# find expression for an angle

• Oct 8th 2012, 02:25 AM
niaren
find expression for one line segment
Attachment 25106

In this geometric setup, I'm trying to find an expression for A dependent on $\phi$, given that $R$ and $\theta$ are known.
My take on this is to write the two expressions for the two sides $DL$ and $DR$

1) $DL^2 = A^2 + R^2 - 2 A R \cos\left(\frac{1}{2}\pi - \phi\right)$
2) $DR^2 = A^2 + R^2 - 2 A R \cos\left(\frac{1}{2}\pi + \phi\right) = A^2 + R^2 + 2 A R \cos\left(\frac{1}{2}\pi - \phi\right)$

and an expression for $DS = 2A$

3) $DS^2 = DL^2 + DR^2 - 2 DL \cdot DR \cos\left(\theta\right)$

Then it seems that I could achieve my goal by eliminating $DL$ and $DR$ in the equation for $DS$.

Summation of 1) and 2) gives

4) $DL^2 + DR^2= 2A^2 + 2R^2$

and multiplication of 1) and 2) gives

5) $DL^2\cdot DR^2 = \left(A^2 + R^2 - 2 A R \cos\left(\frac{1}{2}\pi - \phi\right)\right)\left(A^2 + R^2 + 2 A R \cos\left(\frac{1}{2}\pi - \phi\right) \right)$
$= \left(A^2 + R^2 \right)^2 - 4 A^2 R^2 \cos^2\left(\frac{1}{2}\pi - \phi\right)$

Inserting 4) into 3) and squaring provides

6) $DL^2 \cdot DR^2 = \frac{(R^2 - A^2)^2}{\cos^2\left(\theta\right)}$

Equating 5) and 6) gives an equation dependent on only $A$ and $\phi$

7) $\left(A^2 + R^2 \right)^2 - 4 A^2 R^2 \cos^2\left(\frac{1}{2}\pi - \phi\right) = \frac{(R^2 - A^2)^2}{\cos^2\left(\theta\right)}$

Rearranging

8) $\cos^2\left(\theta\right)\left(A^4 + R^4 + 2 A^2R^2\right) - 4 A^2 R^2 \cos^2\left(\frac{1}{2}\pi - \phi\right)\cos^2\left(\theta\right) = A^4 + R^4 - 2 A^2R^2$

9) $A^4(1-\cos^2\left(\theta\right)) - A^2(2 R^2 + 2 R^2\cos^2\left(\theta\right) - 4 R^2 \cos^2\left(\frac{1}{2}\pi - \phi\right)\cos^2\left(\theta\right)) + R^4(1-\cos^2\left(\theta\right)) = 0$

10) $A^4 - A^2 \frac{2 R^2(1 + \cos^2\left(\theta\right) - 2 \cos^2\left(\frac{1}{2}\pi - \phi\right)\cos^2\left(\theta\right)^2)}{1-\cos^2\left(\theta\right)} + R^4 = 0$

I have a hard time believing this expression to be true because of the following reasoning:
For $\phi=0$ then $A = R\tan\left(\frac{1}{2}\theta\right)$
As $\phi$ is increased I would assume that $A$ should also increase (relative to $R\tan\left(\frac{1}{2}\theta\right)$). However, for the numerical values that I have used to test the expression in 10) I find that $A$ decreases.

Looking back, I can't find out where I go wrong.