# Trigonometry problem - eliminating $\theta$

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!

• +1 for this challenging question. My idea was to convert eveything in $tan$ format, and equate to m. Still solving though.. – MonK Jul 3 '14 at 9:48
• 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? – draks ... Jul 3 '14 at 11:53
• i haven't been able to eliminate $\theta$, but I did eliminate $\alpha$ and found that $m=\csc{3\theta}+\sec{3\theta}$ – 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 $\displaystyle\cos\alpha,\sin\alpha,$

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

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

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

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

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

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

• @PranavArora, Please have a look – lab bhattacharjee Jul 4 '14 at 17:03
• 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. :( – Pranav Arora Jul 4 '14 at 17:51
• @PranavArora, I think proper method should dictate answer, not the reverse:). – lab bhattacharjee Jul 4 '14 at 18:24
• I agree, thank you for the help. :) – Pranav Arora Jul 4 '14 at 18:32