heureka

Do arctan(2) in rad - modus. You get $$1.10714871779\ rad$$

Do arctan(2) in degree - modus. You get $$63.4349488229\ensurement{^{\circ}}$$

$$\\ 63.4349488229\ensurement{^{\circ}} \cdot \frac{\pi}{180} = 1.10714871779\ rad \\\\
1.10714871779\ rad \cdot \frac{180}{\pi} =63.4349488229\ensurement{^{\circ}} \\$$

Given that z=-√2-√2i, calculate z^8

$$\small{\text{
$z= -\sqrt2-\sqrt2\cdot i \qquad
\boxed{z= -\sqrt2\cdot (1+i)}
$}}\\\\
\small{\text{
$
z^8 = \left[-\sqrt2 \cdot(1+i) \right] ^8
$}}\\
\small{\text{
$
z^8 = (-\sqrt2)^8\cdot (1+i)^8
$
}} \\
\small{\text{
$
z^8 = (-1)^8\cdot 2^{\frac82}\cdot (1+i)^8
$
}} \\
\small{\text{
$ z^8 = 1\cdot 2^4\cdot (1+i)^8 $
}} \\
\small{\text{
$
\begin{array}{l|l|l}
z^8 = 2^4\cdot (1+i)^8 \quad & \quad (1+i)^2 =1+2\cdot i+ i^2 \quad & \quad \boxed{\ i^2=-1 \ } \\
& \quad (1+i)^2 = 1+2\cdot i -1 \\
& \quad (1+i)^2 = 2\cdot i \\
z^8 = 2^4 \cdot \left[(1+i)^2\right]^4 \\
z^8 = 2^4 \cdot \left[ 2\cdot i\right]^4 \\
z^8 = 2^4 \cdot 2^4 \cdot i^4 \\
z^8 = 2^8\cdot i^4 \quad & \quad i^4=i^2\cdot i^2\\
& \quad i^4=(-1)\cdot (-1)\\
& \quad i^4 = (-1)^2 \\
& \quad i^4 = 1 \\
\boxed{\ z^8 = 2^8 = 256 \ }
\end{array}
$
}} \\$$

e^i*1.5708 ?

$$\boxed{
\mathrm{e}^{\mathrm{i}\,\varphi} = \cos\left(\varphi \right) + \mathrm{i}\,\sin\left( \varphi\right)
}$$

$$\varphi= \frac{\pi}{2}=1.57079632679$$

$$\mathrm{e}^{\mathrm{i}\,\cdot1.57079632679 }=
\mathrm{e}^{\mathrm{i}\,\cdot\frac{\pi}{2} } = \cos\left(\frac{\pi}{2}\right) + \mathrm{i}\,\cdot\sin\left( \frac{\pi}{2}\right)\\
= 0 + \mathrm{i}\,\cdot1\\
= \mathrm{i}\\$$

e^i*1.5708  = i

Find an angle x where sin x = cos x

$$\cos(x)-\sin(x)=0 \\\\
a\cdot \cos(x) + b\cdot \sin(x) = c \qquad | \qquad a=1 \qquad b=-1 \qquad c= 0\\
a\cdot \cos(x) + b\cdot \sin(x) = c \quad | \quad : a \\
\cos(x) + \frac{b}{a} \cdot \sin(x) = \frac{c}{a} \quad | \quad c = 0 \\\\
\cos(x) + \frac{b}{a} \cdot \sin(x) = 0 \\
\small{\text{
$
\text{We set } \tan{(\varepsilon)} = \frac{b}{a} \quad \text{ we have } a = 1 \text{ and } b = -1 \text{ so } \varepsilon = \arctan{(-1)} = -\frac{\pi}{4}
$
}}\\\\
\cos(x) + \frac{b}{a} \cdot \sin(x) = 0 \quad | \quad \tan{(\varepsilon)} = \frac{b}{a}\\\\
\cos(x) + \tan{(\varepsilon)}\cdot \sin(x) = 0 \\
\cos(x) + \frac{ \sin{(\varepsilon)} } { \cos{(\varepsilon)} } \cdot \sin(x) = 0 \quad | \quad \cdot \cos{(\varepsilon)} \\
\small{\text{
$
\cos{ (\varepsilon)} \cdot \cos(x) + \sin{(\varepsilon)} \cdot \sin(x) = 0 \quad | \quad \cos{ (x-\varepsilon )} = \cos{ (\varepsilon)} \cdot \cos(x) + \sin{(\varepsilon)} \cdot \sin(x)
$}}\\
\cos{ (x-\varepsilon )} =0 \quad | \quad \pm \arccos{}\\
x-\varepsilon = \pm \arccos{(0)} = \pm \frac{\pi}{2}\\
x-\varepsilon = \pm \frac{\pi}{2} \\
x= \varepsilon \pm \frac{\pi}{2} \quad | \quad \varepsilon = -\frac{\pi}{4}\\
x= -\frac{\pi}{4} \pm \frac{\pi}{2} \\\\
x_1= -\frac{\pi}{4} +\frac{\pi}{2} = \frac{\pi}{4} \\
x_1= 45\ensurement{^{\circ}} \pm k\cdot 360\ensurement{^{\circ}}$$

$$\\x_2= -\frac{\pi}{4} -\frac{\pi}{2}
= -\frac{3}{4} \cdot \pi
= -\frac{3}{4} \cdot \pi + 2\pi
= \frac{5}{4} \cdot\pi \\
x_2 = 225\ensurement{^{\circ}} \pm k\cdot 360\ensurement{^{\circ}}$$

$$k=0,1,2\cdots$$

5√32 ?

$$5\sqrt{32} \\
=5\sqrt{16\cdot 2} \\
=5 \underbrace{\sqrt{16}}_{=4} \cdot \sqrt{2} \\
=5\cdot 4 \cdot \sqrt{2} \\
=20 \cdot \sqrt{2} \quad | \quad
\boxed{\ \sqrt{2}= 1.41421356237 \ }\\
= 20 \cdot 1.41421356237\\
=28.2842712475$$

If the radius of a circle is 4cm, what is that area ?

$$\mathrm{Radius} \quad r = 4\ cm \qquad \mathrm{Area}_{Circle}\quad A =\ ?$$

$$\boxed{ A = \pi \cdot {r^2} }$$

$$A = \pi \cdot 4^2 \ cm^2\\
A = \pi \cdot 16\ cm^2 \\
A = 3.14159265359 \cdot 16\ cm^2 \\
A = 50.2654824574 \ cm^2 \\
A \approx 50.27 \ cm^2$$

2 1/3 + 1 1/3

$$2 \frac13 + 1 \frac13 \\\\
= 2 + \frac13 + 1 + \frac13 \\\\
= 2 + 1 + \frac13 + \frac13 \\\\
= 3 + \frac13 + \frac13 \\\\
= 3 + 2\cdot \frac13 \\\\
= 3 + \frac23 \\\\
= 3 \frac23$$

What is the last digit of pi ?

What Question

What to the power of 10 equals 149597870700?

$$\small{\text{$
10^{ \log{(149\ 597\ 870\ 700)} } = 10^{11.1749254120}=149\ 597\ 870\ 700.
$
}}$$