Trigonometric and Polar Functions

A relationship is periodic 周期性 if its output pattern repeats at regular input steps. The period 周期 is the smallest positive $k$ with $f(x+k)=f(x)$ for all $x$. You can build the whole graph by copying a single cycle 周期段, and estimate the period by finding how far apart the pattern repeats. Within each cycle a periodic function still has intervals of increase/decrease and maxima/minima.

An angle is in standard position 标准位置 when its vertex is at the origin and its initial ray lies on the positive $x$-axis. Its radian 弧度 measure is the arc length on a unit circle. For the point $P$ where the terminal ray meets a circle of radius $r$:

On a circle of radius $r$, $x=r\cos\theta$ and $y=r\sin\theta$. Exact values at the special angles ($0,\tfrac{\pi}{6},\tfrac{\pi}{4},\tfrac{\pi}{3},\tfrac{\pi}{2},\dots$) come from isosceles right and equilateral triangle 等边三角形 geometry, with signs set by the quadrant 象限 of the angle.

On the unit circle 单位圆 ($r=1$), $\sin\theta$ is the $y$-coordinate and $\cos\theta$ is the $x$-coordinate. As $\theta$ increases, both oscillate 振荡 smoothly between $-1$ and $1$ with period $2\pi$. Sine starts at $0$ (rising); cosine starts at $1$. They are the same wave shifted by $\tfrac{\pi}{2}$.

A sinusoidal function 正弦型函数 is any transformation of sine (or cosine). Its key features:

• Period and frequency 频率 are reciprocals; $\sin\theta$ has period $2\pi$.
• Amplitude 振幅 = half the distance between the maximum and minimum output.
• Midline 中线 $y=d$ = the average of the maximum and minimum (the horizontal center line).

The graph alternates concave up and concave down each half-cycle.

Worked example. For $f(\theta)=3\sin(2\theta)+1$: the amplitude is $3$, the period is $\dfrac{2\pi}{2}=\pi$, and the midline is $y=1$. So the maximum output is $1+3=4$ and the minimum is $1-3=-2$.

The general sinusoid is

To model periodic data (tides, daylight, temperature): read the period from where the pattern repeats, get the amplitude and midline from the max and min, and set the phase shift so a peak lands at the right input. Technology fits a sinusoidal regression. Such a model is only trustworthy over its contextual domain.

Worked example: a Ferris wheel. A rider's height $h$ (metres) on a wheel varies from $2$ m at the bottom to $50$ m at the top, and one turn takes $12$ minutes. At $t=0$ the rider is at the bottom. Build a model.

• Amplitude $=\tfrac{50-2}{2}=24$; midline $y=\tfrac{50+2}{2}=26$.
• Period $12$, so $b=\tfrac{2\pi}{12}=\tfrac{\pi}{6}$.
• At $t=0$ the height is a minimum. $\cos$ starts at a maximum, so $-\cos$ starts at a minimum and no phase shift is needed:

Check: $h(6)=-24\cos\pi+26=50$ (the top, after half a turn). After $2$ minutes, $h(2)=-24\cos\tfrac{\pi}{3}+26=-12+26=14$ m.

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}$ is the slope of the terminal ray. Its slope values repeat every half-revolution, so its period is $\pi$. It has vertical asymptotes where $\cos\theta=0$ (at $\theta=\tfrac{\pi}{2}+k\pi$), and it is always increasing between consecutive asymptotes, switching concavity at each zero.

The inverse trigonometric functions 反三角函数 – arcsine 反正弦, arccosine, arctangent – reverse the trig functions, so they take a ratio and return an angle. Because sine, cosine, and tangent repeat, their domains must be restricted (e.g. sine to $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$) before an inverse can exist.

So each inverse returns an angle in one fixed range: $\arcsin$ in $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$, $\arccos$ in $[0,\pi]$, $\arctan$ in $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$. For example, $\arcsin\!\left(-\tfrac12\right)=-\tfrac{\pi}{6}$, but $\arccos\!\left(-\tfrac12\right)=\tfrac{2\pi}{3}$ (not $-\tfrac{\pi}{3}$, which is outside $[0,\pi]$).

Use inverse trig functions to solve trig equations. Because the functions are periodic, there are usually infinitely many solutions – write the general solution by adding multiples of the period, then keep those in the required (often contextual) domain.

Worked example. Solve $2\sin\theta=1$ on $[0,2\pi)$. Then $\sin\theta=\tfrac12$, and on the unit circle sine is positive in the first and second quadrants, so $\theta=\dfrac{\pi}{6}$ or $\dfrac{5\pi}{6}$. Over all reals you would add $2\pi k$ to each.

These are the reciprocals 倒数 of the main three:

The Pythagorean identity 毕达哥拉斯恒等式 comes from $x^2+y^2=1$ on the unit circle:

Worked example. Find the exact value of $\sin\tfrac{5\pi}{12}$ (that is, $\sin 75^{\circ}$). Write $\tfrac{5\pi}{12}=\tfrac{\pi}{4}+\tfrac{\pi}{6}$:

Polar coordinates 极坐标 locate a point by $(r,\theta)$ – distance $r$ from the origin at angle $\theta$. Convert with

Worked example. Write the point $(-1,\sqrt3)$ in polar form. $r=\sqrt{(-1)^2+(\sqrt3)^2}=2$ and $\tan\theta=\tfrac{\sqrt3}{-1}=-\sqrt3$. A calculator's $\arctan(-\sqrt3)=-\tfrac{\pi}{3}$ points into the fourth quadrant, but the point is in the second quadrant ($x<0$, $y>0$), so add $\pi$: $\theta=\tfrac{2\pi}{3}$. The polar form is $\left(2,\tfrac{2\pi}{3}\right)$. Always check the quadrant before trusting $\arctan$.

The graph of $r=f(\theta)$ is all points $(r,\theta)$ as $\theta$ sweeps around. Restricting the domain of $\theta$ draws only part of the curve. As $\theta$ increases, $r$ grows or shrinks, tracing spirals, circles, roses, and limaçons.

As $\theta$ increases, the distance from the origin $r$ changes:

• $r$ positive and increasing (or negative and decreasing) $\Rightarrow$ moving away from the origin;
• $r$ positive and decreasing (or negative and increasing) $\Rightarrow$ moving toward the origin.

Where $r$ switches between increasing and decreasing, the distance reaches a relative extreme. The average rate of change of $r$ with respect to $\theta$, $\dfrac{\Delta r}{\Delta \theta}$, estimates how fast the curve moves in or out over an interval.

Worked example. For $r=3+2\cos\theta$ on $\left[0,\tfrac{\pi}{2}\right]$: $r(0)=5$ and $r\!\left(\tfrac{\pi}{2}\right)=3$. The average rate of change is $\dfrac{3-5}{\pi/2-0}=-\dfrac{4}{\pi}\approx-1.27$. Here $r$ is positive and decreasing, so the curve moves toward the origin, on average by about $1.27$ units of distance per radian.

• Use radians and the unit circle: $\cos\theta$ and $\sin\theta$ are the coordinates, always between $-1$ and $1$.
• For a sinusoid $a\sin(b(\theta+c))+d$: $|a|$ is the amplitude, $\tfrac{2\pi}{|b|}$ the period, $d$ the midline, and the phase shift is $-c$ (a $+c$ shifts left).
• Model periodic data by reading period, amplitude, and midline from the max and min.
• Inverse trig functions need a restricted domain and return an angle; trig equations have infinitely many solutions (add the period).
• Convert polar↔rectangular with $x=r\cos\theta$, $y=r\sin\theta$.