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I was working on this problem I came up with where you have a unit circle with a square inscribed in it. Finding the side lengths of this square is trivial ($\sqrt{2}$). However, I wanted to extend this problem. I thought, what if I placed a square adjacent to the previous one such that its left side touches the previous square's right side, it is centered on the $x$-axis, and it has its two right corners touching the circle? Here's a diagram I made of the problem below.

A unit circle with an inscribed square and adjacent square also inscribed

I've come as far as setting up the equation, but I cannot see a way to go about solving it. If I traced a line from the center of the circle to the top right corner of the square, I have a right triangle. I can use trigonometry to find the side lengths of such a triangle.

$$ x = \cos \theta \\ y = \sin \theta $$

I can deduce that the side length $s$ of the square is equal to twice the height of this triangle and also equal to the difference between the base of this triangle and half the side length of the previous square.

$$ s = \cos \theta - \frac{\sqrt{2}}{2} \\ s = 2 \sin \theta $$

By setting these two equations equal to each other, I can attempt to solve for $\theta$, thereby allowing me to solve for the side length of the square.

$$ \cos \theta - \frac{\sqrt{2}}{2} = 2 \sin \theta $$

After reaching this point, I am not sure how to continue. I can immediately see that I cannot solve for $\theta$ easily by just rearranging the equation. I've thought about substituting $\cos \frac{\pi}{4}$ back in for $\frac{\sqrt{2}}{2}$ to make use of some obscure trigonometric identity, but nothing I've tried gets me anywhere near solving the equation. Any hints or solutions would be greatly appreciated. Thank you for taking the time to look at this.

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    $\begingroup$ You can solve $$\cos\theta - 2\sin\theta = \frac{\sqrt{2}}{2}$$ by means of the identity $$p \cos\theta - q \sin\theta = r \cos(\theta + \phi)$$ where $r^2 = p^2 + q^2$ and $\tan\phi = q/p$. (See this trigonograph.) $\endgroup$ – Blue Jul 31 '16 at 10:47
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From the figure below, using Pythagoras' Theorem:

$$\begin{array}{rcl} \left(\dfrac{\sqrt2}2+x\right)^2+\left(\dfrac x2\right)^2&=&1^2\\ (\sqrt2+2x)^2+x^2&=&2^2\\ 2+4\sqrt2x+4x^2+x^2&=&2^2\\ 5x^2+4\sqrt2x-2&=&0\\ x&=&\dfrac{-4\sqrt2\pm\sqrt{32+40}}{10}\\ x&=&\dfrac{\sqrt{72}-4\sqrt{2}}{10}&\mbox{(one solution is rejected)}\\ x&=&\dfrac{6\sqrt2-4\sqrt2}{10}\\ x&=&\dfrac{\sqrt2}{5} \end{array}$$

Modified diagram

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\begin{align*} \left( \frac{\sqrt{2}}{2}+a \right)^{2}+\left( \frac{a}{2} \right)^{2} &= 1 \\ \frac{5a^2}{4}+a\sqrt{2}-\frac{1}{2} &= 0 \\ 5a^2+4a\sqrt{2}-2 &= 0 \\ a &= \frac{-4\sqrt{2} \pm \sqrt{(16)(2)-4(5)(-2)}}{2(5)} \\ a &= \frac{-2\sqrt{2} \pm 3\sqrt{2}}{5} \\ a &= \frac{\sqrt{2}}{5} \quad \text{or} \quad -\sqrt{2} \quad \text{(rejected)} \end{align*}

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If you insist in your trigonometric approach, you can solve your equation as follows: solve for $\cos\theta=2\sin\theta+\sqrt2/2$ and then plug this into the identity $\sin^2\theta+\cos^2\theta=1$. Solve for $\sin\theta$ and discard the spurious solution.

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  • $\begingroup$ I think you meant $\sin^2\theta+\cos^2\theta=1$. $\endgroup$ – Kenny Lau Jul 31 '16 at 15:00
  • $\begingroup$ Yes, of course! Thank you for spotting that, I'll correct now. $\endgroup$ – Aretino Jul 31 '16 at 15:02

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