# Would this be a reasonable way of solving some trig functions by hand?

Hello Math StackExchange,

Today I was reviewing some trigonometry in Algebra 2 (as I am currently doing the course), and I ended up down a rabbit-hole of though. It was simply the idea of being able to do trig functions by hand which got me thinking, and then writing. So, pretty much all day today I have been doing this (although I am not very innovative).

These are my ways of solving some of the functions so far: $x^2 + y^2 = 1$ is a unit circle

Because this is 2d algebra, I can say $\fracπ4$ radians is going to have the line equivalent to the slope of 1. This would work, however I don't know a way without a calculator to get the what the slope would be to reach certain radians.

So, the only thing I can do from here is use the pythagorean theorem to find the hypotenuse, and then make a ratio with that and the arc length. This ratio is $\sqrt2:\fracπ2$; or, when finding slope, $\frac{4}{π\sqrt2}$. I realized later that this wasn't going to be useful for transforming the hypotenuses, rather the cosine was revealing more promising results when used with this and ONLY when the angle was less than $\fracπ4$. Here are some examples of how I work this:

$\fracπ4 * \frac4{π\sqrt2} = \frac1{\sqrt2} = \frac{\sqrt2}2$(that's the cos), $sin^2 + cos^2 = 1$, so $sin^2 = 1 - cos^2$, and therefor $1-\frac24 = \frac24$, which is true, and when y = cos and x = sin, then $\frac24 = m\frac24$, where the slope is 1 (obviously).

$\fracπ8 * \frac4{π\sqrt2} = \frac1{2\sqrt2} = \frac{\sqrt2}4$ (cos, again), $sin^2 = 1 - \frac18, sin = \sqrt\frac78, sin = \frac{\sqrt7}{2\sqrt2}, \frac{\sqrt2}4 * \frac{2\sqrt2}{\sqrt7} = x = \frac{\sqrt7}7$(the slope),

$\frac3{16} * \frac4{π\sqrt2} = \frac{3\sqrt2}8$(cos) $sin^2 = 1 - \frac{18}{64}, sin = \sqrt\frac{23}{32}, sin = \frac{\sqrt46}8, \frac{3\sqrt2}8 * \frac8{\sqrt{46}} = \frac{3\sqrt2}{\sqrt{46}} = \frac3{\sqrt{23}} = \frac{3\sqrt{23}}{23}$

Every angle's slope from $\fracπ4$ to $\fracπ∞$ can, at least I think, be estimated with this method. I struggled further to find something for $\fracπ4$ to $\fracπ2$, but I ultimately failed. The only thing I could use this process for in that sort of angle is to find where a certain slope landed in the process.

For example: If you complete the system $x^2 + y^2 = 1, y=2x$, then you will get $\frac{\sqrt5}5 (sin) and \frac{2\sqrt5}5$ (cos). To convert this into radians, you just need to multiply the cos by the reciprocal of $\frac4{π\sqrt2}$. $\frac{2\sqrt5}5 * \frac{π\sqrt2}4 = \frac{π\sqrt{10}}{10}$, which makes about a 56.920997883º angle. You can do this for almost anything, and it will work well (from where I can see.)

Is this a valid way to do anything? I know I am not using any trig functions, and I know that I can't really do anything without knowing calculus (which I am soon to be learning). I don't even know what I am doing. If this is a half-way valid way to solve things, how do I make this better? To me, I simply just feel better knowing that my trig function can be found and expressed as a simplified thing rather than an insane run-on number. What do you think? Here is a desmos sheet https://www.desmos.com/calculator/amrs1zcpgk

• math.meta.stackexchange.com/q/5020/306553 mathjax references. – Siong Thye Goh Aug 3 '17 at 4:22
• @SiongThyeGoh I fixed that – Arence Aug 3 '17 at 4:57
• There is no proportionality between arc length and sin or cos. When you find the intersection between the unit circle and the line $y=2x$ you actually find $\sin \alpha$ and $\cos\alpha$, where $\alpha$ is the angle the line forms with $x-$axis, but $\dfrac{\sqrt 5}{5}$ is not proportional with the arc length. If so if $\sin 45°=\frac{1}{\sqrt 2}$ then $\sin 90°$ should be the double, which is not because $\sin 90°=1\quad$. Hope this helps – Raffaele Aug 3 '17 at 10:29