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I came across this equation while working out how to hang a rising gate by offsetting one hinge.

$$A = (B + C) \cdot \sin(2 \cdot \tan^{-1}(C / D))$$

I know A, B and D and have to find C. Having no idea how to approach an algebraic solution, I implemented a successive approximation numerical solution, which converges very well. However, I'd like to know if there is any way to start solving this algebraically, where C is used both inside and outside the trig functions. How could I begin?

For the curious, the working to reach this point, and the finished rising gate calculator, are here.

Mike Earnest
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emrys57
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    A great place to start would be to use $\sin(2x) = 2 \sin x \cos x = 2 \tan x / \sec^2 x = 2 \tan x / (1 + \tan^2 x)$. This eliminates the trig functions completely. – Erick Wong Jun 03 '15 at 15:31
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    I think this is the best context provided for a question that I've seen in a long time! Not only did you provide full motivation and specify exactly what you're looking for and where you got stuck, you also blogged your entire work so far, including source code to an alternative approach. What a great question :). – Erick Wong Jun 03 '15 at 15:53
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    Why, Erick, thank you! But it doesn't stop there - I hung the gate when I'd finished! It actually does work. – emrys57 Jun 03 '15 at 16:08

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we will use the fact that $$\sin (2t) = \frac{2\tan t}{1 + \tan^2}. $$

let $\tan^{-1}(c/d) = t.$ then $$-\pi/2 \le t \le t, \tan t = c/d. $$ you have $$a = (b+c)\sin (2t) = \frac{2(b+c)c/d}{1+c^2/d^2} $$ that is $$a =\frac{ 2cd(b+c)}{c^2 + d^2} \to (2d-a)c^2+(2bd)c - ad^2= 0$$ is quadratic in $c.$ you can use the quadratic formula to get $c.$

abel
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  • Is anyone else getting a strange behavior in MathJax where \tan shows up as $\sin$ initially, but then corrects itself on refresh? I saw this for my comment above as well as this answer. – Erick Wong Jun 03 '15 at 15:48
  • @ErickWong, i have not noticed that. – abel Jun 03 '15 at 15:50
  • Hmm, that looks right, but the result differs from my rising calculator's numeric solution by a fraction of a mm. I'm still trying to work out whether my numeric approximation is too approximate, or I've made a mistake in solving your quadratic. More in a bit. Thanks! – emrys57 Jun 03 '15 at 16:37
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    I made a mistake in solving the quadratic. Now I've fixed that, this solution matches my earlier iterative result. Erick, I'm afraid abel gets the tick, but thank you both very much! – emrys57 Jun 03 '15 at 16:47