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I came up with this formula for computing an area $A$ as a function of $x \in (0,6)$: $$A(x) = 2x\sin\left(\frac{3 \pi}{4}-\arctan\left(\frac{10-x}{6-x}\right)\right)\sqrt{x^2-16x+68}$$

And I want to show that, for $x \in (0,6)$: $$A(x) = -2x^2+16x$$

As you can see in the picture below, it seems to be true.

enter image description here

However, I don't know how to prove it.

Lava
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    Please show your efforts in trying to answer the question. – Adam Rubinson Oct 01 '23 at 18:28
  • There are none. I'm really sorry. I wish I had a few ideas, but I don't know how to start. – Lava Oct 01 '23 at 18:54
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    Try using the sine difference formula. $\sin(\arctan(t))$ and $\cos(\arctan(t))$ are algebraic functions of $t$ on some intervals; see e. g. this related question. By the way, how did you get the formula? – Amateur_Algebraist Oct 01 '23 at 19:07
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    Thank you, I'll try with these formulas. I got it by trying to calculate the area of a rectangle inscribed in another rectangle. I wanted to see how to do it without using vectors. – Lava Oct 01 '23 at 19:17
  • Can you please upload a picture of those rectangles? It's very helpful as it would provide motivation for your question. – Benjamin Wang Oct 01 '23 at 22:37

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We first note that $$\sin(\arctan x) = \frac{x}{\sqrt{1+x^2}} \quad \text{and} \quad \cos(\arctan x) = \frac{1}{\sqrt{1+x^2}}.$$ (See here for the $\sin$ case; $\cos$ follows similarly.)

Also recall that $\sin(A-B) = \sin A \cos B - \cos A \sin B$ (see here). Now let $y = \frac{10-x}{6-x}$. Thus, \begin{align*} A(x) &= 2x \sin\left(\frac{3\pi}{4} - \arctan y\right) \sqrt{x^2 - 16x + 68} \\ &= 2x \sqrt{x^2 - 16x + 68} \left(\sin\left(\frac{3\pi}{4}\right)\cos(\arctan y) - \cos\left(\frac{3\pi}{4}\right)\sin(\arctan y)\right) \\ &= 2x\sqrt{x^2-16x+68} \frac{\sqrt 2}{2} \Big(\cos(\arctan y) + \sin(\arctan y)\Big) \\ &= \sqrt 2x\sqrt{x^2 - 16x + 68} \left(\frac{1}{\sqrt{1+y^2}} + \frac{y}{\sqrt{1+y^2}}\right) \\ &= \sqrt 2x\sqrt{x^2 - 16x + 68} \left(\frac{\frac{10-x}{6-x}+1}{\sqrt{\left(\frac{10-x}{6-x}\right)^2 + 1}}\right) \\ &= \sqrt 2x\sqrt{x^2 - 16x + 68} \left(\frac{\frac{16-2x}{6-x}}{\sqrt{\frac{2x^2-32x + 136}{(6-x)^2}}}\right) \\ &= \sqrt 2x\sqrt{x^2 - 16x + 68} \left(\frac{16-2x}{\sqrt 2 \sqrt{x^2-16x+68}}\right) \\ &= 16x - 2x^2, \end{align*} as required.

kipf
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