heureka

how do you solve 2sin(theta) = cos(theta/3)  ?

$$\small{\text{$
\boxed{
~~ 2\sin{(\theta)} = \cos \left( \frac{ \theta}{3}\right) \quad \mathrm{~we~set~} \theta = 3\alpha \quad
2\sin(3\alpha ) = \cos{( \alpha )}
~~}
$}}$$

$$\small{\text{$
\begin{array}{rcl|rrcl|l}
&&&&&&& \mathrm{Formula:}\\
2\sin(3\alpha ) &=& \cos{( \alpha )}
&(1)&\sin{(3\alpha)}&=&
\sin{(\alpha+2\alpha)}=
\sin{(\alpha)}\cos{(2\alpha)} +\cos{(\alpha)}\sin{(2\alpha)}
&\cos{2\alpha} = 1-2\sin^2{(\alpha)} \\
&&&&\sin{(3\alpha)}&=&
\sin{(\alpha)}(1-2\sin^2{(\alpha)}) +2\sin{(\alpha)}\cos^2{(\alpha)}
&\sin{2\alpha} = 2\sin{(\alpha)}\cos{(\alpha)} \\
&&&&\sin{(3\alpha)}&=&
\sin{(\alpha)}(1-2\sin^2{(\alpha)}) +2\sin{(\alpha)}( 1-\sin^2{(\alpha)} )
&\cos^2{\alpha} = 1-\sin^2{(\alpha)} \\
&&&&& \cdots && \\
&&&&\sin{(3\alpha)}&=&
\sin{(\alpha)} \left[(3-4\sin^2{(\alpha)} \right]
&\\
2\sin{(\alpha)} \left[(3-4\sin^2{(\alpha)} \right] &=& \cos{( \alpha )}
&&&&&\\
2\left[(3-4\sin^2{(\alpha)} \right] &=& \cot{( \alpha )}
&&&&&\\
& \cdots &
&&&&&\\
8\sin^2{(\alpha)} &=& 6-\cot{( \alpha )}
&&&&& \frac{1}{ \sin^2{(\alpha)} } = 1+\cot^2{(\alpha)}\\
8 &=& \left[ (6-\cot{( \alpha )} \right] \left[1+\cot^2{(\alpha)} \right]
&&&&&\\
& \cdots &&&&&&\\
\cot^3{(\alpha)}-6\cot^2{(\alpha)}+\cot{(\alpha)}+2 &=& 0 &&&&&\\
& \alpha=\frac{\theta}{3} &&&&&&\\
\cot^3{(\frac{\theta}{3})}-6\cot^2{(\frac{\theta}{3})}+\cot{(\frac{\theta}{3})}+2 &=& 0 &&&&&\\
\end{array}
$}}$$

$$\small{\text{$
\boxed{~~
\cot^3{\left(\frac{\theta}{3}\right)}
-6\cot^2{\left(\frac{\theta}{3}\right)}+\cot{\left(\frac{\theta}{3}\right)}+2 = 0
~~}
$}}\\\\\\
\small{\text{$
\mathrm{substitute:~~}
\boxed{~~u = \cot{ \left(\frac{\theta}{3} \right) }=\frac{1}{\tan{ \left(\frac{\theta}{3} \right) }}
\qquad \theta = 3\arctan{\left(~\frac{1}{u}~\right)} \pm 3\pi \cdot k \quad k=0,1,2,3\cdots ~~ }
$}}\\ \\
\small{\text{$
\boxed{~~u^3 - 6u^2 + u +2 = 0 ~~}
$}}\\\\
\small{\text{$
\begin{array}{rclcc}
u_1 &=& 5.766 435 484
& \theta_1=3\arctan{(\frac{1}{u_1})}
& \theta_1=0.515 128 919 \pm3\pi\cdot k\\\\
u_2 &=& -0.483 611 621
& \theta_2=3\arctan{(\frac{1}{u_2})}
& \theta_2=-3.361 035 503 \pm3\pi\cdot k\\\\
u_3 &=& 0.717 176 136
& \theta_3=3\arctan{(\frac{1}{u_3})}
& \theta_3=2.845 906 583 \pm3\pi\cdot k\\\\
\end{array}
$}}$$

how do you solve 2sin(theta) = cos(theta/3) ?

$$\boxed{~~ 2\sin(\theta) = \cos \left( \frac{ \theta}{3} \right) ~~}$$

 $$\small{\text{$\mathmf{Formula:~~}
\boxed{\cos (3x) = 4 \cos^3 (x) - 3 \cos (x) \qquad 3x=\theta \qquad \cos{( \theta )}=4\cos^3{( \frac{\theta }{3} )}-3\cos{ ( \frac{\theta }{3}) } }
$}}\\\\
\small{\text{$
\begin{array}{rcl}
2\sin(\theta) &=& \cos \left( \frac{ \theta}{3} \right)\\\\
2\sqrt{ 1-\cos^2{\theta } } &=& \cos \left( \frac{ \theta}{3} \right)\\\\
4(\sqrt{ 1-\cos^2{\theta } })^2 &=& \cos^2 \left( \frac{ \theta}{3} \right)\\\\
4( 1-\cos^2{\theta } ) &=& \cos^2 \left( \frac{ \theta}{3} \right)\\\\
4(~ 1-
[4\cos^3{( \frac{\theta }{3} )} - 3\cos{ ( \frac{\theta }{3}) } ]^2~) &=& \cos^2 \left( \frac{ \theta}{3} \right)\\\\
&\cdots &\\
64\cos^6{ (\frac{ \theta}{3}) } - 96 \cos^4{ (\frac{ \theta}{3}) } + 37\cos^2{ (\frac{ \theta}{3}) } -4 &=& 0 \\\\
\end{array}
$}}$$

$$\small{\text{$
\mathrm{substitute:~~} \boxed{~~u = \cos^2{ \frac{\theta}{3} } \qquad \theta_{1\dots 4} = \pm~3\arccos(~\pm\sqrt{u}~) \pm 6k\pi\quad k=0,1,2,3\cdots ~~ }
$}}$$

$$\small{\text{$
\boxed{~~64u^3 - 96u^2 + 37u -4 = 0 ~~}
$}}\\\\
\small{\text{$
\begin{array}{rcl}
u_1 &=& 0.970804435482 \\
u_2 &=& 0.189548547332 \\
u_3 &=& 0.339647017333
\end{array}
$}}$$

Solutions:

$$\\ \small{\text{
$
\begin{array}{lrcl}
& u_1 &=& 0.970804435482 \\
\mathbf{okay} & \mathbf{\theta} &\mathbf{=}& \mathbf{0.515128919784\pm 6k\pi} \\
false &\theta &=& 8.909649040986 \\
false &\theta &=& -0.515128919784 \\
\mathbf{okay} & \mathbf{\theta} &\mathbf{=}& \mathbf{-8.909649040986\pm 6k\pi} \\
\\
& u_2 &=& 0.189548547332 \\
false & \theta &=& 3.361035503365 \\
\mathbf{okay} & \mathbf{\theta} &=& \mathbf{6.063742457405\pm 6k\pi} \\
\mathbf{okay} & \mathbf{\theta} &=& \mathbf{-3.361035503365\pm 6k\pi} \\
false & \theta &=& -6.063742457405 \\
\\
& u_3 &=& 0.339647017333 \\
\mathbf{okay} & \mathbf{\theta} &\mathbf{=}& \mathbf{2.845906582419\pm 6k\pi} \\
false & \theta &=& 6.578871378351 \\
false & \theta &=& -2.845906582419 \\
\mathbf{okay} & \mathbf{\theta} &\mathbf{=}& \mathbf{-6.578871378351\pm 6k\pi}
\end{array}
$}}$$

$$\mathbf{\theta ~in~ rad}$$

A beam of light from a sodium street lamp is found to have a frequency of 5.09 x 1014 Hz. What is the wavelength?

$$\boxed{\; c = f\cdot \lambda \qquad \text{ or }\qquad \lambda=\dfrac{c}{f} \qquad \text{ or }\qquad f=\dfrac{c}{\lambda} \qquad \begin{array}{rcl} c &=& \small{\text{ speed of light in vacuum }} \\\lambda &=& \small{\text{ wavelength }} \\ f &=& \small{\text{ wave's frequency }}\end{Array}\; }$$

$$\small{\text{$
\begin{array}{rclcc} \lambda &=& \dfrac{c}{f} \quad & \quad c = 299\,792\,458 ~ \mathrm{\dfrac{m}{s}} \quad & \quad
f = 5.09 \cdot 10^{14} ~ \mathrm{ Hz } \\\\
\lambda &=& \dfrac{ 299\,792\,458 ~ \mathrm{\dfrac{m}{s}} }{ 5.09\cdot 10^{14} ~ \mathrm{ Hz } } \\\\
\lambda &=& \dfrac{ 2.99\,792\,458\cdot 10^{8} ~ \mathrm{\dfrac{m}{s}} }{5.09\cdot 10^{14} ~ \mathrm{\dfrac{1}{s}} } \\\\
\lambda &=& \dfrac{ 2.99\,792\,458 }{5.09}\cdot 10^{8}\cdot 10^{-14} ~ \mathrm{ m } \\\\
\lambda &=& 0.58898321807\cdot 10^{-6} ~ \mathrm{ m } \\\\
\lambda &=& 588.98321807\cdot 10^{-3} \cdot 10^{-6} ~ \mathrm{m} \\\\
\lambda &=& 588.98321807\cdot 10^{-9} ~ \mathrm{m} \\\\
\lambda &=& 588.98321807 ~ \mathrm{nm}
\end{array}$}}$$

wie leite ich wurzeln ab ?

$$\boxed{ ~~ \sqrt[n]{x} &~=~& x^{ \frac{1}{n} } ~~}\\\\
\begin{array}{rcl}
(\sqrt[n]{x})' &=& (x^{ \frac{1}{n} })' \\ \\
(\sqrt[n]{x})' &=& \frac{1}{n} \cdot x^{ \frac{1}{n} -1 }\\\\
(\sqrt[n]{x})' &=& \frac{1}{n} \cdot x^{ \frac{1}{n} }\cdot x^{-1}\\\\
(\sqrt[n]{x})' &=& \frac{1}{n\cdot x} \cdot x^{ \frac{1}{n} }\\\\
(\sqrt[n]{x})' &=& \frac{ \sqrt[n]{x} }{n\cdot x}
\end{array} \\\\\\
\boxed{ ~~(\sqrt[n]{x})' ~=~ \dfrac{ \sqrt[n]{x} }{n\cdot x} ~=~ \dfrac{1}{ n\cdot \sqrt[n]{x^{n-1} } } ~~ }$$

Beispiele:

$$\\(\sqrt{x})' ~=~ \dfrac{ \sqrt{x} }{2\cdot x}
~=~ \dfrac{1}{ 2\cdot \sqrt{x^{2-1} } }
~=~ \dfrac{1}{ 2\cdot \sqrt{ x } } \\\\\\
(\sqrt[3]{x})' ~=~ \dfrac{ \sqrt[3]{x} }{3\cdot x}
~=~ \dfrac{1}{ 3\cdot \sqrt[3]{x^{3-1} } }
~=~ \dfrac{1}{ 3\cdot \sqrt[3]{x^2 } } \\\\\\
(\sqrt[4]{x})' ~=~ \dfrac{ \sqrt[4]{x} }{4\cdot x}
~=~ \dfrac{1}{ 4\cdot \sqrt[4]{x^{4-1} } }
~=~ \dfrac{1}{ 4\cdot \sqrt[4]{x^3 } } \\\\
\cdots$$

2 to the power of x is 2048, what is x

$$\small{\text{$
\begin{array} {rcl}
2^x &=& 2048 \qquad | \qquad \ln()\\\\
\ln{(2^x)} &=& \ln{(2048)} \\\\
x\cdot \ln{(2)} &=& \ln{(2048)} \\\\
x &=& \dfrac{ \ln{(2048)} } { \ln{(2)} } \\\\
x &=& \dfrac{ 7.62461898616 } { 0.69314718056} \\\\
\mathbf{x} &\mathbf{=}& \mathbf{11} \\\\
\end{array}
$}}$$

show that 7^p-6^p-1 for p > 3 prime is multiple of 43

$$\boxed{~~\textcolor[rgb]{0,150,0}{7} ^p - \textcolor[rgb]{0,0,150}{6}^p -1 \equiv 0 \mod 43 ~~, \quad \mathrm{if~~} p \mathrm{~~is~prime~and~~} p > 3.
} \\\\
\begin{array}{lrclcl}
\mathrm{Because~~ } &\textcolor[rgb]{0,150,0}{7}^6 &=& 117649 &\equiv& \textcolor[rgb]{1,0,0}{1} \mod 43\\
\mathrm{~~and~~ }\\
&\textcolor[rgb]{0,0,150}{6}^6 &=& 46656 &\equiv& \textcolor[rgb]{1,0,0}{1} \mod 43\\
\mathrm{~~and~~ }\\
&43 &=& \textcolor[rgb]{0,0,150}{6}\cdot \textcolor[rgb]{0,150,0}{7} + 1\\
\mathrm{~~and~~ }\\
&\textcolor[rgb]{0,0,150}{6} &=& \textcolor[rgb]{0,150,0}{7} - 1
\end{array}$$

If it can be this case generally is valid ?

$$\boxed{~~\textcolor[rgb]{0,150,0}{a} ^p - \textcolor[rgb]{0,0,150}{(a-1)}^p -1 \equiv 0 \mod [\textcolor[rgb]{0,150,0}{a} \textcolor[rgb]{0,0,150}{(a-1)} +1] ~~, \quad \mathrm{if~~} p \mathrm{~~is~prime~and~~} p > 3.
} \\\\
\begin{array}{lrclcl}
\mathrm{Because~~ } &\textcolor[rgb]{0,150,0}{a}^6 &&&\equiv& \textcolor[rgb]{1,0,0}{1} \mod [\textcolor[rgb]{0,150,0}{a} \textcolor[rgb]{0,0,150}{(a-1)} +1]\\
\mathrm{~~and~~ }\\
&\textcolor[rgb]{0,0,150}{(a-1)}^6 &&&\equiv& \textcolor[rgb]{1,0,0}{1} \mod [\textcolor[rgb]{0,150,0}{a} \textcolor[rgb]{0,0,150}{(a-1)} +1]\\
\end{array}$$

Examples:

$$\begin{array}{|c|c|l|l|c|}
\hline
\textcolor[rgb]{0,150,0}{a}
& \textcolor[rgb]{0,0,150}{a-1}
&\mathrm{~~multiple ~of~~ }&&\\
\hline
2 & 1 & 2*1+1 = 3 & 2^6 = 64 \equiv 1 \mod 3 & 2^p - 1^p - 1 \equiv 0 \mod 3 \\
& & & 1^6 = 1 \equiv 1 \mod 3 &\\
\hline
3 & 2 & 3*2+1 = 7 & 3^6 = 729 \equiv 1 \mod 7 & 3^p - 2^p - 1 \equiv 0 \mod 7 \\
& & & 2^6 = 64 \equiv 1 \mod 7 &\\
\hline
4 & 3 & 4*3+1 = 13 & 4^6 = 4096 \equiv 1 \mod 13 & 4^p - 3^p - 1 \equiv 0 \mod 13 \\
& & & 3^6 = 729 \equiv 1 \mod 13& \\
\hline
5 & 4 & 5*4+1 = 21 & 5^6 = 15625 \equiv 1 \mod 21 & 5^p - 4^p - 1 \equiv 0 \mod 21 \\
& & & 4^6 = 4096 \equiv 1 \mod 21& \\
\hline
6 & 5 & 6*5+1 = 31 & 6^6 = 46656 \equiv 1 \mod 31 & 6^p - 5^p - 1 \equiv 0 \mod 31 \\
& & & 5^6 = 15625 \equiv 1 \mod 31& \\
\hline
7 & 6 & 7*6+1 = 43 & 7^6 = 117649 \equiv 1 \mod 43 & \textcolor[rgb]{1,0,0}{ 7^p - 6^p - 1 \equiv 0 \mod 43 } \\
& & & 6^6 = 46656 \equiv 1 \mod 43 &\\
\hline
8 & 7 & 8*7+1 = 57 & 8^6 = 262144 \equiv 1 \mod 57 & 8^p - 7^p - 1 \equiv 0 \mod 57 \\
& & & 7^6 = 117649 \equiv 1 \mod 57 &\\
\hline
\cdots & \cdots & & & \cdots\\
\hline
\end{array}$$

What is the area of the largest Kepler Triangle that can be inscribed in the circle whose equation is x^2 + y^2  = 9   ????

{A Kepler Triangle is a right triangle whose sides are in the ratio of 1 : √Phi : Phi....where Phi = [1 + √5] / 2 }

$$\boxed{a:b:c = 1:\sqrt{\varphi}:\varphi}$$

$$r_{circle} = \sqrt{9} = 3\qquad c = 2\cdot r_{circle} \\\\
\small{\text{$
\begin{array}{rcl}
\sqrt{ \varphi } & \cdot & \dfrac{ \varphi }{ \sqrt{ \varphi } } = \varphi \\\\
b & \cdot & \dfrac { \varphi } { \sqrt{ \varphi } } = c \\\\
\textcolor[rgb]{1,0,0}{b} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{c \cdot \dfrac{ \sqrt{ \varphi } } { \varphi }}\\\\
\hline
&\\
\end{array}
$}}\\\\\\
\small{\text{$
\begin{array}{rcl}
1 & \cdot & \dfrac{ \varphi }{ 1 } = \varphi \\\\
a & \cdot & \dfrac{ \varphi }{ 1 } = c \\\\
\textcolor[rgb]{1,0,0}{a} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{\dfrac{ c } { \varphi } }\\\\
\hline
&\\
\end{array}
$}}\\\\$$

$$\small{\text{$
\begin{array}{rcl}
A &=& \dfrac{a\cdot b}{2} \\\\
A &=& \dfrac{\dfrac{ c } { \varphi }\cdot c \cdot \dfrac{ \sqrt{ \varphi } } { \varphi } }{2} \\\\
A&=& \dfrac{c^2}{\varphi^2} \cdot \dfrac{ \sqrt{ \varphi } } { 2}\\\\
A &=& 2\cdot r^2 \cdot \dfrac{ \sqrt{ \varphi } } { \varphi^2 } \\\\
A &=& 18 \cdot \dfrac{ \sqrt{ \varphi } } { \varphi^2 } \\\\
A &=& 8.74562889162 \\\\
\end{array}
$}}$$

Eric is standing 6 ft away from the base of the Empire State Building, which is 1453 ft tall. What is the angle of elevation when he looks at the top of the building? Round to the nearest hundredth.

$$\tan{ ( \mathrm{elevation\ensurement{^{\circ}} } ) } = \dfrac{ 1453 ~ft. }{ 6 ~ft.} = 242.1\overline{6} \\\\
\mathrm{elevation\ensurement{^{\circ}}} = \arctan{ ( 242.1\overline{6} ) }\\\\
\mathrm{elevation\ensurement{^{\circ}}} = 89.76 \ensurement{^{\circ}}$$

Factor 8x(3x+1)-(3x+1)

$$\mathbf{8x(3x+1)-(3x+1) = (3x+1)(8x-1)}$$