Let $x=\displaystyle\prod_{n=1}^{180}\left(\cos{\left(\dfrac{n\pi}{180}\right)}+2\right)$ and $y=\displaystyle\sum_{n=0}^{89}\binom{180}{2n+1}\left(\dfrac{3}{4}\right)^n$. How to find $\frac{x}{y}$ ?  I have no idea to find it.  Thanks in advance.
122128
Prove the following:  $\displaystyle\prod_{n=1}^{180}\left(\cos{\left(\dfrac{n\pi}{180}\right)}+2\right)=\displaystyle\sum_{n=0}^{89}\binom{180}{2n+1}\left(\dfrac{3}{4}\right)^n$.  We can state this as $\displaystyle \binom{180}{1}\left(\dfrac{3}{4}\right)^0+ \binom{180}{3}\left(\dfrac{3}{4}\right)^1+\cdots+\binom{180}{179}\left(\dfrac{3}{4}\right)^{89} = \left(\cos\left(\dfrac{\pi}{180}\right)+2\right)\left(\cos\left(\dfrac{2\pi}{180}\right)+2\right) \cdots \left(\cos\left(\dfrac{180\pi}{180}\right)+2\right)$   The problem I am encountering is that there is no general result we can prove about this for $n$ it seems. I tried using induction on $n$ to prove this but it fails for the base case. Also, combinations and cosines seem unrelated, so how could I prove the equation is true?
71858
$y\frac{\text{d}^2y}{\text{d}x^2}-2\left(\frac{\text{d}y}{\text{d}x}\right)^2+xy^3\frac{\text{d}y}{\text{d}x}=0$. I want to get solution in elementary functions in order to do asymptotic analysis. I tried the problem with Mathematica. It did not give any solutions.
48257
Show that: $$(1)\sum_{n=1}^\infty\ln\left(\cos \frac{x}{2^n}\right)=\ln\frac{\sin x}x$$ $$(2). \sum_{n=1}^\infty\frac1{2^n}\tan \frac{x}{2^n}=\frac1x-\cot x$$ Thank you in advance. NOTE: The origial problem (1) I stated is $\sum_{n=1}^\infty\lim_{x\to \infty}\left(\cos \frac{x}{2^n}\right)=\ln\frac{\sin x}x$, which is a mistake as noted by the comments below.
57644
Finding sum of series $\displaystyle \sum^{\infty}_{k=0}\frac{(k+1)(k+2)}{2^k}$   what i try  Let $\displaystyle x=\frac{1}{2},$ Then series sum is $\displaystyle \sum^{\infty}_{k=0}(k+1)(k+2)x^k$  $\displaystyle \sum^{\infty}_{k=0}x^k=\frac{1}{1-x}\Rightarrow \sum^{\infty}_{k=0}x^{k+1}=\frac{x}{1-x}=\frac{1-(1-x)}{1-x}$  differentiating both side  $\displaystyle \sum^{\infty}_{k=0}(k+1)x^{k}=+\frac{1}{(1-x)^2}$  How do i solve it help me please
102527
If  $\displaystyle f(x) = x^3-\frac{3}{2}x^2+x+\frac{1}{4}.$ then the value of $\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))\mathrm dx$  $\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}(f(x))^3-1.5(f(x))^2+f(x)+0.25 \mathrm dx$  could some help me with this, thanks
57873
Is there an easy way to prove the identity?     $$\cos \left ( \frac{\pi}{7} \right ) - \cos \left ( \frac{2\pi}{7} \right ) + \cos \left ( \frac{3\pi}{7} \right ) = \cos \left (\frac{\pi}{3} \right )$$   While solving one , I am stuck, which looks obvious but without any feasible way to approach.   Few observations, not sure if it would help $$ \begin{align} \dfrac{\dfrac{\pi}{7}+\dfrac{3\pi}{7}}{2} &amp;= \dfrac{2\pi}{7}\\\\ \dfrac{\pi}{7} + \dfrac{3\pi}{7} + \dfrac{2\pi}{7} &amp;= \pi - \dfrac{\pi}{7} \end{align} $$
75580
I was hoping to show that  $$\left(1-\frac{x}{n}+o\left(\frac{2x}{n}\right)\right)^n \xrightarrow{n\to\infty} e^{-x}$$  which would be just fine without the little-$o$.  Trying binomial formula:  $$\left(1-\frac{x}{n}+o\left(\frac{2x}{n}\right)\right)^n = \sum_{k=0}^n \binom{n}{k} \left(1-\frac{x}{n}\right)^{n-k} \left( o\left( \frac{2x}{n} \right)  \right)^k$$  Yet it doesn't look like it leads me somewhere where I could use  $$\frac{o\left( \frac{2x}{n} \right)}{\frac{2x}{n}}\xrightarrow{n\to\infty} 0$$  What is the proper way to handle it?  Edited after the comment by Claude Leibovici.
95063
Find the value of:$$\frac{\sqrt{45} + \sqrt{18}} {\sqrt{7+2\sqrt{10}}}$$      I need help solving this question. Every reply is appreciated.  Thanks!
19470
The left hand side has terms involving  $\binom{n}{m}= \dfrac{n!}{(n-m)!m!}$    $$1+\dfrac{1}{2}\binom{n}{1} +\frac{1}{3}\binom{n}{2}+...........+\frac{1}{n+1}\binom{n}{n} = \dfrac{2^{n+1}-1}{n+1}$$  I've done induction and proved $P(0)$ holds and also $P(1),P(2)$ holds (Just in case)  Now I've found $P(K+1)$ so the sum on the left is going to equal $\dfrac{2^{n+1}-1}{n+1}$ + $\dfrac{1}{n+2}$  and the right hand side is going to equal $\dfrac{2^{n+2}-1}{n+2}$  My problem is coming with the algebra though. I can't get those sides to equal. I can't get rid of the $n+1$ in the denominator.   Anyone have any ideas for me? They'd be much appreciated!  P.S. Someone gave me a hint and said that you can use the binomial theorem to solve this.
113912
If $\displaystyle a,b,c\in \left(0,\frac{\pi}{2}\right)\;,$ Then prove that $\displaystyle \frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}&lt;1$   $\bf{My\; Try::}$ Using $$\sin(a+\underbrace{b+c}) = \sin a\cdot \cos (b+c)+\cos a\cdot \sin (b+c)$$  $$ = \sin a\cdot (\cos b\cos c-\sin b\sin c)+\cos a(\sin b\cos c+\cos b\sin c)$$  $$ = \sin a\cos b\cos c-\sin a\sin b\sin c+\cos a \sin b\cos c+\cos a\cos b\sin c$$  Now how can i solve it after  that , Help required, Thanks
93405
Recently, I ran across a product that seems interesting.  Does anyone know how to get to the closed form:  $$\prod_{k=1}^{n}\cos\left(\frac{k\pi}{n}\right)=-\frac{\sin(\frac{n\pi}{2})}{2^{n-1}}$$  I tried using the identity $\cos(x)=\frac{\sin(2x)}{2\sin(x)}$ in order to make it "telescope" in some fashion, but to no avail. But, then again, I may very well have overlooked something.  This gives the correct solution if $n$ is odd, but of course evaluates to $0$ if $n$ is even.   So, I tried taking that into account, but must have approached it wrong.   How can this be shown? Thanks everyone.
43191
Let $x,y,z\in\Bbb R$, and at least one is be a postive integer, and such that  $$x+\frac{1}{y},\;y+\frac{1}{z},\;z+\frac{1}{x}\in\Bbb Z$$ Find the value of $x,y,z$.  My try:  $$\left(x+\dfrac{1}{y}\right)\left(y+\dfrac{1}{z}\right)\left(z+\dfrac{1}{x}\right)=\dfrac{(xy+1)(yz+1)(xz+1)}{xyz}\in Z$$  Then I can't. Thank you.  I konw $(x,y,z)=(1,1,1),(-1,-1,1),(2,0.5,1)$ is such condition,But How prove it?
94432
How would I find the sum below?    $$ \sum_{k=0}^{n} {\frac{1}{k+1} {n \choose k}}=?$$  I need help starting on the problem. Would I first need to utilize the Binomial Theorem proof to get started on it?    Thank you.
88475
let $ y=\displaystyle \sqrt{x+\sqrt{{x}+\sqrt{x+\cdots}}}$, i'm really interesting to know how do I find :  $\displaystyle \frac{dy}{dx}$ ?.  Note: I have used the definition of derivative of the square root function but i don't succed .  Thank you for any help
30871
Is it possible to find $\displaystyle{\cos x \over \cos y}$ if $\displaystyle{\sin x\over \sin y}$ is given? If so, how would one approach this problem? Thank you in advance!
103341
How   $\frac{1}{n}\sum_{i=1}^n X_i^2 - \bar X^2 = \frac{\sum_{i=1}^n (X_i - \bar X)^2}{n}$   i have tried to do that by the following procedure:  $\frac{1}{n}\sum_{i=1}^n X_i^2 - \bar X^2$  =$\frac{1}{n}(\sum_{i=1}^n X_i^2 - n\bar X^2)$  =$\frac{1}{n}(\sum_{i=1}^n X_i^2 - \sum_{i=1}^n\bar X^2)$  =$\frac{1}{n} \sum_{i=1}^n (X_i^2 - \bar X^2)$  Then i have stumbled.
41068
Problem is to find sum  $$\frac{1}{3!}+\frac{4}{4!}+\frac{9}{5!}\cdots$$  What I knew doesn't apply on this problem  Some series are telescoping, some types are solvable using binomial , both look useless here Binomial gives  $$n (n-1) (n-2)x^3=1$$  $$n (n-1)( n-2)(n-3)x^4=4$$  What approach to use here?