Skewed Trigonometric Function

Question

What would be an expression for a periodic function (period $2\pi$) that essentially behaves just like a negative sine function, but it has the following quirk:

It's $0$s lie on the usual places (even integer multiples of $\frac \pi 2$), but it's maximum and minimum values (of $\pm 1$), instead of lying on odd integer multiples of $\frac\pi 2$, lie deviated by an angle $\alpha$ from the even integer multiples of $\frac \pi 2$. These are it's only maximums and minimums.

It's maximums and minimums can be graphically represented as follows:

In Orange we can see the function $-\sin (x)$, and the Red points represent the maximums and minimums of the function (the Gray lines just represent connections between the points, not the actual function).

Much appreciated.

Answer 1

Considering the variables x, y, skew, t, α and k, and setting:

• (x, y, skew, t, α) ∊ ℝ
• k ∊ ℤ
• f = (x, y) = (t + skew × sin t, - sin t), a parametric equation
• α = skew + π/2

We get for skew = (0, -1, -π/2, 2) ⇒ α = (π/2, -1 + π/2, 0, 2 + π/2) and 0 ≤ t ≤ 2π :

Note that ∫f = ∫(x, y) = (t + skew × sin t, (1/2) × (cos t) × (2 + skew × cos t)). For 0 ≤ t ≤ π the area above the curve is always 2, independent of the value of skew.

Note also that the function minima (-1) occurs at t = π/2 + 2kπ (as y = - sin t), for any value of skew. As an example, for skew = 5π/32 and t = π/2 we get:

Answer 2

A possible candidate is

$$y=-\sin\left[x-\left(\frac\pi 2-\alpha\right) y\right]$$

This may not quite meet all the specifications set out in your original question but it does have the following characteristics:

• has values of $0$ at $n\pi$
• minimum points have been displaced from $\;(2n+\frac 12)\pi\;$ to $\; 2n\pi+\alpha\;$
• maximum points have been displaced from $\;(2n-\frac 12)\pi\;$ to $\; 2n\pi-\alpha\;$
• minimum and maximum points retain values of $-1$ and $1$ respectively

This graph created on desmos.com might be helpful.

https://www.desmos.com/calculator/jk52di8qvq

Answer 3

I have figured out the answer. All you need to do is draw an ellipse that is tangent to the lines $y = \pm 1$ at the points of intersection between those lines and a line of slope $\alpha$. The ellipse also needs to pass through $(1,0)$, $(-1,0)$ and $(0,\sin(\alpha))$.

In doing this I figured out the equation needs to be:

$$f(\theta) = \frac{sgn(\sin(\theta))\cdot sgn(\alpha)}{\sqrt{\cot(\theta)^2 - 2\cot(\alpha)\cdot\cot(\theta) + csc(\alpha)^2}}$$

Where $sgn(x)$ is the sign function. It would be nice to be able to get an expression without $sgn(x)$, but alas, I cannot seem to be able to do so.

Answer 4

This should do it with the right values of $a$ and $b$: $$ x\mapsto a\sin x + b \sin (2x) $$ In order to get extreme values at the prescribed points, we need the derivative to be $0$ at those points. So we need \begin{align} -1 & = a\sin\alpha + b \sin(2\alpha) \\ 0 & = a\cos\alpha+2b\cos(2\alpha) \end{align} In matrix form $$ \begin{bmatrix} \sin\alpha, & \sin(2\alpha) \\ \cos\alpha, & 2\cos(2\alpha) \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} -1 \\ 0 \end{bmatrix} $$

You can solve for $a$ and $b$ by Gaussian elimination.