How to type theta as shown below?  Hope image is clear
1659
I know what my symbol or character looks like, but I don't know what the command is or which math alphabet it came from. How do I go about finding this out?
115191
I'm looking for the symbol that's similar to \cdot in math mode, but instead of a small circle, it's a square. Like this:
697
How to type these special letters from European languages in latex? ä, é, and L'?
44031
How to create the following symbol?
13331
I've tried substituting $1+\cos\theta=2\cos^2(\frac{\theta}{2})$ but it didn't give me the answer.
69172
I'm trying to figure out this problem and feel like it's something that must be so simple that I could've done in high school no problem, but for some reason my brain is frozen this morning.  I would really appreciate any help, and want to say thanks in advance. I tried to draw a picture below; I want to find the slope of a line given a point $(x,y)$ and $\theta$.
120589
(1) Evaluation of $\displaystyle \tan \left(\frac{\pi}{7}\right)\cdot \tan \left(\frac{2\pi}{7}\right)\cdot \tan \left(\frac{3\pi}{7}\right) = $ (2) Evaluation of $\displaystyle \tan\left(\frac{\pi}{11}\right)\cdot \tan\left(\frac{2\pi}{11}\right)\cdot \tan\left(\frac{3\pi}{11}\right)\cdot \tan\left(\frac{4\pi}{11}\right)\cdot \tan\left(\frac{5\pi}{11}\right) = $  $\bf{My\; Try::(1):}$ Let $\displaystyle \frac{\pi}{7} = \theta\Rightarrow \pi = 7\theta\Rightarrow (\pi-3\theta) = 4\theta\Rightarrow \tan(\pi-3\theta) = \tan(4\theta)$ So $\displaystyle \tan (3\theta) = -\tan (4\theta)\Rightarrow \frac{3\tan \theta - \tan^3\theta}{1-3\tan^2 \theta} = -\left(\frac{2\tan 2\theta}{1-\tan^2 2\theta}\right) = \frac{4\tan^3 \theta-4\tan \theta}{1+\tan^4 \theta-6\tan^2 \theta}$ So $\displaystyle \left(\frac{3-\tan^2 \theta}{1-3\tan^2 \theta}\right) = \left(\frac{4\tan^2 \theta-4}{1+\tan^4 \theta-6\tan^2 \theta}\right)\Rightarrow \tan^6\theta-21\tan^4 \theta+21\tan^2 \theta-7=0$ Now Let $\displaystyle \tan \theta = y\;,$ then eqn. convert into $\displaystyle y^6-21y^4+21y^2-7=0$ and equation has a roots $\displaystyle y = \tan \left(\frac{\pi}{7}\right)\;,\tan \left(\frac{2\pi}{7}\right)\;,\tan \left(\frac{3\pi}{7}\right)\;,\tan \left(\frac{4\pi}{7}\right)\;,\tan \left(\frac{5\pi}{7}\right)\;,\tan \left(\frac{6\pi}{7}\right)\;$. So Product of Roots is $\displaystyle \tan \left(\frac{\pi}{7}\right)\cdot \tan \left(\frac{2\pi}{7}\right)\cdot \tan \left(\frac{3\pi}{7}\right)\cdot \tan \left(\frac{4\pi}{7}\right)\cdot \tan \left(\frac{5\pi}{7}\right)\cdot \tan \left(\frac{6\pi}{7}\right) = -7$ Now $\displaystyle \tan\left(\frac{4\pi}{7}\right) = \tan\left(\pi-\frac{3\pi}{7}\right)=-\tan\left(\frac{3\pi}{7}\right)$ Similarly $\displaystyle \tan\left(\frac{5\pi}{7}\right) = -\tan\left(\frac{2\pi}{7}\right)$ and $\displaystyle \tan\left(\frac{6\pi}{7}\right) = -\tan\left(\frac{\pi}{7}\right)$ So $\displaystyle \tan\left(\frac{\pi}{7}\right)\cdot \tan\left(\frac{2\pi}{7}\right)\cdot \tan\left(\frac{3\pi}{7}\right)=\sqrt{7}$ I did not understand how can i calculate $(2)$ one using the same method:, bcz it is very lengthy can we solve the $(1)$ and $(2)$ question any other method. If yes then plz explain me, Thanks :: Thanks
52006
How can I Tex this equation?  I try to have exactly the same result but for the moment I only succeeded in having intersecting lines... Thanks!
14634
I've been learning about polar curves in my Calc class and the other day I saw this suspiciously $r=1-\cos \theta$  looking thing in my coffee cup (well actually $r=1-\sin \theta$ if we're being pedantic.) Some research revealed that it's called a caustic. I started working out why it would be like this, but hit a snag. Here's what I did so far:  Consider the polar curves $r=1-\cos \theta$ and $r=1$. Since for a light ray being reflected off a surface (or the inside of my cup) the $\angle$ of incidence =$\angle$ of reflection, a point on the circle $(1,\theta)\to(1,2\theta)$. It looks like this has something to do with the tangent lines, so I pretend there's an $xy$ plane centered at the pole to find the slope of the line connecting the points. Since it's the unit circle the corresponding rectangular coordinates are $(\cos \theta, \sin \theta)$ and $(\cos 2\theta, \sin 2\theta).$ So  $$m={\sin 2\theta-\sin \theta \over \cos 2\theta-\cos \theta}$$ now we see if it matches up with the slope of the lines tangent to the cardioid $$\frac{dy}{dx}={r'\sin \theta+r \cos \theta \over r' \cos \theta - r \sin \theta}={\sin^2\theta - \cos^2 \theta +\cos \theta \over 2\sin \theta \cos \theta -\sin\theta}={\cos \theta - \cos 2\theta\over \sin 2\theta-\sin \theta}$$  They're similar, but not identical. In particular $\frac{dy}{dx}=-\frac1m$. What error have I made, or what have I overlooked conceptually? Thanks in advance.
120849
I don't know which buttons I have to click or what key combination to type to calculate a cube root.
33085
Currently, , ,  and  have MathJax support which turns LaTeX code into equation. At least on math, occasionally there are users that do not know this feature or don't know it's possible to use LaTeX (MathJax), resulting a badly formatted post. I suggest that there should be a common &quot;LaTeX/MathJax typesetting help&quot; page for those sites that enabled MathJax support, like the  that is present on all sites. Preferably, there should also be a line  LaTeX equations $\sin^2 \theta$  in the &quot;How to Format&quot; on the right hand side. See also:
99880
I've used "//" and "\" to create this "W", and have made them into outlines, as I'm trying to give the different layers a colour, as well as having a white space where the lines touch and I would like to remove the top of the second "\".   I'm struggling with the white space at the moment and now I've added a stroke, the fill has decreased. Is there another way I could do this?   I've also tried to use my pen tool to remove the top of the "\" but when using "Path Finder" I cannot find an option to "Minus Forward". I'm a newbie so any advice would be lovely :)
85770
I just spotted someone use \newcommand* in an answer and realised that I'd never quite sorted out in my head what the star (asterisk) was there for.  (This one is practically impossible to search for on the internet so this is a good place to record the answer!)
12149
How can I make these types of lines?
99396
How do I make this in LaTeX? I keep losing track of the brackets.
14408
The symbol I would like to obtain ought to be like the symbol '\pm'.
92779
It seems to be generally well accepted that the correct way to typeset a value with units is to separate the number from the units by a space, and write the units in an upright font.  For example,  One hellameter is \( 1.0\times10^{27} \) m.   The symbol for 'micro-' is the Greek letter mu, but if you compile  At the telecom wavelength of \( 1.55\ \mu \)m ...   then the spacing will be correct and the 'm' will be upright, but the 'mu' won't be.     How do you produce an upright lowercase mu?  More generally, how do you make any Greek letter appear upright?