# Choosing $\alpha$ and $\beta$ so that $f(x) = \frac{\sin (x + \alpha) \sin (x + \beta)}{\cos (x + \alpha)\cos (x + \beta)}$ is independent of $x$

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.

• It's also valid for all $\alpha = \beta \pm \frac{(2n+1) \pi}{2}$ Dec 12 '19 at 16:11

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

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

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

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.

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

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

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