# Finding sine of an angle in degrees without $\pi$

The following is found using a combination of: (a) a polygon with an infinite number of sides is a circle, (b) the perimeter of that polygon is the circumference of the circle that it becomes (of course), (c) the sine theorem, and (d) the ratio between the circumference of a circle and the diameter is $\pi$. $$\pi=\lim_{n\to\infty} n\sin \left(\frac{180}{n}\right)^o$$ I have been trying to rewrite the above expression without $sin$ in it explicitly. So, I researched other expressions for $sin$, and found this: $$\sin (x) = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\frac{x^{11}}{11!}+\cdots$$ Attempting to rewrite this Taylor series in Sigma notation yielded this: $$\sin(x)=\sum_{n=1}^{\infty} \frac{x^{2n-1}}{(2n-1)!}(-1)^{n-1}$$ The issue for me is that I have an angle measure in degrees, but $x$ above is in radians. I would convert the degrees to radians and be done with it just like that, except that I feel that solving for $\pi$ with $\pi$ is like defining a word with itself, and is 'cheating'.

I have been trying to find a way to make this conversion without $\pi$ but to no avail. I also tried using trig identities, but that has just led me in circles.

Is there another way to find the sine of $x^o$ without involving $\pi$ at all?

An even more explicit and blunt form of my question: What is another way to express $sin(x^o)$ without using $\pi$?

To be clear, I am not looking for anything that involves approximations or infinite series that cannot be wholly expressed in a finite amount of space (ex. Taylor series above is okay because it can be expressed in sigma notation).

• How would finding the sine of an angle $x$, where $x$ is measured in degrees, help you to write an exact expression for $\pi$? Feb 19, 2016 at 19:17
• @DavidK - edited. Please tell me of I should elaborate Feb 19, 2016 at 20:19
• Sines and cosines of rational numbers of degrees can be computed as roots of polynomials, without involving $\pi$ at all. For example, $\sin(1 \text{ degree})$ is a root of a polynomial of degree $48$ that's too big to fit in a comment. Feb 19, 2016 at 20:32
• If you want $\sin(180/n \text{ degrees})$ for $n$ a power of $2$, just repeatedly use the half-angle formulas. Feb 19, 2016 at 20:41
• (1)You cannot convert from radians to degrees or vice-versa without using $\pi$ because $180^o= \pi$ radians. (2) A circle is not really an infinite-sided polygon. (3) To prove $\pi=\lim_{n\to \infty} n \sin (180/n)^o$ you need a def'n of $\pi$ to begin. Define $\pi$ as the length of a semi-circle of radius $1$. Then you need to define what "length of a semi-circle" means: It's the limit of the perimeters of polygons that uniformly approximate it This implies $\pi=\lim n\sin (180/n)^o.$ (4)The power series for $\sin x$ is in radians. Feb 19, 2016 at 21:45

## 2 Answers

As noted in a comment, there are equations you can write for the trigonometric functions of rational portions of a right angle (and therefore for the sine of any rational number of degrees) without using $\pi$ or any trigonometric functions. To actually use these equations may prove somewhat cumbersome, however.

Instead of attempting a formula for an arbitrary integer or rational number of degrees, let me just address the desire to evaluate $$\newcommand{\Sin}{\mathop{\mathrm{Sin}}} \newcommand{\Cos}{\mathop{\mathrm{Cos}}} \lim_{n\to\infty} n \Sin\left(\frac{180}{n}\right)$$ where $\Sin$ is the sine function that takes its parameter in degrees, for example, $\Sin(90) = 1$.

It is sufficient to examine the following limit for integer values of $k$: $$\lim_{k\to\infty} 2^k \Sin\left(\frac{180}{2^k}\right).$$ The quantity $\Sin\left(\frac{180}{2^k}\right)$ is easy to compute (at least, it is easy compared to such values as $\Sin(1)$). Let $\Cos(x)$ be the cosine function for $x$ measured in degrees; then apply the half-angle formula for cosines of angles in the interval from $0$ to $180$ degrees, inclusive: $$\Cos\left(\frac{x}{2}\right) = \sqrt{\frac{1 + \Cos x}{2}}.$$

If $x = \dfrac{180}{2^m}$, then $\dfrac x2 = \dfrac{180}{2^{m+1}}$ and the half-angle formula just says that $$\Cos\left(\frac{180}{2^{m+1}}\right) = \sqrt{\frac{1 + \Cos \left(\frac{180}{2^m}\right)}{2}}.$$

You can find the value for any $k$ by starting at $m=1$ and applying the half-angle formula repeatedly for $m=2,3,\ldots, k$: \begin{align} \Cos\left(\frac{180}{2}\right) &= \Cos(90) = 0,\\ \Cos\left(\frac{180}{4}\right) &= \sqrt{\frac{1 + \Cos \left(\frac{180}{2}\right)}{2}} = \sqrt{\frac{1 + 0}{2}} = \frac{1}{\sqrt 2}, \\ \Cos\left(\frac{180}{8}\right) &= \sqrt{\frac{1 + \Cos \left(\frac{180}{4}\right)}{2}} = \sqrt{\frac{1 + \frac{1}{\sqrt 2}}{2}} = \sqrt{\frac{\sqrt 2 + 1}{2\sqrt 2}}, \\ \Cos\left(\frac{180}{16}\right) &= \sqrt{\frac{1 + \Cos \left(\frac{180}{8}\right)}{2}} = \sqrt{\frac{1 + \sqrt{\frac{\sqrt 2 + 1}{2\sqrt 2}}}{2}} = \sqrt{\frac{\sqrt{2\sqrt 2} + \sqrt{\sqrt 2 + 1}}{2\sqrt{2\sqrt 2}}}, \\ \end{align} and so forth to obtain $\Cos\left(\frac{180}{2^k}\right)$ for $k$ as large as desired. Then obtain the sine by $$\Sin\left(\frac{180}{2^k}\right) = \sqrt{1 - \left(\Cos\left(\frac{180}{2^k}\right)\right)^2}.$$

• How would I calculate $Cos(\frac{180}{2^k})$ with a degree measure, then? Wouldn't that involve yet another Taylor series that takes radians and not degrees? (From looking at this page: people.math.sc.edu/girardi/m142/handouts/…) Feb 19, 2016 at 21:15
• No, you do not need to use Taylor series. I have shown all the calculations explicitly for $k = 1,2,3,4$. Just continue like that for higher powers of $2$. You do have to take a square root each time. Feb 19, 2016 at 21:28
• Do I not need to somehow calculate $Cos(\frac{180}{2^k})$, though? How would I calculate that? Feb 19, 2016 at 21:35
• I have given step-by-step instructions. First compute $\newcommand{\Cos}{\mathop{\mathrm{Cos}}}\Cos(180/2)$, then $\Cos(180/4)$, then $\Cos(180/8)$, and so forth using the half-angle formula exactly as shown. The denominator doubles each time, so if you do this enough times the denominator will be $2^k$. Feb 19, 2016 at 21:44
• How do we know that these formulas use degrees and not radians? Feb 19, 2016 at 22:24

In relation to David K's answer, I also note one small problem:

$$\sin(\frac x2)=\pm\sqrt{\frac{1-\cos(x)}2}$$

Obviously, $\sin(\frac x2)$ being negative is fully possible.

However, we know that for $0\le x\le\pi$, $\sin(x)>0$, in other words, it is positive.

On the interval from $-\frac{\pi}2\le x\le\frac{\pi}2$, $\cos(x)>0$.

So, we may start with $\cos(1)\approx0.540302306$.

We have $$\cos(0.5)\approx0.877581562$$

$$\cos(0.25)\approx0.968912422$$

$$\cos(0.125)\approx0.992197667$$

etc. (these values were obtained through the half angle formula)

Similarly, we have: (using half angle formula)

$$\sin(1)\approx0.841470985$$

$$\sin(0.5)\approx0.479425539$$

$$\sin(0.25)\approx0.247403959$$

$$\sin(0.125)\approx0.124674733$$

etc.

Now, knowing our trigonometric identities, one can use the sum of angles formula to approximate the desired value to desired accuracy.