+0

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

0
966
8

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

Guest Jun 12, 2015

#4
+27057
+16

Here's a numerical approach:

.

Alan  Jun 12, 2015
#1
+90027
+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
+93683
+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
0

https://www.desmos.com/calculator

Guest Jun 12, 2015
#4
+27057
+16

Here's a numerical approach:

.

Alan  Jun 12, 2015
#5
+20025
+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
+93683
+5

Thanks Alan        and Heureka,

This was a hard one

Melody  Jun 12, 2015
#7
+20025
+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 :

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

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

heureka  Jun 15, 2015
#8
+93683
0

Thank you Heureka.

Another Heureka special.

Melody  Jun 15, 2015