I have to tell you, I finally figured out how to attach an image. It also enables me to tell you the following story.
For years after high school Trigonometry class, I searched for the Sine of 1°. I guess it was just as a lark. But it is interesting. If you have the Sine of one degree, it enables you to find other values, using the addition formulas.
For a while I thought maybe I could reach it by accident. Or by using other formulas. (As I said, it was just for fun.)
Enter the internet, about 20 years ago. The attached picture gives the formula for the Sine of 1°. (I doubt I would’ve gotten it by accident .)
I don’t think they’re unmatched, they just look wrong because they’re so small.
For example, the very last one should be as tall as the top square root line. It matches the one at the beginning of the third line from the bottom and raises that entire section to a power of 1/3rd.
Similarly, the very first ‘small’ one just after the giant one is closed in the 3rd line just before the 2/3 exponent. A few digits later, there’s another opening parenthesis which is closed at the 2/3 exponent a few lines down.
What’s that I in the last line (just after the 16) (and in a few other spots). I guess it’s a capital i (square root of negative one - but it’s a little odd to see it capitalized).
See here for a different version that is probably equivalent
You can use duplication formula to get sine of 2 degrees. Then by addition formula to get sine of 3 degrees. You also can use half angle formula to get sine 0,5 degrees.
Eventually you get all sines for integer values of degree and fractions having 2’s potenses as denominators.
Sure, but those are hardly the only values of degree for which you can get a “formula”. If you allow non-integral numbers of degrees then it does not stop; e.g.,
That is a nice formula! How did you write it? And is
√3i
supposed to include the i inside the square root? I guess so, but why does it stop after the three? (In your example, in my cut and paste mess it looks even worse, at least on the preview)
First of all: I never use degrees when calculating trigonometric functions.
Secondly: You can calculate any given fraction as a value of (infinite) polynomial of x with integer cofficients and x=1/2. So you can actually use sum formula and half angel formula to get sine of fractional degree.
You can work it out by hand (solution of algebraic equations in terms of radicals) but I cut and pasted it from “sympy” because I was being lazy.
No, better to write \sqrt{-3} or i\sqrt{3} instead of \sqrt{3}i, sorry! That was just computer output.
Not sure what you are trying to do. Sine of 1 degree is \sin(\pi/180), but you (or your computer) can just as easily solve for \sin(2\pi/n) for any natural number n. Of course in general you will need imaginary numbers and higher-order roots (cube roots and so on).
Using square roots, you can get sine (and cosine and tangent) of any multiple of 3º (or power-of-two fractions thereof). And you can get a few other angles, too. But for integer degrees that aren’t a multiple of 3, you’ll always need cube roots in there somewhere. In other words, the ancient problems of Trisecting the Angle and Doubling the Cube are not only both impossible, they’re both impossible for the same reason.
Stated in full, you can construct any angle (or equivalently, calculate it using only square roots) that’s a fraction of a full circle, with a denominator that’s a product of at most one each of the Fermat primes, times any power of 2. The first few Fermat primes are 2, 3, 5, 17, 129… So for instance, you could construct an angle of 29/340 of a full circle, because 340 = 17*5*2^2 . But you couldn’t construct an angle of 1º using only square roots, because that’s 1/360 of a full circle, and 360 = 2^3*3^2*5 , and that 3^2 isn’t allowed.
The rest of this is correct, but I believe the first five (and possibly only) Fermat primes are 3, 5, 17, 257, and 65537, as Fermat numbers are numbers in the form 2^{2^n}+1.
No, I meant how did you manage to write it in the Straight Dope so that it looks like a proper formula. You write 1/6 correctly, I only manage 1/6, which gets when you write not 1, but third root of 7/2 - (21 times sqrt (-3))/2. This is a mess, your formula is clear. How do you write that in this forum?
You can check that putting your formula between $ signs works for in-line formulas: $\sqrt{2}$ yields \sqrt{2}. Displayed equations work between $$ but you may need to leave the rest of the line blank:
.)
This is a mess, your formula is clear. How do you write that in this forum?