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I am working through a situation with trying to fit an equilateral triangle into a square, and I have boiled it down to the following equation: $$\arccos\left(\frac{\frac{L}{2}+x}{S}\right) + \arccos\left(\frac{\frac{L}{2}-x}{S}\right) = \frac{2}{3}\pi\;\mathrm{rad}$$ I need to solve this equation for $S$. After researching it for a couple of hours, I am still lost on how to resolve the two $\arccos$ terms. Could someone please show me how to go about solving this equation for $S$?

I would really appreciate it!

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Notice that: $\cos(A+B)=\cos A \cos B - \sin A \sin B$

In your example $A=\arccos(\frac{\frac{L}{2}+x}{S})$, and $B=\arccos(\frac{\frac{L}{2}-x}{S})$.

The solution follows from the fact that $\cos(\arccos(x))=x$

A specific solution:

$\arccos\left(\frac{\frac{L}{2}+x}{S}\right) + \arccos\left(\frac{\frac{L}{2}-x}{S}\right) = \frac{2}{3}\pi\;\mathrm{rad}$

From here, take the cosine of both sides:

$\cos(\arccos\left(\frac{\frac{L}{2}+x}{S}\right) + \arccos\left(\frac{\frac{L} {2}-x}{S}\right))=\cos(\frac{2π}{3})$

Using the addition formula above, we find that:

$\cos(\arccos\left(\frac{\frac{L}{2}+x}{S}\right))\cdot\cos(\arccos\left(\frac{\frac{L} {2}-x}{S}\right)))-\sin(\arccos\left(\frac{\frac{L}{2}+x}{S}\right))\cdot\sin(\arccos\left(\frac{\frac{L} {2}-x}{S}\right)))=-\frac{1}{2}$

Notice that, by the Pythagorean relationship:

$\sin^2(\arccos(x))+\cos^2(\arccos(x))=1$

Therefore: $\sin(\arccos(x))=\sqrt{1-x^2}$.

Hence, the above equation can be rewritten as:

$(\frac{\frac{L}{2}+x}{S})\cdot(\frac{\frac{L}{2}-x}{S})-\sqrt{1-(\frac{\frac{L}{2}+x}{S})^2}\cdot\sqrt{1-(\frac{\frac{L}{2}-x}{S})^2}=-\frac{1}{2}$

Or, equivalently:

$\frac{L^2-4x^2}{4S^2}-\sqrt{(1-(\frac{\frac{L}{2}+x}{S})^2)(1-(\frac{\frac{L}{2}-x}{S})^2)}=-\frac{1}{2}$

And:

$L^2-4x^2-\sqrt{(-L^2-4Lx+4S^2-4x^2)(-L^2+4Lx+4S^2-4x^2)}=-2S^2$

WolframAlpha tells me (it's probably solvable, but to save time):

$x=±\frac{1}{2}\sqrt{3}\sqrt{S^2-L^2}$

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  • $\begingroup$ Ah, that is the identity that I needed, thank you! What is the name of this identity, so that I may look it up? $\endgroup$ – TheNewGuy Jun 19 '16 at 2:04
  • $\begingroup$ @TheNewGuy Usually it is just referred to as the 'cosine addition formula.' There exist analogous formulas for every trigonometric function, and taken together they may be called 'trigonometric addition formulas.' Basically every identity you will ever encounter can be found on this really helpful page: www2.clarku.edu/~djoyce/trig/identities.html $\endgroup$ – KR136 Jun 19 '16 at 2:16
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Hint: Take cosine (or sine if you please) on both sides. Then use multiple angle formula.(You will need to use pythagoras theorem to find sine of angles)

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    $\begingroup$ Prayas, I am not sure what you mean by the multiple angle formula. After going through List_of_trigonometric_identities#Multiple-angle_formulae I am not sure I see an identity that is readily applicable to the equation. $\endgroup$ – TheNewGuy Jun 19 '16 at 1:53
  • $\begingroup$ multiple= double(in this case), i.e cos(A+B)=cosAcosB−sinAsinB $\endgroup$ – Prayas Agrawal Jun 19 '16 at 1:53

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