+239 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)
+239 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!
+33614 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
+239 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!
+26367 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:
\(\begin{array}{|lrcll|} \hline 1. & \cos(\phi) &=& 0 \\ & \phi &=& 2n\pi\pm\arccos(0) \\ & \phi &=& 2n\pi\pm \frac{\pi}{2} \quad \Rightarrow \quad \mathbf{\phi = (2n-1)\frac{\pi}{2}} \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline 2. & \cos(\frac52\phi) &=& 0 \\ & \frac52\phi &=& 2n\pi\pm\arccos(0) \\ & \frac52\phi &=& 2n\pi\pm \frac{\pi}{2}\quad | \quad \cdot \frac25 \\ & \phi &=& \frac45 n\pi\pm \frac{\pi}{5} \quad \Rightarrow \quad \mathbf{\phi = (2n-1)\frac{\pi}{5}} \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline 3. & \sin(\frac12\phi) &=& 0 \\ & \frac12\phi &=& n\pi+(-1)^n \arcsin(0) \\ & \frac12\phi &=& n\pi+(-1)^n \pi \quad | \quad \cdot 2 \\ & \phi &=& 2n\pi+(-1)^n 2\pi \quad \Rightarrow \quad \mathbf{\phi = 2n\pi} \\ \hline \end{array} \)
