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I am trying to invert the following equation to have it with $\theta$ as the subject:

$y = \cos \theta \sin \theta - \cos \theta -\sin \theta$

I tried both standard trig as well as trying to reformulate it as a differential equation (albeit I might have chosen an awkward substitution). Nothing seems to stick and I keep on ending up with pretty nasty expressions. Any ideas?

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    @draks: thanks, I'll check that (I also think that should be $y$). Just wondering whether there were any more "agile" expressions2012-07-18

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$y = \cos \theta \sin \theta - \cos \theta -\sin \theta$

$y - \cos \theta \sin \theta = -\cos \theta -\sin \theta$

$y^2 - 2y\cos \theta \sin \theta +\cos^2\theta\sin^2\theta= 1+2\cos \theta \sin \theta$

$y^2 - y\sin(2\theta) +\frac14\sin^2(2\theta)= 1+\sin(2\theta)$

This is a quadratic equation for $\sin(2\theta)$ that you can solve.

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    That works! Brilliant, thank you.2012-07-18
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Rewrite the equation as $(1-\cos\theta)(1-\sin\theta)=1+y.$

Now make the Weierstrass substitution $t=\tan(\theta/2)$. It is standard that $\cos\theta=\frac{1-t^2}{1+t^2}$ and $\sin\theta=\frac{2t}{1+t^2}$. So our equation becomes $\frac{2t^2}{1+t^2}\cdot \frac{(1-t)^2}{1+t^2}=1+y.$ Take the square root, and clear denominators. We get $\sqrt{2}t(1-t)=\sqrt{1+y}(1+t^2).$ This is a quadratic in $t$.