Every single time I do anything with circles/triangles I always run into the primary trig ratios. With radians, I found some hope, but it was short-lived, because yet again we needed trig functions to work with it.

Can someone please explain how to do the primary trig functions ie. calculating sin(pi/7) without a calculator or memorization, and more importantly WHY the process is the way it is.

ik a similar question is out there, but the explanations are so damn complicated that I can only understand a few things here and there.

I'm in gr11 so please don't use too much high-level terminology that will take me 20 google searches to understand, I'd rather save the google searches to understand the logic behind the processes.

I'm looking for something like the Taylor series

  • $\begingroup$ Welcome to Math SE! Can you provide some examples of what sorts of problems you're having? As it's written your question is a bit unclear. $\endgroup$ – DMcMor Apr 22 at 3:44
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    $\begingroup$ Historically, they used tables of values and would simply look-up the values (or something very close). There are also polynomial approximations of the trigonometric functions that can be used. $\endgroup$ – Hayden Apr 22 at 3:45
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    $\begingroup$ You have to say what you are using the trig functions for. For many approximate purposes, if you know the values at $30,45,60$ degrees you are in good shape and can approximate the rest. If you are calculating the geometry of shapes, tables can meet your needs. If you are doing substitutions into integrals, you need algebraic relationships, not numeric values. $\endgroup$ – Ross Millikan Apr 22 at 3:47
  • $\begingroup$ The sines and cosines of $30,45,60$ degrees can be understood fairly well using right handed triangles combined with Pythagoras' rule. Is this what you're looking for? $\endgroup$ – John Doe Apr 22 at 3:51
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    $\begingroup$ Ummm... if you have access to a computer to post such a question, then you have access to a computer to calculate the trig values. Moreover, I suspect you have a phone with a calculator in your pocket most of the day. Why would anyone in this day and age want to calculate trig values by hand, knowing that the only way you're going to get the answer to $\sin 29.6^\circ$ and such is with some form of computer. $\endgroup$ – David G. Stork Apr 22 at 4:56

You're gonna have a lot of trouble figuring out weird values like $\sin(29)$ or $\cos(34)$. It's only special angles that are easy to figure out. Then you have to use various formulas to fill in the gaps.

The special angles are: $$\frac{\pi}4, \frac{\pi}3, \frac{\pi}6$$

You can find $\sin(\theta)$ in each of the above cases using the following formulas.

  • $\sin\left(\frac{\pi}4\right)=\frac{1}{\sqrt{2}}$
  • $\sin\left(\frac{\pi}3\right)=\frac{\sqrt{3}}{2}$
  • $\sin\left(\frac{\pi}6\right)=\frac{1}{2}$

Combine these with identities such as $\sin^2(\theta)+\cos^2(\theta)=1$, and you can get many exact values of $\sin(\theta)$, $\cos(\theta)$, and $\tan(\theta)$ without a calculator. There are also double- and triple-angle formulas you can use to figure out angles like $\frac{\pi}{12}$, and there's the CAST rule to figure out negative angles and intermediate angles like $\frac{2\pi}3$. Look those up.

For example, let's say I want to figure out the value of $$x=\cos\left(\frac{\pi}3\right)$$ Well, we know $\sin(\pi/3)=\frac{\sqrt{3}}2$. So using $\sin^2+\cos^2=1$, you can get $$\cos\left(\frac{\pi}3\right) = \sqrt{1-\sin^2\left(\frac{\pi}3\right)} = \sqrt{1 - \frac{3}4} = \sqrt{\frac{1}4} = \frac{1}2.$$

If you're wondering about Taylor series, there's a lot of theory behind that, but basically you need the formula $$\sin(x) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1} = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \frac{1}{7!}x^7 + \cdots$$ You can use these first four terms to approximate $\sin(x)$, but that can still be tough without a calculator.

P.S., re:"too much high-level terminology that will take me 20 google searches to understand" --- you have to be willing to put in the time if you want to gain a better understanding. Twenty google searches is rookie numbers. Math doesn't happen magically over night!

  • $\begingroup$ I'm referring to something like the Taylor Series, I want someone to help me understand it and how it applies to finding side lengths without using a calculator. As for the 20 google searches, I personally have spent quite a while on individual sites, "quality over quantity". $\endgroup$ – Allan Henriques Apr 22 at 4:05

Finding sine or cosine for any angle $\alpha$ can be always reduced to finding a sine or cosine for some angle $\beta \in [0, \pi/2)$.

Finding sine or cosine for such (acute) angle $\beta$ can be always replaced with finding a sine or cosine for some angle $\gamma\in[0, \pi/4)$. For example: $\sin\frac{3\pi}{8}=\sin(\frac\pi2-\frac\pi8)=\cos\frac\pi8$

All you need is a good formula to calculate $\sin x$ and $\cos x$ for $x\in[0, \pi/4)$. For most practical purposes just a few items of the Taylor series will suffice:

$$\sin x\approx x-\frac{x^3}6\tag{1}$$

$$\cos x\approx 1-\frac{x^2}2+\frac{x^4}{24}\tag{2}$$

For the fiven range the maximum error of (1) is 0.00245413 or 0.2%. The maximum error of (2) is 0.000322426, almost neglectable for most practical purposes.


The Taylor series is can be used to find very accurate estimates of trig functions.
However if you want to find $\sin \frac{\pi}{7}$ then it isn't going to get you very far.

$\sin x = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} + \cdots$

If $x$ is an irrational (or trancendental) number like $\frac {\pi}{7}$ you are not going to get very far trying to work this out by hand.

However, I suspect you haven't taken calculus yet, so, teaching you the Taylor series is pointless, as you don't know the math behind it.

The precise values of these trigonometric functions is not so important. It is far less important than it is to understand what these functions represent. And, measuring the lengths of sides of triangles is ultimately one of the less interesting properties, even if this is how they are traditionally introduced.

  • $\begingroup$ Could you give it a try, I can pick up higher level things quite easily as long as they are slightly simplified lol. I'm in advanced placement, so we might even do something like the Taylor Series a little later on. $\endgroup$ – Allan Henriques Apr 22 at 17:42

We say that going all the way around the unit circle is to go "2*pi." That is, going "pi" around the circle is to go half way around and going pi/7 is to go 1/14 of the way around the whole circle.

If we say that our radius is 1, we can solve for our respective x and y values along the circle. Our x value is the "cosine," while our y value is the "sine." Both are made-up, mis-spelled and mistranslated arabic words that have no other meaning; sin and cosine are just 2 of the six ratios we can make by comparing the 3 sides of a right triangle.


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