Here is a trigonometry problem.

Given $$\frac{\cos(\alpha-3\theta)}{\cos^3(\theta)}=\frac{\sin(\alpha-3\theta)}{\sin^3(\theta)} = m$$ Show that $$m^2+m\cos(\theta) = 2.$$

I tried to convert $\sin^3(x)$ into $\sin(3x)$ and similarly to cosine term, but couldn't get the answer. Please tell how to proceed further and any other way to solve it.

  • $\begingroup$ You should typeset that picture, as the picture is likely to disappear over time. $\endgroup$ – zhw. Aug 25 '17 at 3:02
  • $\begingroup$ I am a beginner and don't know how to do that. It would be good if u will help me. Thanks $\endgroup$ – Kamal Aug 25 '17 at 3:05
  • $\begingroup$ I believe I did so @Kamal, let me know if this is the problem you are asking. For future questions, look at math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Isaac Browne Aug 25 '17 at 3:08
  • $\begingroup$ Yes this is the one i am looking for. I will not repeat it in future. Thanks for the help $\endgroup$ – Kamal Aug 25 '17 at 3:10

I think it's wrong.

Try $\theta=45^{\circ}$ and $\alpha=0^{\circ}$.

We obtain $m=-2$ and $m^2+m\cos\theta=2$ is wrong.

By the way, the first condition gives $$\tan\alpha=\frac{3\sin4\theta}{1+3\cos4\theta}.$$

  • $\begingroup$ Maybe it holds good for a specific value. $\endgroup$ – Kamal Aug 25 '17 at 4:10
  • $\begingroup$ @Kamal I think it's another problem already. By the way, I solved your problem. $\endgroup$ – Michael Rozenberg Aug 25 '17 at 4:11
  • $\begingroup$ can you please provide a generalized way without value putting. $\endgroup$ – Kamal Aug 25 '17 at 4:17
  • $\begingroup$ Sorry there is a change in question instead of theta it is alpha $\endgroup$ – Kamal Aug 25 '17 at 4:21
  • $\begingroup$ @Kamal It's another problem already. I solved your first problem It's not fair! Open please another topic. $\endgroup$ – Michael Rozenberg Aug 25 '17 at 4:25

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