Eliminate $\theta$ from $$\sin3\theta=a\cos\theta$$ $$\cos3\theta=b\sin\theta$$
I came to this point while trying to solve this problem: Eliminating $\theta$ from trigonometric system (Remark: Symbols differ from the original question)
We can find $$b+a=\frac{2\cos2\theta}{\sin2\theta}$$ $$b-a=\frac{2\cos4\theta}{\sin2\theta}$$
Though at the moment I can't imagine how to arrange these to get a proper relation between $a$ and $b$. We can replace $\sin$ and $\cos$ with $\tan$ and then proceed, but it is cumbersome.
I suspect that there should be a clever way to end this, giving a beautiful answer. For reference, like in this question, a specific method gives the desired answer.$\leftarrow\small \text{(not strictly relevant)}$
So, what is the happy ending of this problem?
Best Answer
My first idea after seeing the problem is to say that $$ a^2\cos^2\theta+b^2\sin^2\theta=1 $$ From this we get $$ a^2(1+\cos2\theta)+b^2(1-\cos2\theta)=2 $$ whence $$ \cos2\theta=\frac{2-a^2-b^2}{a^2-b^2} $$ But we also have $$ \frac{\sin^23\theta}{a^2}+\frac{\cos^23\theta}{b^2}=1 $$ whence $$ b^2\sin^23\theta+a^2\cos^23\theta=a^2b^2 $$ and, similarly to the above, $$ \cos6\theta=\frac{2a^2b^2-a^2-b^2}{a^2-b^2} $$ Now use the identity for $\cos3x$ in terms of $\cos x$.