4

One of my math professors and I are working on a physics problem involving spinning a chain, and we decided to go as simple as possible and work out the solution explicitly for that case (a long rod hanging from a hinge rotating in a horizontal circle). Then we could hopefully work up from there. In the end, we boiled it down to the point where we had an equation of this form:

$$\frac{1}{\cos \theta} = a \sin \theta - b$$

Depending on the values of $a$ and $b$, there are $0$, $1$, $2$, $3$, or $4$ solutions for $\theta$ in this equation. What I'm curious about is whether there are formulas in terms of $a$ and $b$ that will give these solutions. As an aside, this situation actually reminds me of quadratics - they have $0$, $1$, or $2$ solutions, the solutions are given by the quadratic formula, and the value of $b^2-4ac$ indicates how many real-valued solutions there are. I'm looking for something similar for the equation I've given above, and WolframAlpha is being no help (gasp!).

El'endia Starman
  • 3,988
  • 3
    Square, and substitute $1-\sin^2\theta$ for $\cos\theta$. We get a quartic in $\sin\theta$. (Squaring may introduce extraneous roots.) Quartics are a mess. There are formulas for solving them, dating back to Cardano and Ferrari. But they are very painful to use (less politely, they are kind of useless). – André Nicolas Feb 01 '13 at 00:34
  • @AndréNicolas you have a typo in that first sentence - you mean $\cos^2{\theta}$... +1 for the content though – Glen_b Feb 01 '13 at 02:26
  • Yes I do. Thanks! – André Nicolas Feb 01 '13 at 06:20

2 Answers2

2

You can see why you have up to 4 solutions because you can rearrange the equation to produce a quartic in $\cos{\theta}$:

$$a^2 \cos^4{\theta} + (b^2-a^2) \cos^2{\theta} + 2 b \cos{\theta} + 1 = 0$$

Is there a formula for the roots of this polynomial in terms of $a$ and $b$? Sure, but I imagine it is nasty.

EDIT

I played around with the exact roots in Mathematica, which I can tell you is not the most enlightening exercise I have taken up in this space. That said, there was this square root term that occurred throughout, the radicand of which I imagine acts as a discriminant. That is, the discriminant must be greater than zero for there to be real roots. In case you're curious, the expression for this discriminant is

$$-a^4 \left(a^6-a^4 \left(3 b^2+8\right)+a^2 \left(3 b^4-20 b^2+16\right)-b^6+b^4\right)$$

Ron Gordon
  • 138,521
  • 2
    rlgordonma: I think you have given the wrong quartic equation. For example, let $a$ and $b$ both be zero. El'endia's equation has no solution, but $\theta = 0$ is a solution to your quartic. – Steve Kass Feb 01 '13 at 01:59
  • @SteveKass: thank you, you caught a major goof. It is now fixed. – Ron Gordon Feb 01 '13 at 02:10
  • 1
    Something is still wrong. When $b=0$, your quartic is positive for all $\theta$, but the original equation has have solutions. (When $a\ge 2$, for example.) In any case, a quartic is worth looking at for ideas, though it's worth noting here, as @André-Nicolas did in a separate comment, that the correct quartic may have extraneous solutions because of squaring. This could add to the challenge of answering the original question, even if the discriminant-like expression is manageable. – Steve Kass Feb 01 '13 at 03:35
0

If we put $t=\tan \frac\theta2,\sin\theta =\frac{2t}{1+t^2},\cos\theta=\frac{1-t^2}{1+t^2}$

Then, $$ \frac1{\cos\theta}=a\sin\theta-b--->(1)$$ becomes $$(b-1)t^4-2at^3-2t^2+2at-(b+1)=0--->(2)$$ which is a quartic equation in $t$ hence will definitely have exactly four finite roots if $b-1\ne0$ .

$(1)$ If $b=1,$ the equation reduces to $2at^3+2t^2-2at+2=0--->(3)$

We can make use of this to identify the number of real roots of $(3)$

Clearly, each real root of $(3)$ will correspond to one real of root of of $(1)$ the reason being:

We know, $\tan A=\tan B\implies A=n\pi+B,$ so $\tan(\pi+\frac\theta2)=\tan\frac\theta2$ i.e., or $\tan\left(\frac{2\pi+\theta}2\right)=\tan\frac\theta2$

So, the periods of $\cos\theta,\sin\theta(=2\pi)$ and $\tan \frac\theta2$ are same.

Hence, in each $\in[2n\pi,2(n+1)\pi)$ there will be one-one correspondence between $\cos\theta,\tan \frac\theta2$ and $\sin\theta,\tan \frac\theta2$

$(2)$ If $b\ne1,$ we can write $$t^4-\frac{2a}{b-1}t^3-\frac2{b-1}t^2+\frac{2a}{b-1}t+\frac{b+1}{b-1}=0--->(4)$$

Now, we eliminate the $t^3$ term by putting $y=x-\lambda\implies x=y+\lambda$

$$(y+\lambda)^4-\frac{2a}{b-1}(y+\lambda)^3-\frac2{b-1}(y+\lambda)^2+\frac{2a}{b-1}(y+\lambda)+\frac{b+1}{b-1}=0--->(5)$$

The coefficient of $y^3$ is $4\lambda-\frac{2a}{b-1}$

If we set this to $0,\lambda=\frac a{2(b-1)}\implies y=x+\frac a{2(b-1)}$

Now, we can utilize this to identify the number of real roots of $(5),$ hence of $(4)$

Clearly, each real root of $(4)$ i.e. of $(2)$ will correspond to one real of root $(1)$

lab bhattacharjee
  • 274,582