Unit Circle — Free Trigonometry Review Games.

At 0 radians, the point is (1,0), so cos = 1.

At pi/2, the point is (0,1), so sin = 1.

Reference angle = 180 - 150 = 30 degrees.

In Q2, x is negative (cos) and y is positive (sin).

At pi (180 degrees), the point is (-1,0), so sin = 0.

\(\pi/3 = 60\) degrees. \(\cos(60) = 1/2\).

\(\pi/4 = 45\) degrees. \(\sin(45) = \sqrt{2}/2\).

5*pi/6 is in Q2. Reference angle = pi - 5*pi/6 = pi/6.

\(\sin(\pi/4)/\cos(\pi/4) = (\sqrt{2}/2)/(\sqrt{2}/2) = 1\).

At 3*pi/2 (270 degrees), the point is (0,-1), so cos = 0.

\(7\pi/4\) is in Q4, reference angle \(\pi/4\). sin is negative in Q4: \(-\sqrt{2}/2\).

\(4\pi/3\) is in Q3, reference angle \(\pi/3\). \(\cos(\pi/3)=1/2\), negative in Q3: \(-1/2\).

\(2\pi/3\) is in Q2. \(\sin=\sqrt{3}/2\), \(\cos=-1/2\). \(\tan = -\sqrt{3}\).

\(\sin(\pi/6) = 1/2\), so \(\csc(\pi/6) = 1/(1/2) = 2\).

sin = -1/2 in Q3 and Q4: pi+pi/6=7*pi/6 and 2*pi-pi/6=11*pi/6.

At \(\pi/2\) the terminal point on the unit circle is \((0,1)\), and cosine equals the x-coordinate, giving \(0\). The choice \(1\) is wrong because that is the y-coordinate value at this angle, which corresponds to sine, not cosine. Always remember cosine reads the horizontal coordinate and sine reads the vertical coordinate of the point where the terminal ray meets the unit circle.

At angle \(0\) the terminal point is \((1,0)\), and sine equals the y-coordinate, which is \(0\). The distractor \(1\) is wrong because that is the x-coordinate at this angle, which is the cosine value instead. Starting angles on the unit circle always begin at \((1,0)\), a fact worth memorizing as your anchor point.

Tangent equals \(\sin(\theta)/\cos(\theta)\), and at \(\theta = 0\) this is \(0/1 = 0\). The choice "undefined" is wrong because tangent is only undefined when cosine equals zero, which happens at \(\pi/2\), not at \(0\). Remember tangent is a ratio, so you must check both sine and cosine values before evaluating it.

The reference angle is the acute angle to the nearest x-axis, and since \(210°\) is in Quadrant III, the reference angle is \(210° - 180° = 30°\). The choice \(150°\) is wrong because that would be the reference angle formula for a Quadrant II angle, not Quadrant III. Always identify the quadrant first, since the subtraction rule for finding the reference angle differs by quadrant.

Since \(300°\) lies in Quadrant IV, the reference angle is found by \(360° - 300° = 60°\). The distractor \(30°\) is wrong because that value would result from an incorrect subtraction, not the correct Quadrant IV rule of subtracting from \(360°\). Quadrant IV angles always use \(360°\) minus the angle to find the reference angle.

In Quadrant III both the x-coordinate and y-coordinate on the unit circle are negative, so both cosine and sine are negative there. Quadrant II is wrong because there only cosine is negative while sine remains positive. The mnemonic "All Students Take Calculus" helps track which functions are positive in each quadrant, so its complement tells you when both are negative.

In Quadrant IV the x-coordinate is positive and the y-coordinate is negative, so cosine is positive and sine is negative. Quadrant III is wrong because there both sine and cosine are negative, not just one of them. Knowing the sign pattern by quadrant lets you quickly check whether a computed trig value makes sense.

At \(\theta = \pi\) the terminal point on the unit circle is \((-1, 0)\), so cosine, the x-coordinate, equals \(-1\). The choice \(0\) is wrong because that would be the sine value at this angle, since the y-coordinate is \(0\). The point \((-1,0)\) marking \(\pi\) radians should be memorized alongside the other three axis points.

At \(3\pi/2\) the terminal point is \((0,-1)\), and since sine is the y-coordinate, \(\sin(3\pi/2) = -1\). The choice \(1\) is wrong because it ignores the direction of rotation, placing the point incorrectly at the top of the circle instead of the bottom. The four quadrantal angles \(0, \pi/2, \pi, 3\pi/2\) each have simple coordinate pairs that are essential to memorize.

Since \(\sin(\pi) = 0\) and \(\cos(\pi) = -1\), the tangent is \(0 / (-1) = 0\). The answer "undefined" is wrong because tangent is undefined only when the denominator cosine is zero, which is not the case here. Always compute both sine and cosine numerators before concluding tangent is undefined.

Since \(2\pi/3\) is in Quadrant II, the reference angle is \(\pi - 2\pi/3 = \pi/3\). The choice \(\pi/6\) is wrong because it does not result from subtracting \(2\pi/3\) from \(\pi\) using the correct Quadrant II formula. For Quadrant II angles, always subtract the angle from \(\pi\) to find the reference angle.

The unit circle is defined as the circle centered at the origin with radius exactly \(1\), which is why coordinates on it directly equal cosine and sine values without needing to divide by radius. The choice \(2\) is wrong because a radius of \(2\) would require dividing coordinates by \(2\) to get the trig ratios, contradicting the definition of the unit circle. This radius-1 property is the entire reason the unit circle simplifies trigonometric definitions to just reading coordinates.

An angle of \(2\pi\) is a full revolution, returning to the starting point \((1,0)\), so cosine equals \(1\). The choice \(0\) is wrong because it corresponds to the sine value at this angle, not cosine. Since \(2\pi\) represents a complete rotation, all trig values at \(2\pi\) match those at \(0\).

The angle \(\pi/3\) corresponds to the unit circle point \((\frac{1}{2}, \frac{\sqrt{3}}{2})\), and since sine is the y-coordinate, \(\sin(\pi/3) = \frac{\sqrt{3}}{2}\). The choice \(\frac{1}{2}\) is wrong because that is the x-coordinate at this angle, which is the cosine value instead. The 30-60-90 triangle pattern generates this pair of values, and swapping sine and cosine is a common error to avoid.

At \(\pi/4\), the point on the unit circle is \((\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})\) from the 45-45-90 triangle relationship, so cosine equals \(\frac{\sqrt{2}}{2}\). The choice \(\frac{\sqrt{3}}{2}\) is wrong because that value belongs to the 30-60-90 triangle family used for angles like \(\pi/6\) and \(\pi/3\), not \(\pi/4\). Because \(\pi/4\) lies exactly on the line \(y=x\) within Quadrant I, its sine and cosine values are always equal.

Since \(\sin(\pi/3) = \frac{\sqrt{3}}{2}\) and \(\cos(\pi/3) = \frac{1}{2}\), dividing gives \(\tan(\pi/3) = \sqrt{3}\). The choice \(\frac{\sqrt{3}}{3}\) is wrong because that is actually \(\tan(\pi/6)\), the reciprocal relationship of this angle's cofunction pair. Students should be careful not to confuse \(\pi/6\) and \(\pi/3\) tangent values since they are reciprocals of each other.

The angle \(5\pi/6\) lies in Quadrant II with reference angle \(\pi/6\), and since sine is positive in Quadrant II, \(\sin(5\pi/6) = \sin(\pi/6) = \frac{1}{2}\). The choice \(-\frac{1}{2}\) is wrong because it incorrectly applies a negative sign, ignoring that sine stays positive throughout Quadrant II. Finding the reference angle and then applying the correct quadrant sign is the two-step process needed for any non-Quadrant-I angle.

The angle \(2\pi/3\) has reference angle \(\pi/3\), and since cosine is negative in Quadrant II, \(\cos(2\pi/3) = -\cos(\pi/3) = -\frac{1}{2}\). The choice \(\frac{1}{2}\) is wrong because it omits the necessary negative sign that Quadrant II cosine values require. Always attach the correct sign based on the quadrant after computing the reference angle's value.

Since \(\sin(\pi/6) = \frac{1}{2}\) and \(\cos(\pi/6) = \frac{\sqrt{3}}{2}\), the tangent is \(\frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\) after rationalizing. The choice \(\sqrt{3}\) is wrong because that is the reciprocal value \(\tan(\pi/3)\), not \(\tan(\pi/6)\). Rationalizing the denominator is required to express tangent values in the standard simplified radical form.

Cosecant is the reciprocal of sine, and since \(\sin(\pi/6) = \frac{1}{2}\), we get \(\csc(\pi/6) = \frac{1}{1/2} = 2\). The choice \(\frac{1}{2}\) is wrong because that is simply the sine value itself, not its reciprocal. Remember that csc, sec, and cot are defined as reciprocals of sin, cos, and tan respectively, so always flip the fraction.

Secant is the reciprocal of cosine, and since \(\cos(\pi/3) = \frac{1}{2}\), we get \(\sec(\pi/3) = \frac{1}{1/2} = 2\). The choice \(\frac{2\sqrt{3}}{3}\) is wrong because that is actually \(\sec(\pi/6)\), the reciprocal of a different angle's cosine value. Careful bookkeeping of which angle you're computing the reciprocal for prevents mixing up these standard values.

Since \(\tan(\pi/4) = 1\), cotangent, its reciprocal, is also \(\frac{1}{1} = 1\). The choice \(\sqrt{2}\) is wrong because that is the value of \(\sec(\pi/4)\) or \(\csc(\pi/4)\), not cotangent. Because \(\pi/4\) is the special angle where sine equals cosine, both tangent and cotangent equal exactly \(1\) there.

Since \(11\pi/6\) is in Quadrant IV, the reference angle is \(2\pi - 11\pi/6 = \pi/6\). The choice \(5\pi/6\) is wrong because that value would be obtained using the Quadrant II subtraction rule, which does not apply to a Quadrant IV angle. Always match the correct subtraction formula, \(2\pi\) minus the angle, to Quadrant IV angles specifically.

A negative angle of \(-45°\) is coterminal with \(315°\), which lies in Quadrant IV, giving a reference angle of \(360° - 315° = 45°\). The choice \(135°\) is wrong because it incorrectly treats the angle as if it were in Quadrant II rather than Quadrant IV. When working with negative angles, first find a coterminal positive angle before applying the reference angle rules.

Sine is an odd function, meaning \(\sin(-\theta) = -\sin(\theta)\), so \(\sin(-\pi/6) = -\sin(\pi/6) = -\frac{1}{2}\). The choice \(\frac{1}{2}\) is wrong because it ignores the sign flip that occurs when the angle is negative and sine's odd symmetry is applied. Remembering that sine is odd while cosine is even is essential for quickly evaluating negative angle expressions.

Cosine is an even function, meaning \(\cos(-\theta) = \cos(\theta)\), so \(\cos(-\pi/3) = \cos(\pi/3) = \frac{1}{2}\). The choice \(-\frac{1}{2}\) is wrong because it incorrectly assumes cosine changes sign with a negative angle, which only applies to odd functions like sine. Cosine's even symmetry means reflecting the angle across the x-axis does not change its cosine value.

Since \(\pi < 5\pi/4 < 3\pi/2\), the angle \(5\pi/4\) falls between \(180°\) and \(270°\), placing it in Quadrant III. Quadrant II is wrong because that range only covers angles between \(\pi/2\) and \(\pi\), which \(5\pi/4\) exceeds. Converting radian angles to degrees or comparing them against the quadrant boundary fractions of \(\pi\) helps quickly identify the correct quadrant.

The angle \(135°\) has reference angle \(45°\), and since sine is positive in Quadrant II, \(\sin(135°) = \sin(45°) = \frac{\sqrt{2}}{2}\). The choice \(-\frac{\sqrt{2}}{2}\) is wrong because it incorrectly applies a negative sign that belongs to Quadrant III or IV, not Quadrant II sine values. Quadrant II keeps sine positive while cosine turns negative, a distinction crucial for correctly signing trig values.

The reference angle of \(210°\) is \(30°\), and since cosine is negative in Quadrant III, \(\cos(210°) = -\cos(30°) = -\frac{\sqrt{3}}{2}\). The choice \(-\frac{1}{2}\) is wrong because it uses the sine value of \(30°\) with a sign flip instead of the correct cosine value. Mixing up sine and cosine reference values is a common error, so always double-check which ratio you are computing.

The reference angle of \(300°\) is \(60°\), and since tangent is negative in Quadrant IV, \(\tan(300°) = -\tan(60°) = -\sqrt{3}\). The choice \(\sqrt{3}\) is wrong because it omits the negative sign required in Quadrant IV, where sine is negative and cosine is positive, making their ratio negative. Tangent's sign pattern follows the combined signs of sine and cosine, so it is negative in Quadrants II and IV.

Since \(\sin(3\pi/4) = \frac{\sqrt{2}}{2}\) from the Quadrant II positive sine rule, cosecant is its reciprocal: \(\csc(3\pi/4) = \frac{1}{\sqrt{2}/2} = \sqrt{2}\). The choice \(-\sqrt{2}\) is wrong because sine remains positive in Quadrant II, so its reciprocal cosecant must also be positive. Reciprocal functions always share the same sign as their base function in a given quadrant.

Since \(\cos(5\pi/3) = \frac{1}{2}\) because \(5\pi/3\) is in Quadrant IV where cosine is positive, secant, its reciprocal, is \(\sec(5\pi/3) = \frac{1}{1/2} = 2\). The choice \(-2\) is wrong because cosine does not turn negative in Quadrant IV, so its reciprocal secant should not be negative either. Quadrant IV keeps cosine and secant positive while sine, cosecant, and tangent become negative.

The reference angle of \(2\pi/3\) is \(\pi/3\), and since it lies in Quadrant II where sine is positive but cosine and tangent are negative, we get \(\sin = \frac{\sqrt{3}}{2}\), \(\cos = -\frac{1}{2}\), and \(\tan = -\sqrt{3}\), with reciprocals following the same signs. The last choice is wrong because it applies all-positive Quadrant I signs, ignoring that \(2\pi/3\) actually lies in Quadrant II. Computing all six values requires first finding the reference angle's magnitudes, then correctly signing each function according to the quadrant.

Cosine equals \(-\frac{\sqrt{2}}{2}\) in Quadrants II and III where the x-coordinate is negative, and with reference angle \(\pi/4\), this gives \(\theta = \pi - \pi/4 = 3\pi/4\) and \(\theta = \pi + \pi/4 = 5\pi/4\). The choice \(\pi/4\) and \(7\pi/4\) is wrong because those angles are in Quadrants I and IV where cosine is positive, not negative. Solving trig equations for negative values always requires identifying the two quadrants matching the correct sign before adding or subtracting the reference angle.

Tangent is positive in Quadrants I and III, and with reference angle \(\pi/3\) giving \(\tan(\pi/3) = \sqrt{3}\), the solutions are \(\theta = \pi/3\) and \(\theta = \pi + \pi/3 = 4\pi/3\). The choice \(\pi/3\) and \(2\pi/3\) is wrong because \(2\pi/3\) is in Quadrant II, where tangent is negative, not positive. Since tangent has a period of \(\pi\) rather than \(2\pi\), its solutions in one full circle differ by exactly \(\pi\).

The angle \(11\pi/6\) lies in Quadrant IV with reference angle \(\pi/6\), and since sine is negative in Quadrant IV, \(\sin(11\pi/6) = -\sin(\pi/6) = -\frac{1}{2}\). The choice \(\frac{\sqrt{3}}{2}\) is wrong because it swaps the cosine value of the reference angle into sine's slot, a common mix-up between the two functions. Quadrant IV always yields negative sine and positive cosine, a pattern useful for quickly checking your final sign.

The angle \(7\pi/6\) is in Quadrant III with reference angle \(\pi/6\), and since cosine is negative in Quadrant III, \(\cos(7\pi/6) = -\cos(\pi/6) = -\frac{\sqrt{3}}{2}\). The choice \(-\frac{1}{2}\) is wrong because it incorrectly uses sine's reference value of \(\pi/6\) rather than cosine's. Both sine and cosine are negative throughout Quadrant III, which distinguishes it from Quadrants II and IV where only one function is negative.

The angle \(5\pi/6\) is in Quadrant II with reference angle \(\pi/6\), and since tangent is negative in Quadrant II, \(\tan(5\pi/6) = -\tan(\pi/6) = -\frac{\sqrt{3}}{3}\). The choice \(-\sqrt{3}\) is wrong because it uses the reference angle value for \(\pi/3\) instead of \(\pi/6\), an easy mistake when reference angles are miscalculated. Always double check the reference angle computation itself before applying the quadrant sign rule.

Since \(\cos(4\pi/3) = -\frac{1}{2}\) because \(4\pi/3\) is in Quadrant III where cosine is negative, secant, its reciprocal, is \(\sec(4\pi/3) = \frac{1}{-1/2} = -2\). The choice \(2\) is wrong because it fails to preserve the negative sign of cosine in Quadrant III when taking the reciprocal. Reciprocal identities never change the sign of the original function, so secant must match cosine's sign exactly.

Since \(\sin(5\pi/4) = -\frac{\sqrt{2}}{2}\) because \(5\pi/4\) is in Quadrant III where sine is negative, cosecant, its reciprocal, is \(\csc(5\pi/4) = \frac{1}{-\sqrt{2}/2} = -\sqrt{2}\). The choice \(-\frac{\sqrt{2}}{2}\) is wrong because it simply repeats the sine value rather than correctly inverting the fraction to get cosecant. Taking a reciprocal means flipping the fraction, not just copying the original decimal or radical value.

Using the Pythagorean identity \(\sin^2(\theta) + \cos^2(\theta) = 1\), we find \(\cos^2(\theta) = 1 - \frac{9}{25} = \frac{16}{25}\), giving \(\cos(\theta) = \pm\frac{4}{5}\), and since \(\theta\) is in Quadrant II where cosine is negative, \(\cos(\theta) = -\frac{4}{5}\). The choice \(\frac{4}{5}\) is wrong because it ignores the quadrant condition specifying that cosine must be negative in Quadrant II. Whenever using the Pythagorean identity to find a missing trig value, the quadrant information is essential to select the correct sign.

Using \(\sin^2(\theta) = 1 - \cos^2(\theta) = 1 - \frac{144}{169} = \frac{25}{169}\) and taking the negative root since sine is negative in Quadrant III gives \(\sin(\theta) = -\frac{5}{13}\), so \(\tan(\theta) = \frac{-5/13}{-12/13} = \frac{5}{12}\). The choice \(-\frac{5}{12}\) is wrong because it fails to recognize that both sine and cosine are negative in Quadrant III, making their ratio positive. Dividing two negative numbers always produces a positive result, a detail students often overlook when both trig values share the same negative sign.

Since \(13\pi/6\) exceeds \(2\pi\), subtracting \(2\pi\) gives the coterminal angle \(\pi/6\), which lies in Quadrant I where all trig values are positive, so \(\cos(13\pi/6) = \cos(\pi/6) = \frac{\sqrt{3}}{2}\). The choice with reference angle \(5\pi/6\) is wrong because it fails to first reduce the angle to its coterminal equivalent within $[0, 2\pi)$ before finding the reference angle. Whenever an angle exceeds one full revolution, always subtract multiples of \(2\pi\) first to simplify before applying reference angle rules.

In Quadrant IV, sine is negative while cosine is positive, so their product \(\sin(\theta) \cdot \cos(\theta)\) is the product of a negative and a positive number, which is negative. The choice "Positive" is wrong because it would only be true if both factors shared the same sign, which does not happen in Quadrant IV. Determining the sign of a product of trig functions requires checking the individual signs of each function in that specific quadrant before multiplying.

Solving the equation gives \(\sin(\theta) = -\frac{1}{2}\), which occurs in Quadrants III and IV where sine is negative, and with reference angle \(\pi/6\) this yields \(\theta = \pi + \pi/6 = 7\pi/6\) and \(\theta = 2\pi - \pi/6 = 11\pi/6\). The choice \(\pi/6\) and \(5\pi/6\) is wrong because those angles lie in Quadrants I and II, where sine is positive rather than negative. When solving trig equations, always isolate the trig function first, then use the sign of the result to select the correct pair of quadrants.