4
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ +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
1
$\begingroup$

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 $\displaystyle\cos\alpha,\sin\alpha,$

$\displaystyle\dfrac{\cos\alpha}m=\cos^3\theta\cos3\theta-\sin^3\theta\sin3\theta$

$\displaystyle=\cos^3\theta(4\cos^3\theta-3\cos\theta)-\sin^3\theta(3\sin\theta-4\sin^3\theta)$

$\displaystyle=4(\cos^6\theta+\sin^6\theta)-3(\cos^4\theta+\sin^4\theta)$

$\displaystyle=4\{(\cos^2\theta+\sin^2\theta)^3-3(\cos^2\theta\sin^2\theta)(\cos^2\theta+\sin^2\theta)\}-3\{(\cos^2\theta+\sin^2\theta)^2-2\cos^2\theta\sin^2\theta\}$

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

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

$\endgroup$
  • $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.