Trigonometric Functions — Free Pre-Calculus Review Games.

60 * pi/180 = pi/3.

pi/2 = 90 degrees, sin(90) = 1.

pi = 180 degrees, cos(180) = -1.

The standard cosine function has period 2*pi.

Amplitude = |-5| = 5.

The tangent function has period pi.

Period = 2*pi/4 = pi/2.

\(5\pi/4\) is in Q3, reference angle \(\pi/4\). \(\sin\) is negative in Q3: \(-\sqrt{2}/2\).

Subtracting 4 from the function shifts the graph down 4 units.

\(2\pi/3\) is in Q2, reference angle \(\pi/3\). \(\cos(\pi/3)=1/2\), negative in Q2: \(-1/2\).

Amplitude 3, period pi means b=2, phase shift pi/4 right: y=3*sin(2(x-pi/4)).

tan(x) is undefined where cos(x) = 0: x = pi/2 + n*pi.

3*sin(x) ranges from -3 to 3, minus 2 gives -5 to 1.

\(7\pi/6\) is in Q3, reference angle \(\pi/6\). \(\sin(\pi/6) = 1/2\), negative in Q3: \(-1/2\).

Both sec(x) and tan(x) are undefined where cos(x)=0, i.e., at x = pi/2 + n*pi.

At angle \(0\) the point on the unit circle is \((1,0)\), and sine equals the \(y\)-coordinate of that point, giving \(\sin(0)=0\). The distractor \(1\) is wrong because that value corresponds to \(\sin(\pi/2)\), not \(\sin(0)\). Always remember that sine values come directly from the \(y\)-coordinate of the unit circle point at the given angle.

Cosine equals the \(x\)-coordinate of the unit circle point, and at angle \(0\) that point is \((1,0)\), so \(\cos(0)=1\). The distractor \(0\) is incorrect because that is the sine value at this angle, not the cosine value. This distinction between sine as \(y\)-coordinate and cosine as \(x\)-coordinate is essential for evaluating any unit circle angle.

At \(\pi/4\), \(\sin(\pi/4)=\cos(\pi/4)=\frac{\sqrt{2}}{2}\), so \(\tan(\pi/4)=\frac{\sin(\pi/4)}{\cos(\pi/4)}=1\). The distractor "undefined" is wrong because tangent is only undefined where cosine equals zero, which does not happen at \(\pi/4\). Recognizing that equal sine and cosine values always yield a tangent of \(1\) is a useful shortcut on the unit circle.

Multiplying \(\pi/3\) by the conversion factor \(\frac{180^\circ}{\pi}\) gives \(60^\circ\), since radians and degrees are proportional through \(\pi\) radians equaling \(180^\circ\). The distractor \(90^\circ\) is wrong because that corresponds to \(\pi/2\) radians, not \(\pi/3\). Always use the \(\frac{180^\circ}{\pi}\) conversion factor consistently when switching between radians and degrees.

Multiplying \(90^\circ\) by \(\frac{\pi}{180^\circ}\) gives \(\pi/2\) radians, the standard conversion between the two angle units. The distractor \(\pi\) is wrong because that value represents \(180^\circ\), not \(90^\circ\). Memorizing key benchmark angles like \(90^\circ=\pi/2\) speeds up unit circle work considerably.

The sine function completes one full cycle every \(2\pi\) radians because its graph repeats identically after that interval on the unit circle. The distractor \(\pi\) is wrong because that is only half of one complete sine cycle. The basic period of \(2\pi\) serves as the reference value for calculating periods of transformed sine and cosine functions.

Amplitude is the absolute value of the coefficient in front of the trig function, so for \(y=2\sin(x)\) the amplitude is \(|2|=2\). The distractor \(1\) is wrong because that would be the amplitude of the untransformed \(y=\sin(x)\), not the scaled version. Amplitude always measures the maximum distance the graph reaches from its midline.

At angle \(\pi\) the unit circle point is \((-1,0)\), and since sine is the \(y\)-coordinate, \(\sin(\pi)=0\). The distractor \(-1\) is wrong because that is the \(x\)-coordinate at this angle, which represents cosine, not sine. Knowing the exact coordinates at multiples of \(\pi\) avoids common sign errors.

The unit circle point at \(\pi/2\) is \((0,1)\), and cosine is the \(x\)-coordinate, so \(\cos(\pi/2)=0\). The distractor \(1\) is wrong because that value is the \(y\)-coordinate, which is sine at this angle. This is a classic point where students confuse sine and cosine, so double-checking which coordinate is asked for is important.

Since sine values are \(y\)-coordinates on the unit circle, which has radius \(1\), the output of \(\sin(x)\) never exceeds \(1\) or drops below \(-1\), giving the range \([-1,1]\). The distractor \((-\infty,\infty)\) is wrong because that describes the domain of sine, not its bounded range. Every unscaled sine or cosine function has this same bounded range of \([-1,1]\).

Cosine is defined for every real number input because the unit circle can be traversed indefinitely in either direction without restriction. The distractor "\(x\neq \pi/2+n\pi\)" is wrong because that describes the domain restriction of tangent, not cosine. Unlike tangent, sine and cosine have no domain restrictions since they never involve division by zero.

After a full revolution of \(2\pi\), the unit circle point returns to \((1,0)\), so \(\cos(2\pi)=1\), matching \(\cos(0)\). The distractor \(0\) is wrong because that would only be true at quarter-turn angles like \(\pi/2\). Because sine and cosine are periodic with period \(2\pi\), values repeat exactly after every full rotation.

Adding a constant outside the trig function shifts the entire graph vertically, so the midline moves from \(y=0\) to \(y=3\). The distractor \(y=0\) is wrong because that is the midline of the untransformed \(y=\sin(x)\) before the vertical shift was applied. The vertical shift constant added to a sinusoidal function always equals the new midline value.

The period formula for cosine is \(\frac{2\pi}{|b|}\), and with \(b=3\) this gives \(\frac{2\pi}{3}\). The distractor \(3\pi\) is wrong because it results from multiplying rather than dividing \(2\pi\) by \(3\). Increasing the coefficient \(b\) always compresses the graph horizontally, shortening the period.

In the form \(y=\sin(x-h)\), the graph shifts horizontally by \(h\) units, and here \(h=\pi/4\) produces a shift to the right. The distractor "\(\pi/4\) left" is wrong because a subtraction inside the argument moves the graph right, not left. Remember that \(x-h\) shifts right while \(x+h\) shifts left, which is often the opposite of student intuition.

Amplitude is defined as the absolute value of the leading coefficient, so \(|-3|=3\) regardless of the negative sign. The distractor \(-3\) is wrong because amplitude is always a nonnegative distance, not a signed number. The negative sign instead causes a vertical reflection of the graph, not a change in amplitude.

The constant added at the end of the function, here \(+1\), moves the entire graph vertically upward by \(1\) unit. The distractor "up \(2\)" is wrong because \(2\) is the amplitude coefficient in front of sine, not the vertical shift constant. Vertical shift and amplitude are separate transformations controlled by different parts of the equation.

Since \(\sin(\pi/3)=\frac{\sqrt{3}}{2}\) and \(\cos(\pi/3)=\frac{1}{2}\), dividing gives \(\tan(\pi/3)=\sqrt{3}\). The distractor \(1/\sqrt{3}\) is wrong because that is actually the value of \(\tan(\pi/6)\), a commonly confused reciprocal pair. Memorizing the tangent values at \(30^\circ\), \(45^\circ\), and \(60^\circ\) prevents mixing up these related angles.

On the unit circle, the point at \(\pi/6\) is \(\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)\), and sine equals the \(y\)-coordinate, giving \(\sin(\pi/6)=1/2\). The distractor \(\sqrt{3}/2\) is wrong because that is the \(x\)-coordinate, representing \(\cos(\pi/6)\) instead. Knowing the standard \(30\)-\(60\)-\(90\) coordinate pairs is essential for quick unit circle evaluation.

Cosine is an even function, meaning \(\cos(-x)=\cos(x)\), so \(\cos(-\pi/3)=\cos(\pi/3)=1/2\). The distractor \(-1/2\) is wrong because it incorrectly treats cosine as an odd function, which applies to sine and tangent instead. Recognizing cosine's even symmetry allows quick evaluation of negative angle expressions.

Using the period formula \(\frac{2\pi}{|b|}\) with \(b=1/2\) gives \(\frac{2\pi}{1/2}=4\pi\). The distractor \(2\pi\) is wrong because it ignores the horizontal stretch caused by the coefficient \(1/2\) inside the sine function. A coefficient less than \(1\) inside the argument always stretches the graph, increasing the period beyond \(2\pi\).

Rewriting \(2x+\pi\) as \(2(x+\pi/2)\) shows that the shift is \(\pi/2\) units to the left, since the graph moves opposite the sign inside the parentheses. The distractor "\(\pi\) left" is wrong because it forgets to factor out the coefficient \(2\) before identifying the shift amount. Always factor the coefficient of \(x\) out first before reading off the phase shift value.

Since \(5\pi/3\) lies in the fourth quadrant, its reference angle is found by subtracting from a full rotation: \(2\pi-5\pi/3=\pi/3\). The distractor \(2\pi/3\) is wrong because that formula would apply if the angle were in the second quadrant, not the fourth. Reference angles always measure the acute angle between the terminal side and the nearest \(x\)-axis.

Since \(\sec(x)=\frac{1}{\cos(x)}\) and \(\cos(\pi/3)=\frac{1}{2}\), taking the reciprocal gives \(\sec(\pi/3)=2\). The distractor \(1/2\) is wrong because it mistakenly uses the cosine value itself instead of taking its reciprocal. Secant, cosecant, and cotangent are always found by taking reciprocals of cosine, sine, and tangent respectively.

Since \(\csc(x)=\frac{1}{\sin(x)}\) and \(\sin(\pi/2)=1\), the reciprocal gives \(\csc(\pi/2)=1\). The distractor "undefined" is wrong because cosecant is only undefined when sine equals zero, which is not the case at \(\pi/2\). Watching for where the reciprocal function's denominator becomes zero is key to identifying undefined values correctly.

The angle \(2\pi/3\) lies in the second quadrant with reference angle \(\pi/3\), so cosine is negative and sine is positive, giving the point \((-1/2,\sqrt{3}/2)\). The distractor \((1/2,\sqrt{3}/2)\) is wrong because it keeps cosine positive, which would only be correct in the first quadrant. Always apply the correct sign pattern for each quadrant (ASTC) when finding coordinates from a reference angle.

A rightward horizontal shift of \(\pi/2\) is achieved by subtracting \(\pi/2\) inside the argument, giving \(y=\sin(x-\pi/2)\). The distractor \(y=\sin(x)-\pi/2\) is wrong because subtracting outside the function produces a vertical shift, not a horizontal one. Transformations inside the parentheses affect horizontal position, while transformations outside affect vertical position.

The period formula for tangent is \(\frac{\pi}{|b|}\), and with \(b=2\) this gives \(\pi/2\). The distractor \(\pi\) is wrong because that is the untransformed period of tangent before applying the horizontal compression from \(b=2\). Note that tangent's base period is \(\pi\), unlike sine and cosine which have a base period of \(2\pi\).

Since \(\cot(x)=\frac{\cos(x)}{\sin(x)}\) and both \(\sin(\pi/4)\) and \(\cos(\pi/4)\) equal \(\frac{\sqrt{2}}{2}\), the ratio simplifies to \(\cot(\pi/4)=1\). The distractor "undefined" is wrong because cotangent is only undefined where sine equals zero, which does not occur at \(\pi/4\). When sine and cosine are equal at an angle, both tangent and cotangent evaluate to \(1\).

A right shift of \(\pi/3\) requires subtracting \(\pi/3\) inside the argument, and an upward shift of \(2\) requires adding \(2\) outside the function, producing \(y=\sin(x-\pi/3)+2\). The distractor \(y=\sin(x+\pi/3)+2\) is wrong because adding inside the parentheses shifts the graph left instead of right. Combining transformations requires applying the horizontal shift inside and the vertical shift outside the trig function separately.

Sine equals zero wherever the unit circle point has a \(y\)-coordinate of zero, which occurs at \(x=0\), \(x=\pi\), and \(x=2\pi\) within this interval. The distractor \(\pi/2,3\pi/2\) is wrong because those are the locations of the maximum and minimum values, where sine equals \(\pm1\), not zero. Sine crosses zero at every multiple of \(\pi\), a pattern that repeats throughout its entire domain.

The maximum of sine is \(1\), so substituting into the function gives \(4(1)+1=5\) as the maximum value of the transformed graph. The distractor \(4\) is wrong because it omits adding the vertical shift constant after scaling by the amplitude. The maximum of any sinusoidal function equals the vertical shift plus the amplitude.

A period of \(\pi/2\) requires \(b=\frac{2\pi}{\pi/2}=4\), and factoring the shift correctly as \(4(x-\pi/4)\) ensures the phase shift is exactly \(\pi/4\) right, matching all given specifications. The distractor \(y=2\sin(4x-\pi/4)\) is wrong because it leaves the shift unfactored, which actually produces a phase shift of only \(\pi/16\), not \(\pi/4\). Always factor out the coefficient of \(x\) before identifying the true phase shift in a transformed trig function.

Sine equals \(1/2\) in the first and second quadrants where the reference angle is \(\pi/6\), giving solutions \(\pi/6\) and \(\pi-\pi/6=5\pi/6\). The distractor \(\pi/6,7\pi/6\) is wrong because \(7\pi/6\) lies in the third quadrant, where sine is negative, not positive. Since sine is positive only in quadrants one and two, solutions to \(\sin(x)=\) positive value must come from those two quadrants.

Factoring the argument gives \(3x-\pi/2=3\left(x-\pi/6\right)\), so the phase shift is \(\pi/6\) to the right after dividing \(\pi/2\) by the coefficient \(3\). The distractor \(\pi/2\) right is wrong because it uses the unfactored constant directly, ignoring the necessary division by the coefficient of \(x\). Failing to factor out the horizontal stretch coefficient is one of the most common phase shift errors students make.

Amplitude equals half the distance between max and min, \(\frac{5-(-1)}{2}=3\), and vertical shift equals the average of max and min, \(\frac{5+(-1)}{2}=2\). The distractor "amplitude \(6\), shift up \(2\)" is wrong because it uses the full max-min distance rather than half of it for the amplitude. Extracting amplitude and midline from graph extrema is a core skill for writing sinusoidal equations from data.

Factoring gives \(2x+\pi/3=2\left(x+\pi/6\right)\), so the graph shifts \(\pi/6\) units to the left since the sign inside is positive. The distractor \(\pi/3\) left is wrong because it fails to divide the constant by the coefficient \(2\) before determining the shift. Always divide the added constant by the leading coefficient inside the parentheses to find the true phase shift.

Substituting \(x=\pi\) gives \(y=3\cos(\pi-\pi/2)=3\cos(\pi/2)=3(0)=0\) since cosine of \(\pi/2\) equals zero. The distractor \(3\) is wrong because it incorrectly assumes cosine equals \(1\) at this shifted input rather than correctly evaluating the shifted argument first. Always simplify the argument of the trig function completely before evaluating, since phase shifts change which angle is actually being evaluated.

The period of \(y=\sin(4x)\) is \(\frac{2\pi}{4}=\frac{\pi}{2}\), and dividing the total interval length \(2\pi\) by this period gives \(\frac{2\pi}{\pi/2}=4\) complete cycles. The distractor \(2\) is wrong because it likely results from confusing the coefficient \(4\) with a doubling effect rather than computing the actual period first. To find the number of cycles in an interval, always divide the interval length by the computed period, not by the coefficient directly.

Solving \(\frac{2\pi}{|b|}=4\pi\) gives \(b=1/2\), and pairing this with an amplitude coefficient of \(1/2\) produces \(y=\frac{1}{2}\cos(x/2)\). The distractor \(y=\frac{1}{2}\cos(2x)\) is wrong because \(b=2\) gives a period of \(\pi\), which is much shorter than the required \(4\pi\). Solving the period equation for \(b\) before assembling the full equation prevents mixing up stretch and compression.

The angle \(11\pi/6\) lies in the fourth quadrant with reference angle \(\pi/6\), and since cosine is positive in the fourth quadrant, \(\cos(11\pi/6)=\sqrt{3}/2\). The distractor \(-\sqrt{3}/2\) is wrong because it applies the sign pattern of the second quadrant instead of the fourth. Cosine is positive in quadrants one and four, a fact essential for correctly signing reference angle values.

Tangent has asymptotes where its argument equals \(\pi/2+n\pi\), so solving \(x-\pi/4=-\pi/2\) gives \(x=-\pi/4\), which is closer to the origin than the next asymptote at \(x=3\pi/4\). The distractor \(x=\pi/4\) is wrong because that is the phase shift value itself, not a location where tangent is undefined. Locating tangent asymptotes requires solving the shifted argument equal to \(\pi/2+n\pi\) for multiple integer values of \(n\), then comparing distances.

Since \(\cos(x)\) ranges over \([-1,1]\), multiplying by \(-2\) reverses and scales the range to \([-2,2]\), and adding \(3\) shifts it to \([1,5]\). The distractor \([-2,2]\) is wrong because it stops after the amplitude scaling step and forgets to apply the vertical shift of \(3\). Always apply amplitude scaling first, including any sign flip, and then apply the vertical shift last when finding the range.

Using the period formula \(\frac{2\pi}{|b|}\) with \(b=1/3\) gives \(\frac{2\pi}{1/3}=6\pi\). The distractor \(2\pi/3\) is wrong because it incorrectly multiplies rather than divides by the reciprocal of \(b\). A coefficient smaller than \(1\) inside the argument always stretches the period well beyond the standard \(2\pi\).

Solving gives \(\cos(x)=\sqrt{3}/2\), which occurs where cosine is positive in the first and fourth quadrants, at reference angle \(\pi/6\), yielding \(x=\pi/6\) and \(x=2\pi-\pi/6=11\pi/6\). The distractor \(\pi/6,5\pi/6\) is wrong because \(5\pi/6\) lies in the second quadrant where cosine is negative, not positive. Since cosine is positive only in quadrants one and four, solutions must be drawn from those two regions of the unit circle.