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I have the following type of equation which I wish to solve for $t$:

$$\frac{x}{\cos(t)} - \frac{y}{\sin(t)} = z$$

I have used $c^2 + s^2 = 1$ to get it into the following form:

$$x\sqrt{1-\cos^2(t)} - y \cos(t) = z \cos(t)\sqrt{1-\cos^2(t)}$$

But now I am a little stuck as to how to continue. Is there another identity, e.g. double angle formulae that I should use?

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up vote 4 down vote accepted

First thing, a warning: $\sin(t)$ is not necessarily equal to $\sqrt{1-\cos(t)^2}$, you need $t \in [0, \pi] \pmod {2\pi}$.

As for your problem, I would suggest putting all the $\sqrt{1-\cos(t)^2}$ on the same side of the equation and the other term on the other, factor then square the whole thing. But remember, this only gives you necessary conditions (it's a $\Rightarrow$, not a $\Leftrightarrow$), therefore you need to check all the answers you may find at the end to see if they are in the right range.

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I think this is the right way to do it, but I think the difficulties @Hans points out means I shall have to tackle this a different angle. Incidentally, it is for finding minimum distance points on an ellipse. Thanks for your help. – Bill Cheatham Apr 28 '11 at 15:27
@Bill: Ha, that's funny. My office roommate back when I was a PhD student once tried to find the point on an ellipse closest to a given point, and was surprised to discover that he couldn't get further than reducing the problem to solving a quartic equation... – Hans Lundmark Apr 28 '11 at 18:52

I suspect it will be hard to find a nice expression for the solution. If you do as zulon suggests, you will get an equation of degree 4 in $C=\cos t$. Alternatively, with $T=\tan(t/2)$ you get an equation of degree 4 in $T$ (using $\cos t = (1-T^2)/(1+T^2)$ and $\sin t=2T/(1+T^2)$).

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Thanks. I think this suggests I've formualted the problem wrong. Possibly a numerical solution will do. – Bill Cheatham Apr 28 '11 at 15:25
Here you can find some details on using $\tan(t/2$ substitution:… (And also on simplifying something of the form $a\sin t+b\sin t$, which might be related to your question, but I do not see a way of simplifying your expression using this trick.) – Martin Sleziak Apr 28 '11 at 15:26

$$xsec(t)-ycosec(t)=z$$ $sec(t)=\sqrt{1+T^2}$ and $cosec(t)=\frac {\sqrt {1+T^2}}{T}$ where $T=tan(t)$ . Therefore $$xT\sqrt{1+T^2}+y\sqrt{1+T^2}=z\\=>\sqrt{1+T^2}=\frac{z}{xT+y}\\=>1+T^2=\frac{z^2}{x^2T^2+y^2+2xyT}\\=>x^2T^4+2xyT^3+(x^2+y^2)T^2+2xyT+y^2-z^2=0$$Now I am afraid you will have to solve the equation keeping in mind that the expression $\frac z{xT+y}$ is positive.

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