go to appwiz.cpl/turn win features on/off
turn on telnet
cmd:
Telnet Towel.blinkenlights.nl

http://www.telnet.org/htm/places.htm

# Where to find Lockscreen Spotlight images in Windows 10?

Windows Spotlight is a fancy feature which exists in Windows 10 November Update 1511. It downloads beautiful images from the Internet and shows them on your lock screen! So, every time you boot or lock Windows 10, you will see a new lovely image. However, Microsoft made the downloaded images hidden from the end user. Here is how you can find those images and use them as your wallpaper or somewhere else.
To get access to image files downloaded by the Windows Spotlight feature, follow the instructions below.
1. Press Win + R shortcut keys together on the keyboard to open the Run dialog. Tip: See the complete list of Win key shortcuts available in Windows.
2. Enter the following in the Run box:
%localappdata%\Packages\Microsoft.Windows.ContentDeliveryManager_cw5n1h2txyewy\LocalState\Assets

Press Enter
3. A folder will be opened in File Explorer.
4. Copy all the files you see to any folder you want. This PC\Pictures is suitable.
5. Rename each file you copied to add a ".jpg" extension. You can do so by either selecting each file, and pressing F2, then Tab to quickly rename files.Or an even faster way would be to open a command prompt at the selected folder and type this command:
Ren *.* *.jpg
You are done:
Alternatively, you can use Winaero Tweaker. A tool has been added to find your current lock screen image or grab the entire Spotlight image collection which Windows 10 has downloaded and stored on your drive. Unlike scripts you might be using for this, the tool doesn't collect "garbage" files like provisioned app icons and promoted app tiles. Also, it sorts images according to their screen orientation (Landscape and Portrait). You will find it under Tools\Find Lock Screen Images:
You can get the app here: Download Winaero Tweaker.
Suppose you want cos(nx) and sin(nx) in terms of sin(x).  You'll need cos(x) as well which can be found using
$cos\left(x\right)=±\sqrt{1-\left(sin\left(x\right){\right)}^{2}}$

First note that ${e}^{ix}=cos\left(x\right)+i\ast sin\left(x\right)$
Raising both sides to the nth power:
${e}^{n\ast ix}=\left(cos\left(x\right)+i\ast sin\left(x\right){\right)}^{n}$
And note that
${e}^{n\ast ix}=cos\left(nx\right)+i\ast sin\left(nx\right)$
Therefore, by the transitive property
$\left(\mathrm{cos}\left(x\right)+i\ast \mathrm{sin}\left(x\right){\right)}^{n}=\mathrm{cos}\left(nx\right)+i\ast \mathrm{sin}\left(nx\right)$
So
$cos\left(nx\right)=Re\left(cos\left(x\right)+i\ast sin\left(x\right){\right)}^{n}\right)$
and
$sin\left(nx\right)=Im\left(cos\left(x\right)+i\ast sin\left(x\right){\right)}^{n}\right)$
Expanding using the binomial theorem yields:
$\left(cos\left(x\right)+i\ast sin\left(x\right){\right)}^{n}=$$\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right)\ast \left(cos\left(x\right){\right)}^{n-k}\ast \left(i\ast sin\left(x\right){\right)}^{k}$

Splitting it into real and imaginary parts:
$cos\left(nx\right)=\sum _{k=0}^{⌊\frac{n}{2}⌋}$$\left(\genfrac{}{}{0}{}{n}{2k}\right)\ast \left(-1{\right)}^{k}\ast \left(sin\left(x\right){\right)}^{2k}\ast \left(cos\left(x\right){\right)}^{n-2k}$
And
$sin\left(nx\right)=\sum _{k=0}^{⌊\frac{n-1}{2}⌋}\left(\genfrac{}{}{0}{}{n}{2k+1}\right)\ast$

### armitage archlinux

export MSF_DATABASE_CONFIG=/home/kai/.msf4/database.yml cat  /home/kai/.msf4/database.yml adapter: postgresql  database: databasename  u...