Pre-Calculus: Unit Circle & Periodic Models

How to do it:

1. One radian is the angle that cuts an arc equal to the radius.

2. Convert: $\theta_{\text{rad}} = \theta_{\text{deg}} \cdot \pi/180\text{.}$ A full turn is $2\pi$ radians ($360°\text{).}$

3. Arc length: $\theta = s/r\text{,}$ so $s = r\theta$ with $\theta$ in radians.

Worked example: $180° = \pi$ radians. $90° = \tfrac{\pi}{2}\text{.}$ $30° = \tfrac{\pi}{6}\text{.}$

Check: do not treat "$2\pi$" as about 2. It is about 6.28 radians. Calculators in degree mode will misread $\pi/2\text{.}$

How to do it:

1. Start at $(1,0)$ on the unit circle and walk counterclockwise for positive $\theta\text{.}$

2. The point is $(\cos\theta, \sin\theta)\text{.}$ Negative $\theta$ goes clockwise.

3. Coterminal angles differ by $2\pi\text{;}$ they land on the same point.

Worked example: $\cos(\pi)=-1\text{,}$ $\sin(\pi)=0\text{.}$ $\cos(\tfrac{\pi}{2})=0\text{,}$ $\sin(\tfrac{\pi}{2})=1\text{.}$

Check: Quadrant II sine is positive, cosine negative, matching Grade 6 sign work on the plane.

How to do it:

1. A point $(\cos\theta,\sin\theta)$ is 1 unit from the origin, so $\sin^2\theta+\cos^2\theta=1\text{.}$

2. If you know one coordinate and the quadrant, solve for the other. See Pythagorean identity.

3. Keep the sign that matches the quadrant.

Worked example: If $\sin\theta=3/5$ in Quadrant I, $\cos\theta=4/5$ because $(3/5)^2+(4/5)^2=1\text{.}$

Check: Quadrant II would take $\cos\theta=-4/5\text{.}$ Squaring loses the sign, so you must put it back.

How to do it:

1. Midline $D$ is the average of max and min.

2. Amplitude $A$ is half the range (always positive).

3. Period $T$ is the time for one cycle; $B=2\pi/T$ in $y=A\sin(B(t-C))+D\text{.}$

Worked example: A tide with midline 4 ft, amplitude 2 ft, period 12 hours: $B=2\pi/12=\pi/6\text{,}$ so $y=2\sin(\tfrac{\pi}{6}t)+4$ if you start at the midline going up.

Check: period of $\sin(2t)$ is $\pi\text{,}$ not $2\text{.}$ If $t=0$ is a max, cosine is often cleaner than sine.

How to do it:

1. Learn the first-quadrant values: for $\tfrac{\pi}{6}\text{,}$ $\tfrac{\pi}{4}\text{,}$ $\tfrac{\pi}{3}\text{,}$ sine is $\tfrac{1}{2}\text{,}$ $\tfrac{\sqrt{2}}{2}\text{,}$ $\tfrac{\sqrt{3}}{2}\text{,}$ and cosine runs the same list backwards.

2. For any angle, find its reference angle: the acute angle between the terminal side and the $x$-axis.

3. Copy the first-quadrant value and attach the sign for the quadrant: sine is positive in I and II, cosine in I and IV.

Worked example: $\cos(\tfrac{5\pi}{6})\text{:}$ reference angle $\tfrac{\pi}{6}\text{,}$ quadrant II, cosine negative, so $-\tfrac{\sqrt{3}}{2}\text{.}$ $\sin(\tfrac{7\pi}{4})\text{:}$ reference $\tfrac{\pi}{4}\text{,}$ quadrant IV, sine negative: $-\tfrac{\sqrt{2}}{2}\text{.}$ $\sin(\tfrac{4\pi}{3})\text{:}$ reference $\tfrac{\pi}{3}\text{,}$ quadrant III: $-\tfrac{\sqrt{3}}{2}\text{.}$

Check: the pair must satisfy the identity: $\left(-\tfrac{\sqrt{3}}{2}\right)^2+\left(\tfrac{1}{2}\right)^2=1\text{.}$ "All Students Take Calculus" lists which functions are positive in quadrants I through IV: all, sine, tangent, cosine.

How to do it:

1. $\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{y}{x}\text{,}$ the slope of the terminal ray.

2. It is undefined where $\cos\theta=0\text{,}$ at $\tfrac{\pi}{2}$ and $\tfrac{3\pi}{2}\text{:}$ the graph has vertical asymptotes there.

3. Its period is $\pi\text{,}$ not $2\pi\text{,}$ because a slope repeats every half turn.

Worked example: $\tan(\tfrac{\pi}{4})=1\text{,}$ the slope of the $45°$ line. $\tan(\tfrac{2\pi}{3})=\dfrac{\sqrt{3}/2}{-1/2}=-\sqrt{3}\text{.}$ $\tan(\pi)=\dfrac{0}{-1}=0\text{.}$ $\tan(\tfrac{5\pi}{4})=1$ again, a half turn after $\tfrac{\pi}{4}\text{.}$

Check: tangent is positive in quadrants I and III, where $x$ and $y$ share a sign. The reciprocals are $\sec\theta=\tfrac{1}{\cos\theta}\text{,}$ $\csc\theta=\tfrac{1}{\sin\theta}\text{,}$ and $\cot\theta=\tfrac{1}{\tan\theta}\text{.}$

How to do it:

1. With $\theta$ in radians, arc length is $s=r\theta\text{.}$

2. Sector area is $A=\tfrac{1}{2}r^2\theta\text{.}$

3. If the angle is given in degrees, convert first: multiply by $\tfrac{\pi}{180}\text{.}$ The formulas only work in radians.

Worked example: $r=10$ cm and $\theta=\tfrac{2\pi}{3}\text{:}$ $s=10\cdot\tfrac{2\pi}{3}\approx 20.9$ cm and $A=\tfrac{1}{2}\cdot 100\cdot\tfrac{2\pi}{3}\approx 104.7$ cm$^2\text{.}$ A $60°$ slice of a $12$-inch pizza ($r=6\text{):}$ $\theta=\tfrac{\pi}{3}\text{,}$ so the area is $\tfrac{1}{2}\cdot 36\cdot\tfrac{\pi}{3}=6\pi\approx 18.8$ in$^2\text{.}$

Check: a full turn ($\theta=2\pi\text{)}$ gives $s=2\pi r$ and $A=\pi r^2\text{,}$ the circle formulas you already know. Replay arc length and sector area.

How to do it:

1. In $y=A\sin(B(x-C))+D$ or the cosine version, $|A|$ is the amplitude, $\tfrac{2\pi}{B}$ the period, $C$ the horizontal shift, and $D$ the midline.

2. The maximum is $D+|A|$ and the minimum is $D-|A|\text{.}$

3. Sketch one period from the starting point using five key points: start, quarter, half, three-quarter, end.

Worked example: $y=3\cos(2x)-1\text{:}$ amplitude $3\text{,}$ period $\pi\text{,}$ midline $y=-1\text{,}$ so the maximum $2$ is at $x=0\text{,}$ the minimum $-4$ at $x=\tfrac{\pi}{2}\text{,}$ and the next maximum at $x=\pi\text{.}$ $y=2\sin(x-\tfrac{\pi}{3})$ is a sine wave of amplitude $2$ shifted right $\tfrac{\pi}{3}\text{.}$

Check: factor before reading a shift: $\sin(2x-\pi)=\sin(2(x-\tfrac{\pi}{2}))\text{,}$ so the shift is $\tfrac{\pi}{2}\text{,}$ not $\pi\text{.}$ A negative $A$ flips the graph across the midline.

How to do it:

1. Isolate the trig function.

2. Find the reference angle from the exact-value table or an inverse function.

3. Place it in every quadrant where the sign is right, inside the interval asked. For all solutions, add $2\pi k\text{.}$

Worked example: $2\cos\theta-1=0$ on $[0,2\pi)\text{:}$ $\cos\theta=\tfrac{1}{2}\text{,}$ reference angle $\tfrac{\pi}{3}\text{,}$ cosine positive in I and IV, so $\theta=\tfrac{\pi}{3}$ or $\tfrac{5\pi}{3}\text{.}$ For the tide $y=2\sin(\tfrac{\pi}{6}t)+4\text{,}$ when is $y=5$ in the first $12$ hours? $\sin(\tfrac{\pi}{6}t)=\tfrac{1}{2}\text{,}$ so $\tfrac{\pi}{6}t=\tfrac{\pi}{6}$ or $\tfrac{5\pi}{6}\text{:}$ $t=1$ hour and $t=5$ hours.

Check: substitute back: $2\cos(\tfrac{5\pi}{3})-1=2\cdot\tfrac{1}{2}-1=0\text{.}$ A calculator's $\cos^{-1}$ returns only $\tfrac{\pi}{3}\text{;}$ the quadrant reasoning supplies the second answer.