heureka

simplify tan[ 2*arccos(x/3) ]

$\begin{array}{|rcll|} \hline \cos(\varphi) &=& \frac{x}{3} \\ \varphi &=& \arccos\left(\frac{x}{3} \right) \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \tan(2 \varphi ) &=& \frac{\sin(2 \varphi)}{\cos(2 \varphi)} \quad & | \quad \sin(2 \varphi) = 2 \sin(\varphi)\cos(\varphi) \quad \cos(2\ \varphi) = 2 \cos^2(\varphi)-1 \\ \tan(2 \varphi ) &=& \frac{2 \sin(\varphi)\cos(\varphi)}{2 \cos^2(\varphi)-1} \quad & | \quad \cos(\varphi) = \frac{x}{3} \\ \tan(2 \varphi ) &=& \frac{2 \sin(\varphi)\frac{x}{3}}{2 (\frac{x}{3})^2-1} \quad & | \quad \sin(\varphi) = \sin(\arccos\left(\frac{x}{3} \right)) \\ \tan(2 \varphi ) &=& \frac{2 \sin(\arccos\left(\frac{x}{3} \right))\frac{x}{3}}{2 (\frac{x}{3})^2-1} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \tan(2 \varphi ) &=& \frac{2 \sin(\arccos\left(\frac{x}{3} \right))\frac{x}{3}}{2 (\frac{x}{3})^2-1} \quad & | \quad \sin(\arccos\left(\frac{x}{3} \right)) = \sqrt{1-\left(\dfrac{x}{3}\right)^2} \\ \tan(2 \varphi ) &=& \frac{2 \sqrt{1-\left(\frac{x}{3}\right)^2}\frac{x}{3}}{2 (\frac{x}{3})^2-1} \\ \tan(2 \varphi ) &=& \frac{2 \sqrt{\frac{9-x^2}{9}}\frac{x}{3}}{ \frac{2x^2-9}{9}} \\ \tan(2 \varphi ) &=& \frac{2 \frac{\sqrt{9-x^2}}{3}\frac{x}{3}}{ \frac{2x^2-9}{9}} \\ \tan(2 \varphi ) &=& \frac{2 \frac{\sqrt{9-x^2}}{9}x}{ \frac{2x^2-9}{9}} \\ \tan(2 \varphi ) &=& \frac{2 \sqrt{9-x^2} x}{ 2x^2-9 } \quad & | \quad \sqrt{9-x^2} = \sqrt{3-x}\cdot\sqrt{3+x}\\ \tan(2 \varphi ) &=& \frac{2 \sqrt{3-x}\cdot x \cdot \sqrt{3+x}}{ 2x^2-9 } \\ \mathbf{\tan(2 \arccos\left(\frac{x}{3} \right) ) }& \mathbf{=} & \mathbf{\frac{2 \sqrt{3-x}\cdot x \cdot \sqrt{3+x}}{ 2x^2-9 } } \\ \hline \end{array}$

ein Angesterllter verdient nach einer gehaltserhöhung statt 2 200 euro nunmehr 2 365 euro.
wie viel % betrug die erhöhung?

$\begin{array}{|rcll|} \hline 2200€ + 2200€ \cdot x &=& 2365€ \\ 2200€ \cdot x &=& 2365€ -2200€ \\ x &=& \frac{2365€ -2200€}{2200€} \\ x &=& \frac{165€}{2200€} \\ x &=& 0.075 \qquad (7.5\ \%) \\ \hline \end{array}$

7.5% betrug die Erhöhung.

If r = 9 units and h = 6 units, what is the volume of the cylinder shown above? Use 3.14 for

$V = \pi \cdot r^2 \cdot h \qquad r = 9, \quad h = 6, \quad \pi = 3.14 \\ V=3.14 \cdot 9^2\cdot 6 \\ V = 1526.04 \quad \text{units}^3$

If C^3 + 2*C^2 + (1 - Z)*C + (1 - Z)^ 2 = 0, what values can C and Z have other than C = 0, Z = 1?

$C=-2,\ Z = 1 \\ C=-2,\ Z=-1$

how do you find the area of a circle

$\boxed{area = \pi r^2} \qquad r=\text{radius of the circle} \\ \boxed{area = \pi \cdot \left( \frac{d}{2} \right)^2} \qquad d=\text{diameter of the circle}$

-1 to the power of 2

(-1)² = (-1)*(-1) = 1

Kyle is pulling a box east with a force of 300 newtons at a constant angle of 42° to the horizontal.
Jerome is pushing the box from behind with a force of 350 newtons due east.
Determine the magnitude and direction of the resultant force on the box.

Let m = magnitude
Let D = direction

1. trigonometric:
$\begin{array}{|rcll|} \hline m^2 &=& 300^2+350^2-2\cdot 300 \cdot 350 \cdot \cos(180^{\circ} - 42^{\circ}) \\ m^2 &=& 212500+210000 \cdot \cos(42^{\circ}) \\ m^2 &=& 212500+210000 \cdot \cos(42^{\circ}) \\ m^2 &=& 212500+210000 \cdot 0.74314482548 \\ m^2 &=& 368560.413350 \\ \mathbf{m} & \mathbf{=} & \mathbf{607.091766828\ N } \\\\ \frac{\sin(D)}{300} &=& \frac{\sin(180^{\circ} - 42^{\circ})}{m} \\ \frac{\sin(D)}{300} &=& \frac{\sin(42^{\circ})}{m} \\ \sin(D) &=& \frac{300}{m} \cdot \sin(42^{\circ}) \\ \sin(D) &=& \frac{300}{m} \cdot \sin(42^{\circ}) \\ \sin(D) &=& \frac{300}{607.091766828} \cdot \sin(42^{\circ}) \\ \sin(D) &=& 0.49415922994 \cdot 0.66913060636 \\ \sin(D) &=& 0.33065706517 \\ \mathbf{D} & \mathbf{=} & \mathbf{19.3086615149^{\circ}} \\ \hline \end{array}$

2. vectorial:

$\begin{array}{|rcll|} \hline \vec{K} &=& \dbinom{350}{0} \\ \vec{J} &=& \dbinom{ 300\cdot \cos(42^{\circ}) } { 300\cdot \sin(42^{\circ}) } \\\\ \vec{K}+\vec{J} &=& \dbinom{350}{0} + \dbinom{ 300\cdot \cos(42^{\circ}) } { 300\cdot \sin(42^{\circ}) } \\ \vec{K}+\vec{J} &=& \dbinom{350+300\cdot \cos(42^{\circ})}{300\cdot \sin(42^{\circ})} \\ \vec{K}+\vec{J} &=& \dbinom{572.943447643}{200.739181908} \\\\ m &=& \sqrt{ 572.943447643^2+200.739181908^2} \\ \mathbf{m} & \mathbf{=} & \mathbf{607.091766828\ N } \\\\ \tan(D) &=& \frac{200.739181908}{572.943447643} \\ \tan(D) &=& 0.35036473972 \\ \mathbf{D} & \mathbf{=} & \mathbf{19.3086615149^{\circ}} \\ \hline \end{array}$

The sum of a number and its cube, is 68.

What is the number?

$\begin{array}{|rcll|} \hline n+n^3 &=& 68 \\ \underbrace{n}_{=4}\cdot \underbrace{(1+n^2)}_{=17} &=& 4\cdot 17 \\\\ 1+n^2&=&17\\ n^2 &=& 16 \\ n &=& 4 \checkmark \\ \hline \end{array}$

The number is 4

I have 7 black and 3 white marbles. I will randomly pick 2 of them.

What is the probability that both of them are white?

How to solve this?

$\begin{array}{|rcll|} \hline && \frac{3}{10} \cdot \frac{2}{9} \\ &=& \frac{1}{15} \qquad (6.\bar{6}\ \%) \\ \hline \end{array}$

What is the mean of 80, 92, 77, 83, 72, 91, 84?

$\begin{array}{|rcll|} \hline && 80 + \frac{0+12-3+3-8+11+4}{7} \\ &=& 80 + \frac{12-8+11+4}{7} \\ &=& 80 + \frac{27-8}{7} \\ &=& 80 + \frac{19}{7} \\ &=& 80 + 2\frac{5}{7} \\ &=& 82\frac{5}{7} \\ &=& \frac{579}{7} \\ &=& 82.7142857143 \\ \hline \end{array}$