# Trig Functions

Gold Member
...are fascinating. At least I think so. Sine and cosine are the additive inverses of their respective second derivatives, for example. Astonishing!

Are there any other startling properties of trig functions (not inverse trigs) that would just blow my mind?

Somewhere in the beautiful scale between $$a^2+b^2=c^2$$ and $$e^{i \pi }+1=0$$ is what I'm looking for, I guess.

## Answers and Replies

rochfor1
Check out the Law of Cosines, aka the Pythagorean Theorem for non-right triangles.

snipez90
Well the first property you mentioned is related to a well known theorem. If f is some function that has a second derivative everywhere, and f'' + f = 0, f(0) = a, f'(0) = b. Then f = b*sin + a*cos.

Also, it's well-known that

$$\int_{-\infty}^{\infty}\frac{\sin x}{x}\,dx = \pi.$$

You should try proving the first, since it only requires differentiation. The second is much harder to prove (demonstrating convergence is even a bit tricky), and perhaps less of what you were looking for.

l'Hôpital
One of my personal favorites is

$$2 \prod_{k=1}^{\infty} sec (\frac{\pi}{2^{k+1}}) = \pi$$

or even the elegant

$$\int_{0}^{\pi} cos^n (\theta) cos (n \theta) d\theta = \frac{\pi}{2^n}, n = 0, 1, 2, 3...$$

Gold Member
Nonsense, I'm looking for anything particularly interesting.

On the first theorem presented, I suppose it is intuitive. Since at x=0, a constant multiplied by the sine of x is also 0. Thus, the two constants "switch" in $$f^{(n)}(0)$$ as n increases, changing sign each time they appear.

Right?

Of course, for the second equation, the proof is obvious!

*leaves room to determine proof*

Edit: Gah, took too long to post. Is the giant pi symbol used for the same thing as the giant sigma in summation?

Also, I'm currently trying to remember how to integrate cosine products. I'll try by parts...

l'Hôpital
Nonsense, I'm looking for anything particularly interesting.

On the first theorem presented, I suppose it is intuitive. Since at x=0, a constant multiplied by the sine of x is also 0. Thus, the two constants "switch" in $$f^{(n)}(0)$$ as n increases, changing sign each time they appear.

Right?

Of course, for the second equation, the proof is obvious!

*leaves room to determine proof*

Edit: Gah, took too long to post. Is the giant pi symbol used for the same thing as the giant sigma in summation?

Also, I'm currently trying to remember how to integrate cosine products. I'll try by parts...

The giant pi symbol is "product" as sigma is "sum".

Gold Member
The giant pi symbol is "product" as sigma is "sum".

I see.

Well, for the cosine product function, integration by parts won't work (every two applications, the non-integral part cancels out), I'm pretty sure integration by substitution wouldn't work, and with my limited repertoire (is that how you spell it?), I'm out of options. Same with the sine of x over x function. Infinite products of secants... I doubt it's currently in my ability to understand what's going on, much less prove it.

Und so falle ich.

I am impressed, though. How many series equal some multiple, power, or multiplied power of pi anyways?

l'Hôpital
I see.

Well, for the cosine product function, integration by parts won't work (every two applications, the non-integral part cancels out), I'm pretty sure integration by substitution wouldn't work, and with my limited repertoire (is that how you spell it?), I'm out of options. Same with the sine of x over x function. Infinite products of secants... I doubt it's currently in my ability to understand what's going on, much less prove it.

Und so falle ich.

I am impressed, though. How many series equal some multiple, power, or multiplied power of pi anyways?

The cosine integral is tricky. The product? Not so much.

Hint: Consider $$sin 2\theta = 2 sin \theta cos \theta$$

Gold Member
The cosine integral is tricky. The product? Not so much.

Hint: Consider $$sin 2\theta = 2 sin \theta cos \theta$$

I'm considering it...

Is there a sin(a)/sin(b)= identity anywhere? There's a sin(a)sin(b) identity in this book, but I don't need that...