Trigonometric Functions
Unit 2 of Algebra II. 13 questions below, each with the working. Every answer was checked by a second pass before it was published.
Radian measure, the unit circle, trigonometric functions of any angle, graphing sine and cosine, modelling periodic phenomena, the Pythagorean identity.
How this unit is tested
Start by getting completely fluent with radian-degree conversion and the unit circle values at the special angles (0°, 30°, 45°, 60°, 90° and their multiples). Every other skill in this unit depends on being able to place an angle in a quadrant, find its reference angle, and read off (x,y) = (cosθ, sinθ) without hesitation.
For 'trig functions of any angle' problems, always do the same three steps in order: identify the quadrant, find the reference angle, then attach the correct sign using ASTC (All Students Take Calculus: All positive in QI, Sine in QII, Tangent in QIII, Cosine in QIV). When you're given one ratio (like sinθ) and a quadrant, use the Pythagorean identity to solve for the other ratio, then fix the sign from the quadrant — the identity itself only gives you a magnitude.
For graphing and modeling, translate every equation into four numbers: amplitude, period, midline, and phase shift. Amplitude and midline come from the maximum and minimum values (amplitude = (max−min)/2, midline = (max+min)/2); period comes from how long it takes the pattern to repeat, and B = 2π/period. Decide whether the situation looks like a sine start (midline value at t=0, then rising) or a cosine start (max or min at t=0) before you write the equation — this single choice determines whether you'll need a phase shift at all.
Work problems by hand using exact values (fractions with radicals) until the special angles are automatic, then check messier problems with a calculator in the correct angle mode.
What you have to know
Radian Measure and Conversion
One radian is the measure of a central angle that subtends an arc equal in length to the circle's radius, so $\theta = \dfrac{\text{arc length}}{\text{radius}}$. To convert degrees to radians, multiply by $\dfrac{\pi}{180}$; to convert radians to degrees, multiply by $\dfrac{180}{\pi}$.
Unit Circle Definition of Sine and Cosine
For an angle $\theta$ measured counterclockwise from the positive x-axis, if the terminal side meets the unit circle (radius 1) at point $(x,y)$, then $\cos\theta = x$ and $\sin\theta = y$, and $\tan\theta = \dfrac{y}{x}$ when $x \neq 0$.
Reference Angles and the ASTC Sign Rule
The reference angle is the acute angle between the terminal side of $\theta$ and the x-axis. Any trig value of $\theta$ equals plus-or-minus the same trig value of the reference angle, with the sign determined by quadrant: All ratios positive in QI, Sine (and cosecant) positive in QII, Tangent (and cotangent) positive in QIII, Cosine (and secant) positive in QIV.
Pythagorean Identity
For every angle $\theta$, $\sin^2\theta + \cos^2\theta = 1$. Given one ratio and the quadrant, solve for the other by isolating it and taking the square root, then choose the sign that matches the quadrant.
General Sinusoidal Equation
A sine or cosine graph can be written as $y = A\sin(B(x-C)) + D$ or $y = A\cos(B(x-C)) + D$, where amplitude $= |A|$, period $= \dfrac{2\pi}{|B|}$, phase shift $= C$, and midline is $y = D$.
13 practice questions
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Convert $\dfrac{5\pi}{6}$ radians to degrees.
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Answer. 150°
Multiply by $\dfrac{180}{\pi}$: $\dfrac{5\pi}{6}\times\dfrac{180}{\pi} = \dfrac{5\times180}{6} = 150°$. -
Convert 315° to radians, expressed as a fraction of $\pi$.
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Answer. $\dfrac{7\pi}{4}$
Multiply by $\dfrac{\pi}{180}$: $315\times\dfrac{\pi}{180} = \dfrac{315\pi}{180} = \dfrac{7\pi}{4}$. -
Find a positive angle coterminal with $-50°$.
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Answer. 310°
Add 360° to get a coterminal angle in the standard range: $-50°+360° = 310°$. -
Find the exact value of $\sin(120°)$ using the unit circle.
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Answer. $\dfrac{\sqrt3}{2}$
120° is in Quadrant II with reference angle $180°-120°=60°$. Sine is positive in QII, so $\sin(120°)=\sin(60°)=\dfrac{\sqrt3}{2}$. -
Find the reference angle for 250°, and state its quadrant.
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Answer. Reference angle 70°, Quadrant III
250° lies between 180° and 270°, so it's in QIII. The reference angle is $250°-180°=70°$. -
An angle θ terminates in Quadrant III with a reference angle of 60°. Find $\cos\theta$.
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Answer. $-\dfrac{1}{2}$
$\cos(60°)=\dfrac12$, but cosine is negative in QIII (only tangent is positive there), so $\cos\theta=-\dfrac12$. -
If $\sin\theta = -\dfrac{5}{13}$ and θ terminates in Quadrant IV, find $\cos\theta$.
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Answer. $\dfrac{12}{13}$
Using $\sin^2\theta+\cos^2\theta=1$: $\cos^2\theta = 1-\dfrac{25}{169}=\dfrac{144}{169}$, so $\cos\theta=\pm\dfrac{12}{13}$. Cosine is positive in QIV, so $\cos\theta=\dfrac{12}{13}$. -
Given $\sin\theta=\dfrac35$ and $\cos\theta=-\dfrac45$ (θ in Quadrant II), find $\tan\theta$.
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Answer. $-\dfrac34$
$\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{3/5}{-4/5} = -\dfrac34$, consistent with tangent being negative in QII. -
State the amplitude, period, and midline of $y = 4\sin(2x) - 1$.
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Answer. Amplitude 4, period π, midline y = -1
Here $A=4$ so amplitude is 4; $B=2$ so period $=\dfrac{2\pi}{2}=\pi$; $D=-1$ gives the midline $y=-1$. -
Write the equation of a cosine function with amplitude 3, period $4\pi$, midline $y=2$, and no phase shift.
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Answer. $y = 3\cos\left(\dfrac{x}{2}\right) + 2$
Amplitude gives $A=3$ and midline gives $D=2$. Solve period$=\dfrac{2\pi}{B}=4\pi$ for $B$: $B=\dfrac{2\pi}{4\pi}=\dfrac12$, so the equation is $y=3\cos\left(\dfrac{x}{2}\right)+2$. -
A sinusoidal graph has a maximum at (0, 5) and a minimum at (π, -1). Write its equation as a cosine function.
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Answer. $y = 3\cos(x) + 2$
Amplitude $=\dfrac{5-(-1)}{2}=3$, midline $=\dfrac{5+(-1)}{2}=2$. The max-to-min distance is half a period, so the full period is $2\pi$, giving $B=1$. Since the max occurs at $x=0$, cosine (not sine) is the right shape with no phase shift. -
If $\cos\theta = \dfrac{2}{3}$ and $0 < \theta < \dfrac{\pi}{2}$, find $\sin\theta$.
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Answer. $\dfrac{\sqrt5}{3}$
From $\sin^2\theta+\cos^2\theta=1$: $\sin^2\theta = 1-\dfrac49=\dfrac59$, so $\sin\theta=\pm\dfrac{\sqrt5}{3}$. Since θ is in Quadrant I, sine is positive: $\sin\theta=\dfrac{\sqrt5}{3}$. -
The temperature $T(t)$ in °F, $t$ hours after midnight, is modeled by $T(t) = 15\sin\left(\dfrac{\pi}{12}(t-9)\right) + 60$. Find the maximum temperature and the time of day it occurs.
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Answer. 75°F at t = 15, which is 3:00 PM
The maximum of sine is 1, occurring when $\dfrac{\pi}{12}(t-9)=\dfrac{\pi}{2}$, so $t-9=6$ and $t=15$. Then $T(15)=15(1)+60=75°F$; since $t$ is hours after midnight, $t=15$ is 3:00 PM.
What people get wrong
- Leaving the calculator in the wrong angle mode. If the problem is in radians and the calculator is set to degrees (or vice versa), every trig value comes out wrong even though the keystrokes are correct — always check the mode before computing.
- Reporting the reference angle as if it were the final answer. The reference angle only gives the magnitude of the trig value; you must still attach the correct sign using the quadrant (ASTC) before the answer is complete.
- Using the identity $\sin^2\theta+\cos^2\theta=1$ and forgetting the ± choice. Solving for $\sin\theta$ or $\cos\theta$ from the identity gives a square root, which is only correct once you've picked the sign that matches the given quadrant.
- Confusing amplitude with the maximum value, or midline with zero. Amplitude is half the distance between max and min, and midline is the average of max and min — read both off the graph or context before writing the equation, don't assume midline is always 0.
- Choosing sine when the situation actually starts at a maximum or minimum (or choosing cosine when it starts at the midline). Decide the starting behavior first, then pick the function and sign that matches, rather than defaulting to sine out of habit.
- Inverting the period formula, writing $B = \dfrac{2\pi}{\text{period}}$ as period $=\dfrac{2\pi}{B}$ backwards or dividing degrees by $2\pi$ instead of by 360. Keep the formula $B=\dfrac{2\pi}{\text{period}}$ written down and plug in carefully.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of Algebra II. Free, and no account needed to start.