Mar 29, 2012 · $\begingroup$ Raman: From Wikipedia: "With the advent of the Bulletin board system, or BBS, and later the Internet, typing messages in all caps became closely identified with "shouting" or attention-seeking behavior, and is considered very rude.

Forgive me xD mmm What about trying, in a tedious way, from the power $2$ like $$(\cos x + i\sin x)^2 = \cos^2 x - \sin^2 x + 2i\sin x \cos x$$ And then use some trig identity like $\sin^2 = 1 - cos^2$ or $\cos^2 = \frac{1}{2}(1 + 2\cos)$ et cetera...

Jan 16, 2017 · All you need to do is cancel the I_ns and move the -nI_n to the left hand side: n int cos^n x dx=sin x cos^(n-1)x + (n-1) int cos^(n-2)x dx . Dividing through by n gives the reduction formula. For any particular small positive value of n you can apply this repeatedly to get down to the integral either of 1 or of cos x.

The proof proposed by Winther is the method Euler used to derive $$ \sum_{n=1}^{\infty}\frac{\sin nx}{n}=\frac{\pi-x}{2}\quad (0<x<2\pi). $$ This is often called the first Fourier series, since Euler mentioned it in a letter to Christian Goldbach as early as 1744.

The last two are just quotients of the first two, where a $\cos^nx$ was factored. Share. Cite. Follow ...

Mar 10, 2018 · What is the period of cos nx? Trigonometry. 1 Answer Shiva Prakash M V Mar 10, 2018 ...

Mar 26, 2016 · Your question is: #intcos(mx)cos(nx)dx# To simplify this, use the cosine product-to-sum formula, namely: #cos(A)cos(B)=1/2[cos(A-B)+cos(A+B)]#

Nov 12, 2014 · $\begingroup$ You can use induction: express $\cos((n+1)x)$ in terms of $\cos nx$ and $\cos x$. Also, look up "Chebyshev polynomials" and "cosine". $\endgroup$ – user147263

See the results posted here, where I show that $$\sum_{n=-\infty}^{\infty} \frac{\sin^2{a n}}{n^2} = \pi a$$ ...

Here is the induction step: it comes down to proving \begin{gather*}\frac{\sin \dfrac{nx}{2} \cos\dfrac{(n+1)x}{2}}{\sin\dfrac{x}{2}}+\cos(n+1)x=\frac{\sin\dfrac{(n+1 ...