Trigonometry Reference
Trigonometric Graphs and Equations
Sine and cosine graphs turn unit-circle patterns into periodic models, and exact equations return to those same reference angles.
Solved Examples
Jump directly to a worked example by graph feature or equation type.
Fact Table
| Feature | For $y=a\sin(bx)+k$ or $y=a\cos(bx)+k$ | Meaning |
|---|---|---|
| Amplitude | $|a|$ | Distance from the midline to a peak. |
| Period | $\frac{2\pi}{|b|}$ | Length of one full cycle. |
| Midline | $y=k$ | Vertical center of the graph. |
| Inverse trig | $\arcsin$, $\arccos$, $\arctan$ | Returns a principal angle in a fixed interval. |
Content Formulas
Sine
$$y=a\sin\bigl(b(x-h)\bigr)+k$$
Cosine
$$y=a\cos\bigl(b(x-h)\bigr)+k$$
Exact-Value Strategy
$$\text{find the reference angle, then choose every allowed quadrant}$$
For equations on $[0,2\pi)$, use the unit circle after isolating the trigonometric function. Do not stop after finding only the Quadrant I solution.
Classic Examples
Read Graph Features
Describe the amplitude, period, and midline of $y=2\sin(3x)-1$.
Solution Steps
- Read the amplitude from the absolute value of the outside coefficient.
- Use $\frac{2\pi}{|b|}$ with $b=3$ for the period.
- Read the vertical shift as the midline.
$$\text{amplitude}=|2|=2$$$$\text{period}=\frac{2\pi}{3}$$$$\text{midline: }y=-1$$
Use an Inverse Trig Function
Find $\arccos\left(-\frac12\right)$.
Solution Steps
- Recall that cosine is $-\frac12$ at reference angle $\frac{\pi}{3}$ in Quadrants II and III.
- Use the principal range of arccosine, $[0,\pi]$.
- Select the Quadrant II angle.
$$\cos\left(\frac{2\pi}{3}\right)=-\frac12$$$$\arccos\text{ returns an angle from }0\text{ to }\pi$$$$\boxed{\arccos\left(-\frac12\right)=\frac{2\pi}{3}}$$
Solve a Sine Equation
Solve $\sin x=\frac{\sqrt3}{2}$ on $[0,2\pi)$.
Solution Steps
- Identify the reference angle with sine value $\frac{\sqrt3}{2}$.
- Choose the quadrants where sine is positive.
- State both solutions in the requested interval.
$$\text{reference angle}=\frac{\pi}{3}$$$$\sin x>0\text{ in Quadrants I and II}$$$$\boxed{x=\frac{\pi}{3},\ \frac{2\pi}{3}}$$
Solve a Quadratic Trig Equation
Solve $2\cos^2x-1=0$ on $[0,2\pi)$.
Solution Steps
- Isolate $\cos^2x$.
- Take both square roots.
- Use the unit circle to list every angle with cosine $\pm\frac{\sqrt2}{2}$.
$$2\cos^2x=1\Rightarrow\cos^2x=\frac12$$$$\cos x=\pm\frac{\sqrt2}{2}$$$$\boxed{x=\frac{\pi}{4},\ \frac{3\pi}{4},\ \frac{5\pi}{4},\ \frac{7\pi}{4}}$$