Prove cos2(x)+sin2(x) = 1

Someone needs to catalogue the awesomeness in this thread, perhaps in the style of the STEP solution threads (eg).

wow..........old thread.

Here's another.

$\cos^2 x + \sin^2 x = (\cos x + i\sin x)(\cos x - i\sin x) = e^{ix}e^{-ix} = e^0 = 1$

Maybe diegofrantic digging this thread up was worthwhile

Here's a proof from the power series definition. Certainly not the neatest way but it assumes comparatively little real analysis.

Firstly we need to know how to multiply series:

Assuming absolute convergence (which the series for sin and cos satisfy!):

$\displaystyle \left( \sum_{n=0}^{\infty} a_n \right) \left( \sum_{n=0}^{\infty} b_n \right) = \sum_{n=0}^{\infty} c_n$

where $\displaystyle c_n = \sum_{k=0}^{n} a_kb_{n-k}$

Next we need two sums which crop up in the multiplication of the sin and cos series:

$\displaystyle (1+x)^{2n} = \sum_{k=0}^{2n} {2n \choose k} x^k$

$\displaystyle (1-x)^{2n} = \sum_{k=0}^{2n} {2n \choose k} (-1)^k x^k$

Adding these we get the cancellation when k is odd and doubling up when k is even. Putting x=1 (and being careful when n=0) we have:

$\displaystyle \sum_{k=0}^{n} {2n \choose 2k} = \left\{ {1\ \ \ \ \ \ n=0 \atop 2^{2n-1}\ n>0} \right.$

Doing the same with 2n+2 replacing 2n and subtracting we get the even terms cancelling and the odd terms doubling up and so:

$\displaystyle \sum_{k=0}^{n} {2n+2 \choose 2k+1} = 2^{2n+1}$

(disclaimer: No guarantee those are the easiest proofs. Just the first ones I thought of!!)

$\displaystyle \cos(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}$

$\displaystyle \cos^2 (x) = \sum_{n=0}^{\infty} c_n$

where $\displaystyle c_n = \sum_{k=0}^{n} (-1)^k \frac{x^{2k}}{(2k)!} (-1)^{n-k} \frac{x^{2n-2k}}{(2n-2k)!}$

$\displaystyle = \frac{(-1)^n x^{2n}}{(2n)!} \sum_{k=0}^{n} {2n \choose 2k} = \left\{ {1 \atop \frac{(-1)^n x^{2n}}{(2n)!} 2^{2n-1}} \right.$

And similarly with sin:

$\displaystyle \sin^2(x) = \sum_{n=0}^{\infty} c_n$

where $\displaystyle c_n = \sum_{k=0}^{n} (-1)^k \frac{x^{2k+1}}{(2k+1)!} (-1)^{n-k} \frac{x^{2n-2k+1}}{(2n-2k+1)!}$

which comes to:

$\displaystyle \frac{(-1)^n x^{2n+2}}{(2n+2)!} 2^{2n+1}$

(using the second binomial sum above)

Finally, $\displaystyle \cos^2(x) + \sin^2(x) = 1 + \sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} 2^{2n-1} + \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+2}}{(2n+2)!} 2^{2n+1}$

And if we re-index the second sum we get:

$\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+2}}{(2n+2)!} 2^{2n+1} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{2n}}{(2n)!} 2^{2n-1}$

And the sums cancel giving the result.

(I'll be impressed if anyone bothers reading through this properly!!)

Looks like this thread should be titled "Look how many ways I can prove sin^2(x) + cos^2(x) = 1".

I would be really surprised as well. What is the point in going in for really complicate proofs when really simple ones exist. I would go for a simple right-angled triangle labelled O, A and H. Then apply pythagoras theorem, followed by dividing each term by H^2.

SsEe's post was excellent. why u hatin?