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

Guest Jun 12, 2015

Best Answer 

 #4
avatar+26322 
+16

Here's a numerical approach:

 

 Numerical solution:

.

Alan  Jun 12, 2015
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8+0 Answers

 #1
avatar+78575 
+10

Here's a graphical solution......https://www.desmos.com/calculator/rhbj1xne7n

 

The solutions occur at about   29.5°, 163.1° and 347.4°  on [0, 360] degrees

 

 

CPhill  Jun 12, 2015
 #2
avatar+90988 
+10

Thanks Chris,

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

Here is the Wolfram|Alpha solution 

http://www.wolframalpha.com/input/?i=2*sin%28x%29%3Dcos%28x%2F3%29

Melody  Jun 12, 2015
 #3
avatar
0

https://www.desmos.com/calculator

Guest Jun 12, 2015
 #4
avatar+26322 
+16
Best Answer

Here's a numerical approach:

 

 Numerical solution:

.

Alan  Jun 12, 2015
 #5
avatar+18712 
+15

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:

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

 

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

 

heureka  Jun 12, 2015
 #6
avatar+90988 
+5

Thanks Alan        and Heureka,   

 

This was a hard one  

Melody  Jun 12, 2015
 #7
avatar+18712 
+5

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

 

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

 

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

 

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

 

Compare with wolframalpha.com http://www.wolframalpha.com/input/?i=2*sin%28x%29%3Dcos%28x%2F3%29 :

$$\small{\text{$
\mathrm{we~ have~ seen:~~}
\boxed{~~
\cot^3{ \left( \frac{\theta}{3} \right) }
-6\cot^2{ \left( \frac{\theta}{3} \right) }
+ \cot{ \left( \frac{\theta}{3} \right) }
+2 = 0 ~~}
$}}\\\\\\$$

$$\small{\text{$
\begin{array}{lrcl}
\mathrm{we~set~}&\cot{\left(\frac{\theta}{3}\right)} &=&
\dfrac
{1-\tan^2{ (\frac{\theta}{6}) } }
{ 2\tan{ (\frac{\theta}{6}) }}\\\\
\mathrm{and~use~} x= \tan{ (\frac{\theta}{6}) }\\\\
\mathrm{then~}&\cot{\left(\frac{\theta}{3}\right)} &=&
\dfrac{1-x^2}{2x} \\\\
\mathrm{we~substitute}&
\left(\dfrac{1-x^2}{2x}\right)^3
-6\left(\dfrac{1-x^2}{2x}\right)^2
+\left(\dfrac{1-x^2}{2x}\right)
+2 &=& 0 \\\\
&
\dfrac{ \left( 1-x^2 \right)^3 }{8x^3}
-6\dfrac{ \left( 1-x^2 \right)^2 }{4x^2}
+ \dfrac{ \left( 1-x^2 \right) }{2x}
+ 2 &=& 0 \qquad |\qquad \cdot 8x^3\\\\
&
\left( 1-x^2 \right)^3
- 12x \left( 1-x^2 \right)^2
+ 4x^2 \left( 1-x^2 \right)
+ 16x^3 &=& 0\\\\
&&\cdots\\\\
&
1-3x^2+3x^4-x^6-12x+24x^3-12x^5+4x^2-4x^4+16x^3 &=& 0\\\\
\mathrm{finally}& \mathbf{x^6+12x^5+x^4-40x^3-x^2+12x-1} & \mathbf{=}& \mathbf{0}\\\\
\mathrm{\textcolor[rgb]{1,0,0}{wolframalpha.com~solution:}}& \mathbf{x^6+12x^5+x^4-40x^3-x^2+12x-1} & \mathbf{=}& \mathbf{0}\\\\
\mathrm{with}& \frac{\theta}{6} = \arctan{(x)}\pm\pi\cdot k\\\\
&
\theta = 6[\arctan{(x)}\pm\pi\cdot k]\\\\
&
\theta = 6\arctan{(x)}\pm 6\pi\cdot k\\\\
\end{array}
$}}$$

 

heureka  Jun 15, 2015
 #8
avatar+90988 
0

Thank you Heureka.   

 

Another Heureka special.            

Melody  Jun 15, 2015

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