I'm looking for sufficient (and necessary would be good too) conditions on $a,b,c$ such that

\begin{align} a\cos\phi + b \cos 3\phi + c \cos 5\phi \geq -1 \hspace{20pt} (\forall \phi) \end{align}

Given that $a+b+c=\frac 12$. In practise, $a,b,c$ are actually parametrisations of $\theta$:

\begin{align} a&=\frac 9 {16} + 2\theta \\ b&=-\frac 1 {16} - 3\theta \\ c&=\theta \end{align}

And I'm looking for the condition(s) on $\theta$. Deriving directly to find local minima reduces to solving a quintic (which I don't know how to do). Expressing $\cos n\phi$ as a polynomial in $\cos\phi$ seems numerically promising: Denoting $\alpha=\cos\phi$, this is equivalent to guaranteeing

\begin{align} (a+5c-3b)\alpha + (4b-20c) \alpha ^3 + (16c) \alpha ^5 \geq -1 \hspace{20pt} \forall \alpha\in[-1,1] \end{align}

Deriving here reduces to solving a quartic (which is one degree and a whole world of algebraic concepts easier). My plan was to just throw this into Matlab and graph the results. Alternatively, maybe it's feasible to find the general roots of the derivative which are real, and correspond to local minima, and ensure the condition. That sounds easier than it probably is in practise because of the dependence on $\theta$ which will probably generate more cases than it's practical to deal with.

Lagrange multipliers would have helped but there's not really a constraint here.

I was hoping for a closed-form, more analytical insight. A superposition identity that simplifies this, or some other trigonometric simplification. Thanks.


Numerically, it appears that to double precision, the inequality holds for $-0.284203953109394 \leq \theta \leq 0.145635938534616$. Maybe some kind of monotonicity in $\theta$ can prove this is a unique interval.


The function (governing the inequality) is linear in $\theta$, which means we really only need to solve the 1-dimensional inequality for any fixed $\theta$ (say $\theta=0$), and extrapolate the extrema linearly. This indeed proves that there is a unique interval around $\theta=0$ on which the inequality will hold.

  • $\begingroup$ For whatever it's worth, if you express it in terms of $a$, $b$, and $c$, then those equations with $\theta$ are a constraint. So is the sum being $\frac12$. $\endgroup$ – dfeuer Aug 31 '13 at 22:57
  • $\begingroup$ I can't help thinking those prime numbers ($3$ and $5$) might be useful somehow in determining the range of $\phi\mapsto(\cos\phi,\cos3\phi,\cos5\phi)$. $\endgroup$ – dfeuer Aug 31 '13 at 23:24
  • $\begingroup$ Why not get rid of $a$, $b$, and $c$ and replace them with expressions of $\theta$? That might clarify the problem. $\endgroup$ – Stefan Smith Sep 1 '13 at 0:20

If we use your parametrisation, in the interval $\left(\frac{4n-1}2\pi,\frac{4n+1}2\pi\right)$ for $n\in\Bbb Z$, the inequality is $$\theta\ge\frac{-16-9\cos\phi+\cos3\phi}{16(2\cos\phi-3\cos3\phi+\cos5\phi)}=f(\phi)$$ so we want its maximum value in this interval. Similarly for the interval $\left(\frac{4n+1}2\pi,\frac{4n+3}2\pi\right)$ we want the reverse inequality. Now $$f'(\theta)=\frac{9\sin\phi-3\sin3\phi}{16(2\cos\phi-3\cos3\phi+\cos5\phi)}-\frac{(-16-9\cos\phi+\cos3\phi)(-2\sin\phi+9\sin3\phi-5\sin5\phi)}{16(2\cos\phi-3\cos3\phi+\cos5\phi)^2}$$ and setting to zero gives $$\small 3(3\sin\phi-\sin3\phi)(2\cos\phi-3\cos3\phi+\cos5\phi)=(-16-9\cos\phi+\cos3\phi)(-2\sin\phi+9\sin3\phi-5\sin5\phi)$$ which results in \begin{align}-25\sin\phi\cos3\phi+9\sin\phi\cos5\phi-3\sin3\phi\cos5\phi=&-75\sin3\phi\cos\phi+45\sin5\phi\cos\phi-5\sin5\phi\cos3\phi\\&+32\sin\phi-144\sin3\phi+80\sin5\phi.\end{align} Now the identity $\sin(A-B)=\sin A\cos B-\sin B\cos A$ yields $$\small 50\sin3\phi\cos\phi+2\sin5\phi\cos3\phi-36\sin5\phi\cos\phi=80\sin5\phi+9\sin4\phi-144\sin3\phi-28\sin2\phi+32\sin\phi$$ and the identity $\sin A\cos B=\frac12(\sin(A-B)+\sin(A+B))$ yields $$\text{LHS}=25(\sin2\phi+\sin4\phi)+\sin2\phi+\sin8\phi-18\sin4\phi-18\sin6\phi$$ and denoting $\sin k\phi$ as $\phi_k$, we get the eighth-order polynomial $$\boxed{\phi_8-18\phi_6-80\phi_5-2\phi_4+144\phi_3+54\phi_2-32\phi_1=0}$$ Solving for $\phi\ne0$ gives us two solutions, one of which corresponds to $\theta\le f(\hat\phi_{(1)})$ and the other one to $\theta\ge f(\hat\phi_{(2)})$. This, however, cannot be determined analytically.

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