# Why does computing sine of an angle seem to recursively require sine?

I want some intuitive understanding of the trigonometric functions. One way is to understand ways they can be computed when just an angle in degrees or radians is given. The sine of an angle is defined as the ratio of two sides of a right triangle. If this is the case then I can't find a way to compute it without requiring sine.

To compute the sine of an angle:

1) Create right triangles.

2) Compute their angles.

3) Once a right triangle with the correct angle (or close enough) is found. Compute the opposite/hypotenuse ratio.

The problem: Step 2 requires computing angles of right triangles. The way to do this appears to be to compute the opposite/hypotenuse ratio and use the inverse of sine to find the angle. The inverse of sine seems to require similar steps only this time find a triangle with the correct ratio, instead of correct angle, and then compute the angle of that triangle which seems to be to use...the inverse of sine.

How do I compute the sine of an angle without requiring a sine function within the sine function I'm trying to create?

• And please anyone if you know how to make this question clearer feel free to edit. I'm horrible at writing and obviously math, at least currently. – user1594138 May 28 '14 at 14:53
• What is the definition of angle in the first place? – Hagen von Eitzen May 28 '14 at 14:57
• Construction of the triangle can be done with a protractor and straight edge. The point of the ratio definition is that you just build the triangle like engineers have for millenia (or have it given to you from on high) and you work with it. – ex0du5 May 28 '14 at 14:59
• @JesseMadnick Yes that is my question, sorry for being unclear. But the steps I go through in my question should make what I'm trying to do pretty clear despite me failing to word the question correctly. And so the problem, as I wrote in my question, is how do I create the correct right triangle because to measure if the triangle I created has the correct angle I seem to require the inverse of sine. – user1594138 May 28 '14 at 16:01
• Trigonometric functions are transcendental, meaning that their values (i.e. sin(32.71)) cannot in general be expressed algebraically. Instead, one calculates an angle by approximating. A couple of methods would be: a) Making a binary expansion of the angle (i.e. 16 + 4 + 2 + 1) and recursively bisecting angles until you reach a satisfactory accuracy. b) Numerically solving the differential equation whose solution is sine. (d^2x/dt^2 = - x) – Myridium May 28 '14 at 16:29

If the angle $x$ is in radians, $$\sin x = x-\frac {x^3}{3!}+\frac {x^5}{5!}-\frac {x^7}{7!}+\dots$$with odd powers and alternating signs. This converges pretty quickly, especially for small $x$.

• I don't think this is in the least bit helpful for OP given his/her seemingly limited knowledge.. – Cameron Williams May 28 '14 at 15:44
• @CameronWilliams What OP seems to say is that he wants to compute the sine function directly from the angle - this is easy to implement and works. It isn't so easy to explain or prove without some sophistication, I agree - which is why I asked in a comment about the geometric and other data available. – Mark Bennet May 28 '14 at 15:48
• You're definitely right. I missed your comment above. – Cameron Williams May 28 '14 at 15:49
• @MarkBennet But this is just an estimation? I've searched for what computers use to compute the sine of an angle and this has been one of the things that has popped up. The problem is I still don't learn how to compute the sine of an angle from this since its just an estimation. What if I wanted to prove that the sine of 90 degrees equals 1? This may give a correct answer but what if it doesn't? – user1594138 May 28 '14 at 15:51
• @user1594138: To compute the sine of $30, 45$ or $60$ degrees, there are geometric arguments one can use (based on equilateral triangles or squares). Then one can use the half-angle formula (or related formulas) to compute the sine of $15$. So, geometric reasoning and trig formulas will give you the sine of angles like these. But for an arbitrary angle like, say, $3.27$ degrees, you're basically out of luck: estimates are the best you're going to get. – Jesse Madnick May 28 '14 at 16:01

The OP's question is as follows: given an angle measure (like, say, 57 degrees), how does one actually construct an angle with that measure? The OP would then like to use this angle to create a right triangle in order to find the sine of the angle. The OP is interested in exact values, not approximations.

There are a few issues with this approach:

1. If your method of constructing angles involves only a compass and straightedge, then you're usually out of luck. Sure, you can use a compass and straightedge to construct angles like 30 degrees or 45 degrees, or other nice numbers like those. But for arbitrary angle measures (like 5.7 degrees), a compass and straightedge won't cut it. So, your best best is to use a protractor, which, yes, will be an approximation.

2. Once you have your angle in place, and manage to construct the right triangle, how exactly do you intend to measure the side lengths? Again, if you use a ruler, then you'll have to be satisfied with approximations.

So, what is one to do?

For 30 degrees and 60 degrees, one can find their sines by bisecting an equilateral triangle. For 45 degrees, one can find its sine by bisecting a square (diagonally). Perhaps another user can explain or show you how to do this.

Once one has the sines of 30, 45, and 60 degrees on hand, one can use the trigonometric addition formulas to find the sines of other multiples of 15 degrees (e.g.: 0, 15, 30, 45, 60, 75, 90, etc.).

For angles that are nice fractions of 180 degrees (like 18 degrees, or 36 degrees, say), there are similar formulas (or geometric reasoning) that one can use, if one is a little clever.

But for arbitrary angles, like $3.72$ degrees, say, you're just simply out of luck: you're not going to be able to find an exact expression for such angles -- at least not unless you're really lucky or exceedingly clever.

• Other users can chime in, but the only angles that I know of whose sines are "easy" to calculate explicitly are those whose measures are multiples of $9$ degrees, or multiples of $15$ degrees. With some work I could probably find $\sin(1^\circ)$, but it would take a long time and be very messy... – Jesse Madnick May 28 '14 at 16:31

If one reduces it to the most simple form, an angle is a point on the unit circle. In general it is an equivalence class of pairs of rays originating from the same point, after rotation, the first ray can be made horizontal, the originating point the origin and the second ray is defined by one point on it, which can be chosen as a point on the unit circle.

The coordinates $(x,y)=(c,s)$ on the unit circle are thus the primary object.

To give a measure to the angle, one divides the unit circle into equal parts. This equal-part division is trivial for 4 and 6, 12, 24... parts, and still an easy geometric construction for 5 parts. We know for example that the point $(\tfrac12\sqrt3, \tfrac12)$ is the next point above $(1,0)$ in the division into 12 parts. Now the question is how to mark these points. One system is to mark the full circle as $360°$, one other system by arc length on the unit circle.

Since we can construct with ease the angle for $30°$, one can approximate every other angle of $a°$ by expressing the fraction $a/30$ in binary and proceed by angle bisections.

In the opposite way, every point of the circle is in one of the 12 segments of $30°$ and by angle bisection one can narrow down the angle segment of that point to any desired precision.

And now one may ask for the analytical formula connecting the angle number and the coordinates on the unit circle, which leads to the trigonometric functions and the natural role of the arc length as their argument.

The half angle at the bisector for a point $(c,s)$ in the first or second quadrant on the unit circle is $\sqrt{\frac{1+c}2},\sqrt{\frac{1-c}2}$. Thus for instance the point for $15°$ is the bisector for the angle $30°$ is given as $$c_{15°}=\frac12\sqrt{2+\sqrt3}=\frac1{2\sqrt{2}}\sqrt{3+2\sqrt3+1}=\frac{\sqrt3+1}{2\sqrt{2}}$$ and $$s_{15°}=\frac12\sqrt{2-\sqrt3}=\frac1{2\sqrt{2}}\sqrt{3-2\sqrt3+1}=\frac{\sqrt3-1}{2\sqrt{2}}$$

Multiples of an angle can be computed as the first rows of the powers of the rotation matrices $$\begin{bmatrix}c&-s\\s&c\end{bmatrix}^k$$

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