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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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Wolfram can help: $±cos^{-1}(\frac{\sqrt{\theta+1}}{\sqrt2}-\frac12 \sqrt{-2 \theta-2 \sqrt2 \sqrt{\theta+1}-1}+1/2)$ –  draks ... Jul 18 '12 at 10:50
    
In my view words like "tricky" don't belong in the title -- that's a very subjective assessment. (I like your use of "subject", though :-) –  joriki Jul 18 '12 at 10:56
    
@draks: I think you wanted $y$ there instead of $\theta$? –  joriki Jul 18 '12 at 11:03
    
@joriki: you're absolutely right - edited the title to keep it as factual as possible. –  maruko Jul 18 '12 at 11:07
    
@joriki ah right, sorry can't edit that anymore –  draks ... Jul 18 '12 at 11:08
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2 Answers 2

up vote 4 down vote accepted

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

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

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

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

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

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That works! Brilliant, thank you. –  maruko Jul 18 '12 at 11:15
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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 $t$.

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