cdot r}{T+\phi}}\right)$$

Example 2:

(for the second member)

$$\Delta=\arccos\left(\frac{\arcsin(\sin(\cos(\Delta)\cdot\frac{2\pi\cdot r}{T+\phi}))-\phi}{\frac{2\pi/T}{r}}\right)$$

Chaos
  • 151

1 Answers1

1

Note that $\arcsin (\sin x) = x$. So the original equation simplifies to the following:

$$ \frac{T}{2\pi} \left [ \cos(\Delta)\cdot \frac{2\pi r}{T + \phi} - \phi \right ] = r \cos(\Delta) $$

You can then isolate the cosine in this expression and take the inverse cosine to get $\Delta$. First multiply by $2\pi/T$ to get the following:

$$ \begin{align} \cos(\Delta)\cdot \frac{2\pi r}{T + \phi} - \phi &= \frac{2\pi r}{T} \cos(\Delta) \end{align}$$

Then subtract the term on the right, add $\phi$ to both sides, and factor out $2\pi r \cos(\Delta)$ to get:

$$ 2\pi r \cos(\Delta) \left[ \frac{1}{T+\phi} - \frac{1}{T} \right] = \phi $$

Finally, dividing both sides by the extra factors we get the desired result:

$$ \cos(\Delta) = \frac{\phi}{2\pi r} \left[ \frac{1}{T+\phi} - \frac{1}{T} \right]^{-1} $$

All that's left is to take the inverse cosine of both sides. You can clean this up a little more by adding those fractions together as well.

wgrenard
  • 3,678
  • Yes, I did noted that. Now I have the simplified equation, but I still don´t understand how can I isolate Delta. To me it looks like Delta will continue to fall inside cosines. – Chaos Mar 12 '17 at 07:13
  • From here you should solve the equation for $\cos(\Delta)$. i.e. isolate $\cos(\Delta)$ on one side of the equation. (Can you do that part?) You should have a cosine on one side and the other side should not involve any trig functions at all. Once you have this simply take the inverse cosine of both sides to get $\Delta$. – wgrenard Mar 12 '17 at 07:23
  • But $\Delta$ is in both sides. If I just do that (isolate the second member for example), I can´t calculate the first member, because $\Delta$ is still there. How can I have both on the same side without anything else? – Chaos Mar 12 '17 at 07:32
  • @Chaos See my edit; I've expanded on what I mean. – wgrenard Mar 12 '17 at 07:41
  • Awesome. Yesterday I was too tired to continue, but I´m back. Thanks. You used some tricks I´m not used to. It took me a little to understand it, but now I can see it clearly. Unfortunately, the result doesn´t work well for the application I´m looking for. I get wrong values and complex numbers, when they should be Real instead. – Chaos Mar 13 '17 at 00:38
  • @Chaos Well, what determines if $\Delta$ is real or complex are your input values. Note that $|\cos \Delta| \leq 1$ in order for $\Delta$ to be real. Assuming you've done all the math correctly, you're finding that for your values of $\phi$, $r$, and $T$ there is no real $\Delta$ satisfying the original equation. – wgrenard Mar 13 '17 at 16:38