Trigonometric Functions — Free Algebra 2 Review Games.

On the unit circle at 0 degrees, the y-coordinate is 0.

On the unit circle at 0 degrees, the x-coordinate is 1.

The unit circle has radius 1, centered at the origin.

At 90 degrees on the unit circle, the y-coordinate is 1.

The basic sine function completes one cycle every 2*pi radians.

The amplitude is |a| in y = a*sin(x). Here amplitude = 3.

At 180 degrees, the x-coordinate on the unit circle is -1.

Period = 2*pi/|b| = 2*pi/2 = pi.

180 degrees = pi radians.

\(\tan(45) = \sin(45)/\cos(45) = 1\).

y = sin(x - c) shifts right by c. Phase shift is pi/3 right.

\(5\pi/6\) is in Q2, reference angle \(\pi/6\). \(\sin(\pi/6) = 1/2\), positive in Q2.

cos ranges from -1 to 1, so 2*cos ranges from -2 to 2, plus 1 gives -1 to 3.

Amplitude = |-4| = 4. Period = 2*pi/3.

sin(x) = 0 at x = 0, pi, and 2*pi.

At \(90^\circ\), the terminal point on the unit circle is \((0,1)\), and cosine equals the \(x\)-coordinate of that point, giving \(\cos(90^\circ)=0\). The distractor "\(1\)" confuses cosine with sine at this angle, since \(\sin(90^\circ)=1\) instead. Always remember cosine tracks the horizontal coordinate and sine tracks the vertical coordinate on the unit circle.

The point on the unit circle at \(180^\circ\) is \((-1,0)\), and since sine is the \(y\)-coordinate, \(\sin(180^\circ)=0\). The choice "\(-1\)" wrongly matches the \(x\)-coordinate, which is actually the cosine value at that angle. Recognizing that sine values are zero whenever the terminal point lies on the \(x\)-axis is key for locating zeros of sine.

At \(270^\circ\) the unit circle point is \((0,-1)\), so cosine, the \(x\)-coordinate, equals \(0\). The option "\(-1\)" incorrectly reflects the \(y\)-coordinate, which is the sine value at \(270^\circ\). A useful rule is that cosine equals zero at every angle where the terminal side lies along the vertical axis.

Since \(\tan(\theta)=\frac{\sin\theta}{\cos\theta}\) and \(\sin(0^\circ)=0\) while \(\cos(0^\circ)=1\), dividing gives \(\tan(0^\circ)=0\). The distractor "\(\text{undefined}\)" would apply only if cosine were zero, which occurs at \(90^\circ\), not \(0^\circ\). Always check that the denominator cosine is nonzero before evaluating tangent.

The cosine function repeats its full cycle every \(2\pi\) radians because it traces the \(x\)-coordinate around the entire unit circle exactly once in that interval. The distractor "\(\pi\)" only covers half a cycle, which is not enough to return to the starting value and direction. For the parent function \(y=\cos(x)\), the period is always \(2\pi\) unless a coefficient multiplies \(x\).

Sine is defined for every real-number input because the unit circle allows angles of any size, including negative angles and angles beyond \(2\pi\). The choice "\([-1,1]\)" actually describes the range of sine, not its domain. Students should remember that domain restrictions in trigonometric functions typically come from division (as in tangent), which sine does not involve.

Amplitude measures half the distance between the maximum and minimum values, and since \(y=\sin(x)\) ranges from \(-1\) to \(1\), the amplitude is \(1\). The distractor "\(2\)" would be the full range span, not half of it. Whenever a sine or cosine function has no coefficient in front, its amplitude defaults to \(1\).

The point on the unit circle corresponding to \(270^\circ\) is \((0,-1)\), and because sine gives the \(y\)-coordinate, \(\sin(270^\circ)=-1\). The option "\(0\)" mistakenly treats \(270^\circ\) like an angle on the \(x\)-axis, but it actually lies on the negative \(y\)-axis. Minimum sine values of \(-1\) always occur at \(270^\circ\) plus any multiple of \(360^\circ\).

A full rotation of \(360^\circ\) returns the terminal point to \((1,0)\), the same as \(0^\circ\), so \(\cos(360^\circ)=1\). The distractor "\(0\)" would apply at \(90^\circ\) or \(270^\circ\), not after a complete revolution. This illustrates the periodic nature of cosine, repeating identical values every \(360^\circ\).

By definition, the unit circle is the set of points exactly one unit away from the origin, so it is centered at \((0,0)\). The distractor "\((1,0)\)" is actually a point on the circle, not its center. Keeping the center fixed at the origin is what allows angles to be measured consistently starting from the positive \(x\)-axis.

An angle of \(0\) radians starts along the positive \(x\)-axis, and since the radius is \(1\), the terminal point is \((1,0)\). The option "\((0,1)\)" corresponds instead to an angle of \(\frac{\pi}{2}\), a quarter turn counterclockwise. Recognizing the starting point at \(0\) radians helps anchor the rest of the unit circle's key angles.

By the standard definition of the unit circle, the vertical coordinate of the terminal point equals \(\sin(\theta)\) for any angle \(\theta\). The distractor "\(\cos(\theta)\)" instead corresponds to the horizontal coordinate of that same point. This coordinate relationship is the foundation for deriving all sine and cosine values directly from the unit circle.

Since \(\sin(180^\circ)=0\) and \(\cos(180^\circ)=-1\), the ratio \(\tan(180^\circ)=\frac{0}{-1}=0\). The distractor "\(\text{undefined}\)" only applies when the cosine in the denominator is zero, which is not the case here. Tangent is undefined only at angles like \(90^\circ\) and \(270^\circ\) where cosine equals zero.

Amplitude is defined as the absolute value of the coefficient in front of the trig function, so for \(y=-2\cos(x)\) the amplitude is \(|-2|=2\). The distractor "\(-2\)" ignores the rule that amplitude must always be a nonnegative value representing distance, not direction. The negative sign only flips the graph vertically; it does not change how far the graph stretches from its midline.

The period of a cosine function is found using \(\frac{2\pi}{|b|}\), and with \(b=3\), the period becomes \(\frac{2\pi}{3}\). The distractor "\(3\pi\)" incorrectly multiplies instead of dividing by the coefficient inside the function. A larger coefficient on \(x\) always compresses the graph horizontally, producing a shorter period.

To convert degrees to radians, multiply by \(\frac{\pi}{180^\circ}\), so \(270^\circ \times \frac{\pi}{180^\circ}=\frac{3\pi}{2}\). The distractor "\(\frac{2\pi}{3}\)" comes from an arithmetic slip in simplifying the fraction rather than correctly reducing \(\frac{270}{180}\). Always simplify the fraction \(\frac{\text{degrees}}{180}\) fully before multiplying by \(\pi\) to avoid this kind of error.

To convert radians to degrees, multiply by \(\frac{180^\circ}{\pi}\), giving \(\frac{\pi}{4}\times\frac{180^\circ}{\pi}=45^\circ\). The distractor "\(90^\circ\)" would correspond to \(\frac{\pi}{2}\) radians, not \(\frac{\pi}{4}\). Memorizing that \(\frac{\pi}{4}\) radians equals \(45^\circ\) is essential since it appears frequently in unit circle problems.

On the unit circle, the point at \(30^\circ\) is \(\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)\), and since sine is the \(y\)-coordinate, \(\sin(30^\circ)=\frac{1}{2}\). The distractor "\(\frac{\sqrt{3}}{2}\)" is actually the cosine value at this same angle, showing how the two coordinates can be mixed up. Memorizing the special angle values for \(30^\circ\), \(45^\circ\), and \(60^\circ\) speeds up unit circle problem solving significantly.

The unit circle point for \(60^\circ\) is \(\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\), and cosine equals the \(x\)-coordinate, so \(\cos(60^\circ)=\frac{1}{2}\). The distractor "\(\frac{\sqrt{3}}{2}\)" belongs to the sine value at this angle instead. Notice that \(\sin(30^\circ)=\cos(60^\circ)\), an example of the complementary angle relationship in trigonometry.

Using \(\tan(\theta)=\frac{\sin\theta}{\cos\theta}\) with \(\sin(60^\circ)=\frac{\sqrt{3}}{2}\) and \(\cos(60^\circ)=\frac{1}{2}\), the ratio simplifies to \(\sqrt{3}\). The distractor "\(\frac{1}{\sqrt{3}}\)" is actually the reciprocal, corresponding to \(\tan(30^\circ)\) instead. Remembering that tangent values at \(30^\circ\) and \(60^\circ\) are reciprocals of each other helps avoid this common mix-up.

The angle \(\frac{\pi}{2}\) radians equals \(90^\circ\), where the unit circle point is \((0,1)\), so sine, the \(y\)-coordinate, equals \(1\). The distractor "\(0\)" would apply at \(\pi\) radians instead, where the point lies on the \(x\)-axis. Converting radians to familiar degree benchmarks like \(90^\circ\) makes evaluating these expressions much faster.

Since \(\frac{\pi}{3}\) radians equals \(60^\circ\), and the \(x\)-coordinate on the unit circle at that angle is \(\frac{1}{2}\), cosine equals \(\frac{1}{2}\). The distractor "\(\frac{\sqrt{3}}{2}\)" corresponds instead to the sine value at this same angle. Recognizing \(\frac{\pi}{3}\) as a common radian benchmark equal to \(60^\circ\) is essential for quick unit circle recall.

Using the period formula \(\frac{2\pi}{|b|}\) with \(b=\frac{1}{2}\), the period becomes \(2\pi \div \frac{1}{2}=4\pi\). The distractor "\(2\pi\)" mistakenly assumes the coefficient does not affect the period at all, ignoring the horizontal stretch caused by dividing \(x\) by \(2\). Whenever the coefficient on \(x\) is a fraction less than \(1\), the graph stretches horizontally, producing a longer period.

Amplitude equals the absolute value of the coefficient multiplying the trigonometric function, so for \(y=\frac{1}{2}\cos(x)\), the amplitude is \(\frac{1}{2}\). The distractor "\(1\)" ignores the coefficient entirely and assumes the parent function's amplitude instead. A coefficient with magnitude less than \(1\) compresses the graph vertically toward the midline.

Adding a constant outside the trig function shifts the entire graph vertically, so the midline moves from \(y=0\) to \(y=3\). The distractor "\(y=0\)" describes the midline of the untransformed \(y=\sin(x)\), before the vertical shift was applied. Whenever a constant \(k\) is added to a sine or cosine function, the new midline becomes \(y=k\).

The maximum of \(\sin(x)\) is \(1\), so substituting gives \(2(1)-1=1\) as the maximum value of the transformed function. The distractor "\(2\)" mistakenly treats the amplitude alone as the final maximum, forgetting to subtract the vertical shift. To find the true maximum, always compute (midline) + (amplitude) rather than using the amplitude by itself.

The minimum of \(\cos(x)\) is \(-1\), so plugging in gives \(3(-1)+2=-1\) as the minimum of the shifted function. The distractor "\(-3\)" only accounts for the amplitude and ignores the vertical shift of \(+2\) added afterward. The correct approach is always (midline) \(-\) (amplitude) to find the minimum value of a transformed sine or cosine function.

An angle of \(150^\circ\) falls between \(90^\circ\) and \(180^\circ\), which is the range defining Quadrant II on the unit circle. The distractor "Quadrant I" only covers angles between \(0^\circ\) and \(90^\circ\), which does not include \(150^\circ\). Memorizing the degree ranges for each quadrant helps quickly determine the sign of sine and cosine for any given angle.

The angle \(\frac{2\pi}{3}\) radians equals \(120^\circ\), a reference angle of \(60^\circ\) in Quadrant II, where sine remains positive, giving \(\sin\left(\frac{2\pi}{3}\right)=\frac{\sqrt{3}}{2}\). The distractor "\(\frac{1}{2}\)" is the sine value for the reference angle \(30^\circ\), not \(60^\circ\), showing the importance of correctly identifying the reference angle. Since sine is positive throughout Quadrant II, only the reference angle's value matters, with no sign change needed.

The angle \(\frac{5\pi}{4}\) radians equals \(225^\circ\), which lies in Quadrant III where cosine is negative, and its reference angle of \(45^\circ\) gives a magnitude of \(\frac{\sqrt{2}}{2}\), so \(\cos\left(\frac{5\pi}{4}\right)=-\frac{\sqrt{2}}{2}\). The distractor "\(\frac{\sqrt{2}}{2}\)" ignores the negative sign that Quadrant III requires for cosine values. Always determine both the reference angle's magnitude and the correct sign based on the quadrant before finalizing a trig value.

Applying the period formula \(\frac{2\pi}{|b|}\) with \(b=\frac{1}{3}\) gives \(2\pi \div \frac{1}{3}=6\pi\). The distractor "\(3\pi\)" results from an incomplete division, using the coefficient as a direct multiplier rather than correctly dividing \(2\pi\) by the fraction. A fractional coefficient less than \(1\) inside a trig function always stretches the graph and produces a period longer than \(2\pi\).

Adding a constant outside the sine function moves every point on the graph up by that amount, so \(y=\sin(x)+2\) shifts the parent graph vertically upward by \(2\) units. The distractor "It is stretched vertically by a factor of \(2\)" describes what would happen if \(2\) multiplied \(\sin(x)\) instead of being added to it. Distinguishing between constants added outside the function, which shift the graph, and coefficients multiplying the function, which stretch it, is critical for accurate graphing.

An amplitude of \(3\) requires the coefficient in front to be \(3\), and solving \(\frac{2\pi}{b}=\pi\) gives \(b=2\), so the correct equation is \(y=3\sin(2x)\). The distractor "\(y=3\sin(x)\)" has the correct amplitude but the wrong period, since its period is \(2\pi\) rather than \(\pi\). Building a sine equation from given amplitude and period values requires solving the period formula for \(b\) separately from identifying the amplitude coefficient.

With amplitude \(3\) and midline \(y=-2\), the maximum is \(-2+3=1\) and the minimum is \(-2-3=-5\), giving the range \([-5,1]\); note the negative sign on the amplitude flips the graph but the range calculation still uses \(|{-3}|=3\). The distractor "\([-3,3]\)" incorrectly ignores the vertical shift of \(-2\) entirely. Always compute both midline plus amplitude and midline minus amplitude to find the true maximum and minimum of a shifted trig function.

Cosine equals zero wherever the terminal point on the unit circle lies on the vertical axis, which happens at \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\) within one full rotation. The distractor "\(x=0, \pi\)" instead lists the points where sine equals zero, since those angles fall on the horizontal axis. Distinguishing zeros of sine from zeros of cosine depends on recognizing which axis the terminal point touches.

Rewriting the argument as \(x-\left(-\frac{\pi}{4}\right)\) shows the phase shift is \(-\frac{\pi}{4}\), meaning a shift left, and the constant \(-1\) outside the function gives a vertical shift of \(-1\). The distractor "Phase shift \(\frac{\pi}{4}\)" mistakenly reads the shift as rightward because it overlooks that addition inside the parentheses corresponds to a leftward shift. Always rewrite the argument in the form \(x-h\) before identifying whether the phase shift is left or right.

The amplitude is the coefficient \(5\); the period comes from \(\frac{2\pi}{2}=\pi\); and factoring the argument as \(2\left(x-\frac{\pi}{4}\right)\) reveals a phase shift of \(\frac{\pi}{4}\) to the right. The distractor "phase shift \(\frac{\pi}{2}\) right" fails to factor out the coefficient of \(2\) before reading the shift, mistakenly using the raw constant \(\frac{\pi}{2}\) instead. Whenever the coefficient on \(x\) is not \(1\), always factor it out of the parentheses before identifying the true phase shift.

Using the period formula \(\frac{2\pi}{|b|}\) with \(b=\frac{\pi}{2}\), the period simplifies to \(\frac{2\pi}{\pi/2}=4\). The distractor "\(\pi\)" incorrectly treats the coefficient as if it were a plain number rather than one containing \(\pi\), causing the \(\pi\) terms to cancel incorrectly. When the coefficient inside a trig function itself contains \(\pi\), the resulting period is often a rational number rather than a multiple of \(\pi\).

The angle \(\frac{11\pi}{6}\) radians equals \(330^\circ\), which lies in Quadrant IV where cosine is positive, and its reference angle of \(30^\circ\) gives a magnitude of \(\frac{\sqrt{3}}{2}\), so the value is \(\frac{\sqrt{3}}{2}\). The distractor "\(-\frac{\sqrt{3}}{2}\)" applies the wrong sign, as if the angle were in Quadrant III instead of Quadrant IV. Knowing that cosine is positive in both Quadrant I and Quadrant IV prevents this common sign error.

Since sine is negative in Quadrants III and IV, and the reference angle for \(\sin^{-1}\left(\frac{1}{2}\right)\) is \(\frac{\pi}{6}\), the solutions are \(\pi+\frac{\pi}{6}=\frac{7\pi}{6}\) and \(2\pi-\frac{\pi}{6}=\frac{11\pi}{6}\). The distractor "\(x=\frac{\pi}{6}, \frac{5\pi}{6}\)" gives the solutions for \(\sin(x)=\frac{1}{2}\) in Quadrants I and II, the wrong sign entirely. Solving trig equations always requires first identifying which quadrants match the required sign before applying the reference angle.

Shifting sine right by \(\frac{\pi}{2}\) produces the identity \(\sin\left(x-\frac{\pi}{2}\right)=-\cos(x)\)... wait, correction needed—verify actual identity: \(\sin(x-\pi/2) = -\cos(x)\), so they're not identical. The distractor "reflection over the \(x\)-axis" is actually correct here, not a distractor.

The maximum of any sine function occurs when the sine term equals \(1\), so the maximum value here is \(2(1)+4=6\). The distractor "\(2\)" only reports the amplitude itself and forgets to add the vertical shift of \(4\) that raises the entire graph. Regardless of the phase shift or period, the maximum of a sine or cosine function always equals midline plus amplitude.

Amplitude is the absolute value of the coefficient in front of the trig function, so \(y=-5\cos(x)\) has amplitude \(|-5|=5\), which is greater than the amplitude \(4\) of \(y=4\sin(2x)\). The distractor "\(y=4\sin(2x)\), with amplitude \(4\)" correctly identifies its own amplitude but fails to compare it accurately against the larger value of \(5\). Comparing amplitudes always involves comparing absolute values of the leading coefficients, regardless of any negative signs or coefficients inside the function affecting the period.

An amplitude of \(2\) sets the coefficient in front to \(2\), solving \(\frac{2\pi}{b}=\pi\) gives \(b=2\), and cosine naturally reaches its maximum at \(x=0\) with no phase shift needed, producing \(y=2\cos(2x)\). The distractor "\(y=2\sin(2x)\)" has the correct amplitude and period but reaches zero, not its maximum, at \(x=0\), since sine starts at the midline. Recognizing that cosine graphs naturally start at a maximum while sine graphs start at the midline is key to matching equations to graph descriptions.

The angle \(\frac{5\pi}{4}\) radians equals \(225^\circ\), lying in Quadrant III where both sine and cosine are negative, so their ratio, tangent, becomes positive, giving \(\tan\left(\frac{5\pi}{4}\right)=1\). The distractor "\(-1\)" would apply in Quadrant II or IV where sine and cosine have opposite signs, but that is not the case here. Since tangent is the ratio of sine to cosine, it is positive whenever both sine and cosine share the same sign, as they do throughout Quadrant III.