On high school, I was taught that I could obtain any sine value with some basic arithmetic on the values of the following image:

enter image description here

But I never really understood where these values where coming from, some days ago I started to explore it but I couldn't discover it. After reading for a while, I remembered that the sine function is:


Then I thought that I just needed to calculate $\frac{1}{x}$ where $0 \leq x \leq 1$ but it gave me no good results, then I thought that perhaps I could express not as a proportion of the opposite and hypotenuse, I thought I could express it as the ratio between slices of the circumference, for example: circumference $=\pi$, then divided it by $4$ (to obtain the slice from $0$ to $90$ degrees) then I came with: $x/ \frac{\pi}{4}$ where $0\leq x \leq \frac{\pi}{4} $ but it also didn't work, the best guess I could make was $\sqrt{x/ \frac{\pi}{4}}$, the result is in the following plot:

enter image description here

The last guess I made seems to be (at least visually) very similar to the original sine function, it seems it needs only to be rotated but from here, I'm out of ideas. Can you help me?

  • $\begingroup$ are you interessted in the values of $\sin$ on your image, or for the value of $\sin$ for any $x$ ? $\endgroup$ – Dominic Michaelis Feb 17 '13 at 17:19
  • $\begingroup$ @DominicMichaelis Values for any $x$. $\endgroup$ – Billy Rubina Feb 17 '13 at 17:32
  • $\begingroup$ Is your question answered? If yes, could you be so kind to accept an answer? $\endgroup$ – Dominic Michaelis Feb 17 '13 at 18:51
  • 15
    $\begingroup$ In respectful disagreement with @Dominic, I think a questioner should wait a full day before accepting any answer, to judge among the various answers to identify the most helpful one. $\endgroup$ – Lubin Feb 17 '13 at 19:11

As in Alfonso Fernandez's answer, the remarkable values in your diagram can be calculated with basic plane geometry. Historically, the values for the trig functions were deduced from those using the half-angle and angle addition formulae. So since you know 30°, you can then use the half-angle formula to compute 15, 7.5, 3.25, 1.125, and 0.5625 degrees. Now use the angle addition formula to compute 0.5625° + 0.5625° = 2*0.5625°, and so on for 3*0.5625°, 4*0.5625°...

These would be calculated by hand over long periods of time, then printed up in long tables that filled entire books. When an engineer or a mariner needed to know a particular trig value, he would look up the closest value available in his book of trig tables, and use that.

Dominic Michaelis points out that in higher math the trig functions are defined without reference to geometry, and this allows one to come up with explicit formulae for them. You may reject this as mere formalist mumbo-jumbo, but conceptually I find that the university-level definitions for the trig functions make much more sense than the geometric ones, because it clears the mystery on why these functions turn up in situations that have nothing to do with angles or circles. So eventually you may lose your desire to have the values computed from the geometrical definition.

Of course, if you're going to be using the geometrical definition anyway, you could also just grab a ruler and a protractor and measure away all night, and compute a table of trig values that way.

One final note: you're still using the "ratio of sides of a triangle" definition for the trig functions. I strongly recommend you abandon this definition in favor of the circular definition: $sin(\theta)$ is the height of an angle $\theta$, divided by the length of the arm of the angle, $cos$ is the same for the width of an angle, and $tan$ is the slope of the arm of the angle. The reason why I recommend you use this definition is because, while it's as conceptually meaningful as the triangular one (once you think about it for a second), it allows you to easily see where the values for angles greater than 90° are coming from. The triangular definition is so limited that I personally find it destructive to even bother teaching in school, I wonder if it wouldn't be easier to jump right in with the circular definition. I know it held me back for years.

  • $\begingroup$ Yes, it seems you got some time stuck like me. For your comment on formalism, I really have no problem with them - I wanted to understand the source of the values at any cost. $\endgroup$ – Billy Rubina Feb 17 '13 at 18:47

All of the values in your picture can be deduced from two theorems:

  1. The Pythagorean theorem: If a right triangle has sides $a,b,c$ where $c$ is the hypotenuse, then $a^2+b^2=c^2$
  2. If a right triangle has an angle of $\frac{\pi}{6} = 30^\circ$, then the length of the side opposite to that angle is half the length of the hypotenuse.

Both can be proven with elementary high school geometry.

Let's see how this works for Quadrant I (angles between $0$ and $90^\circ$), as the rest follows from identities.

  1. $\sin 0 = 0$; this is clear from the definition.
  2. $\sin 90^\circ = 1$; less intuitive because it breaks the triangle, but $\sin 90^\circ =\cos0$, and $\cos 0 = 1$ because the adjacent side and the hypotenuse coincide when the angle is $0$.
  3. $\sin 45^\circ = \frac{1}{\sqrt{2}}$; this corresponds to an isosceles triangle, and if we set the sides to be $1$, then by the Pythagorean theorem, the hypotenuse is $\sqrt{2}$.
  4. $\sin30^\circ = \frac{1}{2}$; this follows immediately from theorem 2 above.
  5. $\sin60^\circ = \frac{\sqrt{3}}{2}$; if we take a $30^\circ-60^\circ-90^\circ$ triangle, and set the side opposite to the $30^\circ$ angle to be $1$, then the hypotenuse is $2$ and the side opposite to the $60^\circ$ angle satisfies $x^2 + 1 = 2^2$, so its length is $\sqrt{3}$ - and thus $\sin 60^\circ = \frac{\sqrt{3}}{2}$.
  • $\begingroup$ This is petty much how I would have answered the question. $\endgroup$ – Lubin Feb 17 '13 at 19:08
  • 3
    $\begingroup$ Alternatively, you can take an equilateral triangle, break it in half and find $\sin\left(30^\circ\right)$ and $\sin\left(60^\circ\right)$ using the Pythagorian theorem. $\endgroup$ – Ian Mateus Feb 17 '13 at 19:17
  • $\begingroup$ The hypotenuse is $\sqrt{2}$? I thought I should consider that the hypotenuse is always one, due to this page on Stillwell's Mathematics and It's history. Now I'm confused. $\endgroup$ – Billy Rubina Feb 24 '13 at 17:05
  • $\begingroup$ @GustavoBandeira For a given triangle, say the one with angles $(45^\circ, 45^\circ, 90^\circ)$, we can "change the units" however we like. If we call the lengths of the sides $1$, then the length of the hypotenuse is $\sqrt{2}$, and if we call the length of the hypotenuse $1$, the lengths of the sides would be $\frac{1}{\sqrt{2}}$. It is true that when defining trigonometric functions on a circle, the hypotenuse corresponds to the radius and it usually makes sense to call it $1$, but the results are the same either way. $\endgroup$ – Alfonso Fernandez Feb 24 '13 at 21:09

The values of the sine function can be calculated by $$\sin(x)=\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}=x-\frac{x^3}{1\cdot 2\cdot 3}+\frac{x^5}{1\cdot 2 \cdot 3 \cdot 4 \cdot 5} \mp \dots$$ In university the $\sin$ function is introduced without geometric motivation, it is introduced as a function from the exponential function (i hope you know this one) and $$\sin(x)=\frac{1}{2i} \cdot (e^{ix}-e^{-ix})$$ where $i$ is the imaginary unit.
Maybe another formula is more motivationed, but it's really complicated to make it rigorous, the main idea is, that you can write a polynomial as a product of its zeros and a number -- for example $$ 3(x-1)(x+2)=3x^2 +3x-6$$ Now you can try the same with the $\sin$ function but as $$x \cdot (x-\pi) (x+\pi) \cdot$$ is $\infty$ for nearly all $x$ you write it a bit different. In fact $$\sin(x)=x\cdot \prod_{k=1}^\infty \left(1-\frac{x^2}{k^2\pi^2}\right)$$ (note i used $(a+b)\dot (a-b)=a^2-b^2$. This one is more motivationed by the geometric interpretation, from the $0$ of the $\sin$ function.

As the comments were right, there is a connection. The Eulerformula is $$\exp(ix)=\cos(x)+i \sin(x)$$ and a number in the complex plane can be written as $$z=|z|\cdot e^{i \varphi}$$ where $$\varphi= \arctan\left(\frac{\Im(z)}{\Re(z)}\right)$$ and $\Im(z)$ is the imaginary part of $z$ and $\Re(z)$ is the real part of $z$.

  • $\begingroup$ It's impossible to obtain the sin values with the geometric motivation then? $\endgroup$ – Billy Rubina Feb 17 '13 at 17:06
  • 1
    $\begingroup$ The factorial should be outside the parentheses in your summation formula. $\endgroup$ – Clayton Feb 17 '13 at 17:10
  • 5
    $\begingroup$ I have no idea how you crafted "In university the sin function is introduced without geometric motivation" – I admit, that there is no word "triangle" or "hypotenuse" in those formulae, but I cannot agree with you. Surely $e^{it}$ is a circle and $\sin t = \Im(e^{it})$ is exactly the same definition that children learn at school (just using different names). $\endgroup$ – dtldarek Feb 17 '13 at 17:22
  • 1
    $\begingroup$ @dtldarek: Had this been in an answer I would have upvoted it. The connection with the complex unit circle is surely the perspective that unifies this topic. $\endgroup$ – DWin Feb 17 '13 at 21:27
  • 1
    $\begingroup$ @JackM: I don't think the sin is the only solution that that differential equation. Case in point, $f(x) = 0$, $f'(x) = 0$, $f''(x) = 0$, and $-f''(x) = 0$ is also $f = -f''$. It would not be a good way to define a function. $\endgroup$ – Lie Ryan Feb 18 '13 at 1:41

[Images from Wikipedia. ] $$\sin x = \frac ah\quad \cos x = \frac bh \quad \tan x = \frac ab$$

$\qquad\qquad\qquad\qquad\qquad\qquad$ enter image description here

And see how this relates to the unit circle, where $$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac y1 = y$$ $$ \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac x 1 = x$$

$\qquad\qquad\qquad\qquad\qquad\qquad$enter image description here

So in your image, the values $(x, y)$ on the unit circle depends on the angle of interest: in particular, $(x, y) = (\cos \theta, \sin \theta)$:

These values of $x = \cos\theta, y =\sin\theta$ and $h = 1$ (the radius of the unit circle, and the hypotenuse of the corresponding right triangle) can then be computed using the Pythagorean Theorem: $$x^2 + y^2 = h^2 = 1 $$ just as $$\sin^2x + \cos^2x = 1$$

You're on your way if you combine this theorem with the fact that when a right triangle has an angle of $\theta = \frac{\pi}{6} = 30^\circ$, then the length of the side opposite to that angle is half the length of the hypotenuse (the y-value of the triangle with hypotenuse 1, in the unit circle = 1/2). Hence, $y = \sin\left(30^\circ\right) = 1/2$.

For example, knowing this, we can find $$x^2 + y^2 = 1 \iff \cos^2(30^\circ) + \sin(30^\circ) = 1 \iff \cos^2(30) + (1/2)^2 = 1$$ $$\iff \cos^2(30) = 3/4 \implies \cos(30) = \sqrt 3/2$$


This question is really a classic question for me.

The first thing you want to know is how do you measure the angles. I will talk in terms of a "clock" to ease textual explanation, with 1pm - 12pm denoting the position of, let's say a second hand.

Consider that, at 3pm, the angle is $0^\circ$. To move up a few degrees, go anti-clockwise. The first angle you will see is $30^\circ$. $\sin30^\circ = \frac 1 2.$ Since the hypotenuse is also the radius of the circle, it takes the value of $1$. Since $\sin = \frac{\text{opposite}}{\text{hypotenuse}}$, the blue dot's vertical distance from the $x$-axis is definitely 0.5

Continue this in an anti-clockwise walk and you can see, at 12pm, you would have moved $90^\circ$. Continue down this track, at 9pm, it would be a half-turn, which explains the $180^\circ$...and so on. We know that the $\sin$ graph is cyclical...so if you have an angle greater than a full revolution ie $360^\circ$, just subtract revolutions until the range $0^\circ$ to $360^\circ$ is reached.

That is how you read this circle.

  • 3
    $\begingroup$ What's with the "Haha..." at the beginning? Is the question funny? $\endgroup$ – Pedro Tamaroff Feb 17 '13 at 23:13
  • 1
    $\begingroup$ the question isnt funny, but the good old days when i was learning this 10 years ago were. $\endgroup$ – bryanblackbee Feb 18 '13 at 1:38

If we go to the minor detail of how we get the values of sin,cos, tan it is the function of numerical analysis is which we have different methods to get the values of these one important method i remember is newton ralphson.

i studied numerical analysis 2years back. howeever if you need to really find out you can search in this area because this is the origin.

See Newton's method (Wikipedia)


$cos(\phi)$ is the $x$ value of a point at angle $\phi$, and $sin(\phi)$ is the $y$ value of a point at angle $\phi$; on the unit circle (at radius 1).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.