# Solution to trigonometric problem...

• Nov 25th 2010, 09:44 AM
Guffmeister
Solution to trigonometric problem...
Hi all!

I have a problem which I need to solve, but I can't be sure that there is a simple solution.

I have:

$\frac{cos^2(\phi(x))}{(1-cos(\phi(x)))} = f(x)$

and I want to write $\phi(x)$ as:

$\phi(x) = g(x)$.

The problem looks quite nice, and slightly familiar, but I can't for the life of me work it out. My trigonometry has deserted me.
• Nov 25th 2010, 08:25 PM
Soroban
Hello, Guffmeister!

I'll simplify the wording of the problem . . .

Quote:

$\text{Given: }\:\dfrac{\cos^2\!\phi}{1-\cos\phi} \:=\: f$

$\text{Solve for }\phi$

Cross-multiply:. $\cos^2\!\phi \;=\;f\!\cdot\!(1-\cos\phi)
$

. . . . . . . . . . . . $\cos^2\!\phi \;=\;f - f\!\cdot\!\cos\phi$

. . $\cos^2\!\phi + f\!\cdot\!\cos\phi - f \;=\;0$

Quadratic Formula: . $\cos\phi \;=\;\dfrac{-f \pm \sqrt{f^2 + 4f}}{2}$

Therefore: . $\phi \;=\;\cos^{-1}\!\bigg(\dfrac{-f \pm \sqrt{f^2 + 4f}}{2}\,\bigg)$

• Nov 26th 2010, 04:06 AM
Guffmeister
(Rofl) Waaaaagh!

Great! That was very simple, and not even a trigonometric identity in sight. I thought it looked like a nice equation to solve.

Thanks very much.