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I've encountered the following function while working on a project:

$$ f(\theta) = a \sin \theta + b \cos \theta +c \sin \theta \cos \theta + d \sin^2 \theta + e \cos^2 \theta $$

where a through e are all real non-zero numbers and $0 \leq \theta < 2 \pi$.

I know this function has at least two roots. For reasons I won't bore you with, finding the roots numerically isn't ideal for the project I'm working on.

Applying the tangent half-angle substitution described in this answer results in a fourth order polynomial, which is troublesome to find the roots of without numerical methods.

$$ 0 = (e-b)t^4 + (2a-2c) t^3 - 2et^2 + (2a+2c+4d)t + (b+e) $$

Is there a better non-numerical option for finding the roots?

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    $\begingroup$ I don't think the statement "I know this function has at least two roots" is true. For $a=b=c=d=1$ and $e=2$, the only root in $[0,2\pi)$ is equal to $3\pi/2$. $\endgroup$
    – jasnee
    Sep 29, 2021 at 22:14
  • $\begingroup$ @jasnee To clarify, there are additional constraints on a through e that I haven't detailed here. This root finding problem arose when I was trying to find the maximum and minimum of a function. $f(\theta)$ is part of that function's derivative. $\endgroup$
    – Jack Elsey
    Sep 29, 2021 at 22:23
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    $\begingroup$ First thought: the first two terms add to something of the form $A_1 \sin (\theta + \phi_1)$ where the amplitude $A$ and phase shift $\phi_1$ can be calculated with phasor diagrams (I think $A_1 = \sqrt{a^2 + b^2}$ and $\phi_1 = \arctan (b/a)$ if $a$ is positive), and the other three terms likewise add to something of the form $A_2 \sin (2\theta + \phi_2)$. Though I'm not sure this will help a lot: Wolfram Alpha nopes out of giving me closed-form solutions even to simple equations like $\sin \theta = 2 \sin (2\theta + \pi/4)$. $\endgroup$ Sep 29, 2021 at 22:24
  • $\begingroup$ @ConnorHarris haha, yes I ended up here after WolframAlpha showed me the door... $\endgroup$
    – Jack Elsey
    Sep 29, 2021 at 22:25
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    $\begingroup$ Is the original function any nicer (especially with the constraints on the constants)? There are sometimes non-calculus optimization methods. $\endgroup$ Sep 29, 2021 at 22:54

2 Answers 2

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I don’t know whether this helps you at all, but you might consider making the substitution $y=\sin\theta$, $x=\cos\theta$, to get the equation $ay+bx+cxy+dy^2+ex^2=0$, since you’re looking for roots of $f$. The resulting curve in the plane is a conic of some shape or other, depending on the coefficients, and you want to intersect it with the circle $x^2+y^2=1$.

You thus have two conics, which intersect in four points in the complex projective plane, counting multiplicity, by Bézout’s Theorem. So the problem is, far as I can see, unavoidably quartic. Seems to me that with numerical inputs $\{a,b,c,d,e\}$ the only plausible method of finding the roots is something numerical, like Newton-Raphson. (There is a formula for the general quartic, but you Do Not want to try to use it.)

(And: I wouldn’t have gone via the tangent half-angle formulas, I would have just written $\cos\theta=\sqrt{1-\sin^2\theta\,}$ and manipulated the radicals away. You still get a quartic, likely not the same one.)

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Hint

A)

$$ ex^2 + dy^2 + 2\left( {\frac{c}{2}} \right)xy + bx + ay = f(x,y) = 0 $$ is a conic passing through the origin.

B)

$$ \left\{ \begin{array}{l} x = \cos \theta \\ y = \sin \theta \\ \end{array} \right.\quad \Rightarrow \quad x^2 + y^2 = 1 $$ is a circle of unit radius with center at the origin

C)

the zeros of your function $$ f(\theta) = 0 $$

correspond to the crossing of the conic with the circle, which may consist of none, two ("most probably"), or four points.

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    $\begingroup$ Three points is a possibility too, if the two conics are suitably tangent. $\endgroup$
    – Lubin
    Sep 29, 2021 at 22:37
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    $\begingroup$ @ Lubin It is "normal" practice then to consider that case as a $4$ points. $\endgroup$
    – G Cab
    Sep 29, 2021 at 22:43
  • $\begingroup$ Depending on whether the associated quartic touches the $x$-axis in two lobes, one, or none. Makes better sense, @GCab , I admit. $\endgroup$
    – Lubin
    Sep 30, 2021 at 0:46

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