Eliminate $\theta$ from the equations $$\frac{\cos(\alpha-3\theta)}{\cos^3\theta}=\frac{\sin(\alpha-3\theta)}{\sin^3\theta}=m$$

Ans: $m^2+m\cos\alpha-2=0$.

I tried using the following two identities: $$\cos(3\theta)=4\cos^3(\theta)-3\cos(\theta)$$ $$\sin(3\theta)=3\sin(\theta)-4\sin^3(\theta)$$ but these didn't help much. I am sure that this is a simple problem but I am unable to figure out the right approach to solve it. :(

Any help is appreciated. Thanks!

  • $\begingroup$ +1 for this challenging question. My idea was to convert eveything in $tan$ format, and equate to m. Still solving though.. $\endgroup$ – MonK Jul 3 '14 at 9:48
  • $\begingroup$ Shouldn't then be $\frac{\cos(\alpha-3\theta)}{\cos^3\theta}-\frac{\sin(\alpha-3\theta)}{\sin^3 \theta}=0$? W|A doesn't support that in general, or what did I get wrong? $\endgroup$ – draks ... Jul 3 '14 at 11:53
  • $\begingroup$ i haven't been able to eliminate $\theta$, but I did eliminate $\alpha$ and found that $m=\csc{3\theta}+\sec{3\theta}$ $\endgroup$ – John Joy Jul 3 '14 at 13:31

Using $\displaystyle\sin(A-B),\cos(A-B)$ and on rearrangement we have $$\cos\alpha\cos3\theta+\sin\alpha\sin3\theta-m\cos^3\theta=0\ \ \ \ (1)$$

$$\cos\alpha\sin3\theta-\sin\alpha\cos3\theta+m\sin^3\theta=0\ \ \ \ (2)$$

Solving for $\displaystyle\cos\alpha,\sin\alpha,$





$\displaystyle\implies\dfrac{\cos\alpha}m=1-6(\sin\theta\cos\theta)^2\ \ \ \ (3)$

Similarly, $\displaystyle\dfrac{\sin\alpha}m=-3(\sin\theta\cos\theta)\ \ \ \ (4)$

Compare the values of $(\sin\theta\cos\theta)^2$ from $(3),(4)$ to eliminate $\theta$ and simplify

  • $\begingroup$ @PranavArora, Please have a look $\endgroup$ – lab bhattacharjee Jul 4 '14 at 17:03
  • $\begingroup$ Thank you @lab bhattacharjee. This looks nice but I am not able to reach the answer from $(3)$ and $(4)$. I squared $(4)$ and then substituted it in $(3)$ but this doesn't simplify to the given answer. :( $\endgroup$ – Pranav Arora Jul 4 '14 at 17:51
  • $\begingroup$ @PranavArora, I think proper method should dictate answer, not the reverse:). $\endgroup$ – lab bhattacharjee Jul 4 '14 at 18:24
  • $\begingroup$ I agree, thank you for the help. :) $\endgroup$ – Pranav Arora Jul 4 '14 at 18:32

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