All Answers 358129 Answers

Thanks Alan        and Heureka,   

This was a hard one  

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:

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

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

$$tan(-x) cos(-x)=-tanxcosx=\frac{-sinx}{cosx}*cosx = -sinx$$

100% of 20 = 20

10% of 20 = 2

40% of 20 = 4*10% of 20 = 4*2 = 8

2.5% of 20 = 1/4 * 10% of 20 = 1/4 * 2 = 0.5

Add up the numbers in bold to get 142.5% of 20 = 20+8+0.5 = 28.5

.

I don't know that mathematics has a father; but it has a prince!  

Here's a numerical approach:

.

Very nice, Melody......

@@ End of Day Wrap  Fri 12/6/15   Sydney, Australia Time   12:05am  (Yes it is Sat)   ♪ ♫

Good evening, (It is for me anyway  )

Today our brilliant mathematicians were Radix, Bertie, CPhill, ThisGuy Alan and worstsubjectmath.

(There you go worstsubjectmath, I bet you never thought you would ever be referred to a a 'brilliant mathematician'  )

Sat 13/6/15

If you would like to comment on other site issues please do so on the Lantern Thread.  Thank you.    

Interest Posts: 

FTJ means 'For the juniors' 

Yep......I used fiora's  approach  [more or less ].......but.....I like the way heureka did it, too.....

Did you already know the answer and how to get there ?

https://www.desmos.com/calculator

LOL!!!!!......nah.......it's just something I thought up that might make a nice problem.......

Was it a test Chris ?

Thanks Heureka  

This has been included in our reference material thread :)

Oh yea I forgot to do that bit, thanks Chris.    

Thanks Chris,

I was playing with it too but I didn't really get anywhere.

Here is the Wolfram|Alpha solution 

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?

Hz is wavelengths per sec

so I use the units in a rather unusual way. I am sure others would use a formua.

we have

 $$\\frequency =\frac{5.09*10^{14}\;\;\lambda}{second}\qquad speed\;of\;light=\frac{299792458\;m}{sec}\qquad we\;want\;\; \frac{m}{\lambda}\\\\
\frac{299792458\;m}{sec}\times \frac{sec}{5.09*10^{14}\;\;\lambda}\qquad $The seconds cancel out$\\\\
=\frac{299792458\;m}{5.09*10^{14}\;\;\lambda}$$

$${\frac{{\mathtt{299\,792\,458}}}{\left({\mathtt{5.09}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{14}}}\right)}} = {\mathtt{0.000\: \!000\: \!588\: \!983\: \!218\: \!1}}$$

So the wavelength is    $$\approx 589*10^{-9}\;\;metres = 589\;nanometres$$    

so if 11+7=18 and 1/2+1/2=1 18+1=19

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?

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

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

92.7254901961

$$4$$  is rational and  $$\sqrt{11}$$   is irrational.