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I have a computer problem that I was able to reduce to an equation in quadratic form, and thus I can solve the problem, but it's a little messy. I was just wondering if anybody sees any tricks to simplify it?

$$\sin^2\beta ⋅ d^4 + c^2\left(\cos^2\beta⋅\cos^2\alpha-\frac{\cos^2\beta}{2}-\frac12\right)d^2 + sin^2\beta⋅\frac{c^4}{16} = 0$$

Obviously I am using the quadratic formula to solve for $d^2$.

$\beta$, $\alpha$, and $c$ are known.

That middle term is so ugly. Perhaps this could even be simplified and solved without the quadratic formula?

By dividing through with $\sin^2\beta$ I came up with (assuming I did it correctly, I didn't double check it):

$$d^4 + \left[\cot^2\beta\cos^2\alpha - \frac{\csc^2\beta}{2}-\frac{\cot^2\beta}{2}\right]c^2d^2 + \frac{c^4}{16} = 0$$

Which is a little less ugly. Am I missing a cool trick to simplify this?

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You can simplify the middle term somewhat.

$$c^2\left(\cos^2(\beta)\cos^2(\alpha)-\frac{\cos^2(\beta)}{2}-\frac{1}{2}\right)$$

$$=\frac{c^2}{2}\left(2\cos^2(\beta)\cos^2(\alpha)-\cos^2(\beta)-1\right)$$

$$=\frac{c^2}{2}\left(\cos^2(\beta)\left[2\cos^2(\alpha)-1\right]-1\right)$$

$$=\frac{c^2}{2}\left(\cos^2(\beta)\cos(2\alpha)-1\right)$$

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  • $\begingroup$ This cleaned up the root term in the quadratic quite a bit. $\endgroup$ – alfreema Oct 9 '14 at 22:51
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Going back to your first equation you can write the middle term in parentheses as:

$$\left(\cos^2\beta⋅\cos^2\alpha-\frac{\cos^2\beta}{2}-\frac12\right) \\ = \frac{1}{2}\left(\cos^2\beta (2 \cos^2\alpha- 1) -1 \right) \\ = \frac{1}{2}\left(\cos^2\beta \cos 2 \alpha - 1 \right)$$

Whether that's simpler or not is a matter of taste, I suppose.

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