# Eliminate $\theta$ from $\sin3\theta=a\cos\theta$ and $\cos3\theta=b\sin\theta$

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?

• Hint: we have $\tan3\theta=\frac{a}{b}\cot\theta$ which, defining $t:=\tan\theta$, rearranges to a quadratic in $t^2$.
– J.G.
Dec 17, 2021 at 8:18
• Another interesting equality is $\frac{a b -1}{2} = \cos 4 \theta$, and with the first of your equalities produces a relation between $a$, $b$. Dec 18, 2021 at 1:30

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 $$\cos x$$.

I used WolframAlpha to get the following relation between $$a$$, $$b$$

$$4 - 3 a^2 - 2 a b + a^3 b - 3 b^2 + 2 a^2 b^2 + a b^3=0$$ which is equivalent to $$(a-b)^2 = \frac{ ((a+b)^2 -4)^2}{(a+b)^2 + 4}$$ or $$v^2 = \frac{ (u^2 -1)^2}{u^2 + 1}$$ where $$u=\frac{a+b}{2}$$ and $$v=\frac{a-b}{2}$$.

The contour curve is the union of two graphs $$v = \pm \frac{ u^2 -1}{\sqrt{u^2 + 1}}$$

$$\bf{Added:}$$ With trigonometry, one checks that for $$a = \frac{\sin 3 \theta}{\cos \theta}\\ b= \frac{\cos 3 \theta}{\sin \theta}$$ we have $$\frac{a+b}{2} = \cot 2 \theta \\ \frac{a b -1}{2} = \cos 4 \theta$$

and from here we get a relation between $$a$$, $$b$$

• I didn't mention in my question that (because it is irrelevant), but if you are interested, in the above problem (associated with the other), $a, b$ are $\frac{c(\sqrt3+1)}{(x+y)}$ and $\frac{c(\sqrt3-1)}{(x-y)}$ where $c$ is constant. Though don't bother about this :)
– ACB
Dec 17, 2021 at 10:56
• @Flagged: You are very welcome. I am surprised that the relation turns out to have a neat form. Let me examine the other question now. Dec 17, 2021 at 11:05
• Please wait and sorry... there should be a $4$ in the denominator of both fractions. (e.g. $\frac{c(\sqrt3+1)}{4(x+y)}$) And since you are going to check the other question, $c$ is actually $a$ there.
– ACB
Dec 17, 2021 at 11:07
• @Flagged: About the other question, solving the system gets a solution in terms of $b=\frac{\sin 3 \theta}{\sin\theta}$ and $a=\frac{\sin 3 \theta}{\cos \theta}$ ( a bit different from this problem). With trig, we get $\frac{a^2}{b^2} = \frac{3-b}{1+b} = \tan^2 \theta$. Your problem seems to be inspired, but not directly connected to that problem. Cheers! Dec 18, 2021 at 4:01
• Thanks, BTW, Here's my method ------ We have $$\frac{b+a}2=\frac{\cos2\theta}{\sin2\theta}=u$$ $$\frac{b-a}2=\frac{\cos4\theta}{\sin2\theta}=v$$ Squaring the second equation gives, $$v^2=\frac{\cos^2 4\theta}{\sin^2 2\theta}$$ $$v^2=\frac{(\cos^2 2\theta-\sin^2 2\theta)^2}{\sin^2 2\theta}$$ $$v^2=\frac{\sin^42\theta\left(\dfrac{\cos^2 2\theta}{\sin^2 2\theta}-1\right)^2}{\sin^2 2\theta}$$ $$v^2=\frac{\sin^2 2\theta(u^2-1)^2}{\cos^2 2\theta+\sin^2 2\theta}$$ $$v^2=\frac{(u^2-1)^2}{\dfrac{\cos^2 2\theta}{\sin^2 2\theta}+1}$$ $$v^2=\frac{(u^2-1)^2}{u^2+1}$$
– ACB
Dec 27, 2021 at 1:31