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

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!

Solution 1:

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

Solution 2:

We have, $$\frac{\cos(\alpha-3\theta)}{\cos^3\theta}=\frac{\sin(\alpha-3\theta)}{\sin^3\theta}=m$$

Then, $$m=\frac{\sin\theta\cos(\alpha-3\theta)+\cos\theta\sin(\alpha-3\theta)}{\sin\theta\cos^3\theta+\cos\theta\sin^3\theta}=\frac{\sin(\alpha-2\theta)}{\sin\theta\cos\theta}$$

Expanding $\sin(\alpha-2\theta)$ and using $2\sin\theta\cos\theta=\sin2\theta$, we get, $$\frac m2=\sin\alpha\cot2\theta-\cos\alpha\tag1$$

Similarly, $$m=\frac{\cos(\alpha-2\theta)}{\cos^4\theta-\sin^4\theta}=\frac{\cos(\alpha-2\theta)}{\cos2\theta}$$

which gives, $$m=\cos\alpha+\sin\alpha\tan2\theta\tag2$$

From $(1)$ and $(2)$,

$$\left(\frac m2+\cos\alpha\right)\left(m-\cos\alpha\right)=\sin^2\alpha$$ $$\frac{m^2}2+\frac m2\cos\alpha-\cos^2\alpha=\sin^2\alpha$$ $$m^2+m\cos\alpha-2=0.$$

$\square$