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Consider the function:

$$f(x) = \frac{\sin (x + \alpha) \sin (x + \beta)}{\cos (x + \alpha)\cos (x + \beta)}$$

Choose the parameters $\alpha$ and $\beta$ such that $f(x)$ does not depend on $x$.

Using Werner's formulas, I obtained:

$$f(x) = \frac{\cos(\alpha - \beta) - \cos(2x + \alpha + \beta)}{\cos(2x + \alpha + \beta) + \cos(\alpha - \beta)}$$

If $\alpha = \beta \pm \frac{\pi}{2}$, $f(x) = -1$.

Is this the only way to obtain a constant value from $f(x)$? I can't figure out if there are some alternatives, neither I am able to prove that there are not.

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  • $\begingroup$ It's also valid for all $\alpha = \beta \pm \frac{(2n+1) \pi}{2}$ $\endgroup$ Dec 12 '19 at 16:11
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$f(x) = \frac{\sin(x + \alpha)\sin(x + \beta)}{\cos(x + \alpha)\cos(x + \beta)} = \frac{\cos(\alpha-\beta) - \cos(2x + \alpha + \beta)}{\cos(\alpha-\beta) + \cos(2x + \alpha + \beta)}$.

We need $f(x; \alpha, \beta) = const$, which is equivalent to $f'(x; \alpha, \beta) = 0, \forall x$.

$f'(x; \alpha, \beta) = \frac{4\sin(2x+\alpha + \beta)\cos(\alpha - \beta)}{\left(\cos(\alpha-\beta) + \cos(2x + \alpha + \beta)\right)^2} = 0, \forall x \Leftrightarrow \cos(\alpha-\beta) = 0 \Leftrightarrow \alpha-\beta = \frac{\pi}{2} + \pi n, n = 0, \pm 1, \pm 2, \ldots$.

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You have that there is a constant $k$ such that $$ \frac{\cos(\alpha - \beta) - \cos(2x + \alpha + \beta)}{\cos(2x + \alpha + \beta) + \cos(\alpha - \beta)} = k$$ Some algebra turns this into $$(1 - k)\cos(\alpha - \beta) = (k+1)\cos(2x + \alpha + \beta)$$ This has to hold for every $x$. Since $\cos(2x + \alpha + \beta)$ is a nonconstant function and $(1 - k)\cos(\alpha - \beta)$ is constant, the coefficient $k + 1$ must be zero. So $k = -1$. Plugging this into the above gives $$2\cos(\alpha - \beta) = 0$$ Hence $\cos(\alpha - \beta) = 0$. The zeroes of $\cos x$ are at ${\pi \over 2} + 2n\pi$ for any integer $n$. Hence in order for $\cos(\alpha - \beta) = 0$ we require that $\alpha = \beta + {\pi \over 2} + 2n\pi$ for some integer $n$. And going backwards in the above, you can see that any $\alpha$ and $\beta$ satisfying $\alpha = \beta + {\pi \over 2} + 2n\pi$ do in fact satisfy your desired condition.

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$$f(x)=-1+\dfrac{\cos(\alpha-\beta)}{\cdots}$$

So,we need $\cos(\alpha-\beta)=0$ to keep $f(x)$ constant

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  • $\begingroup$ Sorry, I can't follow you. How did you obtain that expression for $f(x)$? $\endgroup$
    – BowPark
    Dec 12 '19 at 16:51
  • $\begingroup$ @BowPark,$$\cos(x+\alpha-(x+\beta))=?$$ $\endgroup$ Dec 12 '19 at 18:03
  • $\begingroup$ Got it, thank you, this is a clever way to separate the constant part of the function from the $x$-dependent one. $\endgroup$
    – BowPark
    Dec 12 '19 at 19:18

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