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avatar+142 

Find the general solution of the equation:

 

Sin Ø  - Sin 2Ø = Sin 4Ø - Sin 3Ø

 

Answer :2np,(2n - 1)p/2,(2n + 1)p/5   (p=pye)

 Dec 8, 2018
 #3
avatar+28065 
+5

Here's my solution:

 

 Dec 8, 2018
 #4
avatar+142 
+1

Hi Alan...great job thanks!

Prior to posting question and after long hours of problem solving I did obtain 36 deg and 108 degrees. My main issue was applying the standard formulas for the general angle...in this case cosine. Therefore I am now trying to see how I can reconcile those answers....hopefully I can solve !

 

All the best for Xmas of course and New Year....thanks again! 

 Dec 9, 2018
edited by OldTimer  Dec 9, 2018
 #5
avatar+28065 
+1

36° is pi/5 so put n = 1 in (2n-1)pi/5

 

108° is 3pi/5 so put n = 2 in (2n-1)pi/5

 Dec 9, 2018
 #6
avatar+142 
0

Hi Alan thanks for your follow up......If I may bother you for a bit more clarification

 

>>The formulas i am following are as below:

 

Therefore, I have understood solutions obtained as far as angles go...

but

the logic of how you derived  extensions (as highlighted below in your answer) escapes me.

Thanks for your patience!

 

   

 Dec 10, 2018
edited by OldTimer  Dec 10, 2018
 #7
avatar+28065 
+2

Perhaps these graphs will help (note, I've rearranged your equality to create a function f that is zero when the equality is true):

 

Alan  Dec 10, 2018
 #8
avatar+22550 
+9

General Solution - Trigonometry

Sin Ø  - Sin 2Ø = Sin 4Ø - Sin 3Ø

 

\(\begin{array}{ rcl } \sin(\phi) - \sin(2\phi) &=& \sin(4\phi) - \sin(3\phi) \\ &\text{or} & \\ \sin(3\phi)+\sin(\phi) &=& \sin(4\phi) + \sin(2\phi) \\ \end{array} \)

 

Formula:

\(\begin{array}{|rcll|} \hline \underbrace{\sin(3\phi)+\sin(\phi)}_{=2\sin(2\phi)\cos(\phi)} &=& \underbrace{\sin(4\phi) + \sin(2\phi)}_{=2\sin(3\phi)\cos(\phi)} \\\\ 2\sin(2\phi)\cos(\phi) &=& 2\sin(3\phi)\cos(\phi) \\\\ \sin(2\phi)\cos(\phi) &=& \sin(3\phi)\cos(\phi) \\\\ \sin(3\phi)\cos(\phi) -\sin(2\phi)\cos(\phi) &=& 0 \\\\ \cos(\phi)\Big(\sin(3\phi)-\sin(2\phi) \Big) &=& 0 \\ \hline \end{array} \)

 

Formula:

\(\begin{array}{|rcll|} \hline \cos(\phi)\Big(\underbrace{\sin(3\phi)-\sin(2\phi)}_{=2\cos(\frac52\phi)\sin(\frac12\phi)} \Big) &=& 0 \\\\ \cos(\phi)2\cos(\frac52\phi)\sin(\frac12\phi) &=& 0 \quad | \quad :2 \\\\ \large{\mathbf{\cos(\phi)\cos(\frac52\phi)\sin(\frac12\phi)} }& \large{\mathbf{=}} & \large{\mathbf{0}} \\ \hline \end{array}\)

 

General Solution:

\(\begin{array}{|lrcll|} \hline 1. & \cos(\phi) &=& 0 \\ & \phi &=& 2n\pi\pm\arccos(0) \\ & \phi &=& 2n\pi\pm \frac{\pi}{2} \quad \Rightarrow \quad \mathbf{\phi = (2n-1)\frac{\pi}{2}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline 2. & \cos(\frac52\phi) &=& 0 \\ & \frac52\phi &=& 2n\pi\pm\arccos(0) \\ & \frac52\phi &=& 2n\pi\pm \frac{\pi}{2}\quad | \quad \cdot \frac25 \\ & \phi &=& \frac45 n\pi\pm \frac{\pi}{5} \quad \Rightarrow \quad \mathbf{\phi = (2n-1)\frac{\pi}{5}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline 3. & \sin(\frac12\phi) &=& 0 \\ & \frac12\phi &=& n\pi+(-1)^n \arcsin(0) \\ & \frac12\phi &=& n\pi+(-1)^n \pi \quad | \quad \cdot 2 \\ & \phi &=& 2n\pi+(-1)^n 2\pi \quad \Rightarrow \quad \mathbf{\phi = 2n\pi} \\ \hline \end{array} \)

 

laugh

 Dec 12, 2018
edited by heureka  Dec 12, 2018
edited by heureka  Dec 12, 2018
 #9
avatar+142 
+1

Wow...food for thought! Thanks.

 Dec 12, 2018
 #10
avatar+142 
0

Dear all thanks for all your help! What a great team you are! 

 Dec 14, 2018

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