Unit Circle Values - Algebra 3

Find the exact value of $\cos\frac{5\pi}{6}\sin\frac{4\pi}{3}$ .
Answer: $\frac{3}{4}$ $\cos\frac{5\pi}{6}=-\frac{\sqrt{3}}{2}$ and $\sin\frac{4\pi}{3}=-\frac{\sqrt{3}}{2}$ . The product of two negative values is $\frac{3}{4}$ . The negative choices come from using only one negative sign; $\frac{1}{4}$ comes from using $\frac{1}{2}$ instead of $\frac{\sqrt{3}}{2}$ for one factor.
For an angle $\theta$ in standard position with $0 \le \theta < 2\pi$ , suppose $\tan\theta = \sqrt{3}$ and $\cos\theta < 0$ . What is $\theta$ ?
Answer: $\frac{4\pi}{3}$ When you're given a trigonometric condition like $\tan\theta=\sqrt{3}$ plus a sign restriction such as $\cos\theta<0$ , start by recalling the signs of trig functions in each quadrant. Tangent is positive in Quadrants I and III, but since cosine must be negative, you can only be in Quadrant III. That's the key move: the sign restriction eliminates one of the possible quadrants. The reference angle with tangent $\sqrt{3}$ is $\frac{\pi}{3}$ , because $\tan\frac{\pi}{3}=\sqrt{3}$ . In Quadrant III, the angle is $\pi + \frac{\pi}{3} = \frac{4\pi}{3}$ , so that is the correct angle. You can verify: cosine is negative in Quadrant III, and tangent is positive because sine is also negative there. Why are the others traps? The choice $\frac{\pi}{3}$ has the right reference angle but is in Quadrant I, where cosine is positive, so it fails the condition. The choice $\frac{2\pi}{3}$ has negative tangent, not positive, so it fails the value of tangent entirely. Similarly, $\frac{5\pi}{3}$ gives negative tangent and positive cosine, so it also cannot satisfy the conditions. Study tip: whenever a trig-value problem includes a sign restriction, first determine which quadrant the angle must be in, then find the reference angle and place it in that quadrant. This two-step process prevents most quadrant errors.
What is the exact value of $\sin\left(\frac{11\pi}{6}\right) + \cos\left(\frac{2\pi}{3}\right)$ ?
Answer: $-1$ When you see trig values at special angles, your first move is to locate each angle on the unit circle and find its reference angle. For $\sin\left(\frac{11\pi}{6}\right)$ : $\frac{11\pi}{6}$ is just $\frac{\pi}{6}$ short of $2\pi$ , so it lies in quadrant IV. The reference angle is $\frac{\pi}{6}$ , and sine is negative in quadrant IV; since $\sin\left(\frac{\pi}{6}\right)=\frac12$ , you get $\sin\left(\frac{11\pi}{6}\right)=-\frac12$ . For $\cos\left(\frac{2\pi}{3}\right)$ : $\frac{2\pi}{3}$ is $\frac{\pi}{3}$ short of $\pi$ , so it lies in quadrant II. The reference angle is $\frac{\pi}{3}$ , and cosine is negative in quadrant II; since $\cos\left(\frac{\pi}{3}\right)=\frac12$ , you get $\cos\left(\frac{2\pi}{3}\right)=-\frac12$ . Adding these gives $\left(-\frac12\right)+\left(-\frac12\right)=-1$ . If you picked $0$ , you may have assumed the two terms cancel, but both are negative here, so they add together. If you picked $-\frac12$ , you likely evaluated only one term and forgot to add the second. If you picked $\frac12$ , the sign error probably came from forgetting that cosine in quadrant II is negative and sine in quadrant IV is negative. Study tip: draw a quick quadrant sign chart or use a memory aid for signs, but always write down both reference-angle values first, then apply the quadrant sign before summing.
Find the exact value of $\sin\frac{5\pi}{4}\cos\frac{7\pi}{6}$ .
Answer: $\frac{\sqrt{6}}{4}$ $\sin\frac{5\pi}{4}=-\frac{\sqrt{2}}{2}$ and $\cos\frac{7\pi}{6}=-\frac{\sqrt{3}}{2}$ . Their product is $\left(-\frac{\sqrt{2}}{2}\right)\left(-\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{6}}{4}$ . The negative choices come from dropping one or both quadrant signs; $\frac{\sqrt{3}}{4}$ uses the wrong reference value.
Find the exact value of $\tan\frac{5\pi}{4} + \sec\frac{5\pi}{4}$ .
Answer: $1-\sqrt{2}$ At $\frac{5\pi}{4}$ , the reference angle is $\frac{\pi}{4}$ . In quadrant III, tangent is positive, so $\tan\frac{5\pi}{4}=1$ . Cosine is negative, so $\sec\frac{5\pi}{4}=-\sqrt{2}$ . Thus the sum is $1-\sqrt{2}$ . The choice $-1-\sqrt{2}$ uses a negative tangent; $-1+\sqrt{2}$ uses both signs incorrectly; $1+\sqrt{2}$ takes secant positive.
Which angle $\theta$ in the interval $(-\pi, \pi]$ has $\cos\theta = -\frac{\sqrt{2}}{2}$ and $\tan\theta = -1$ ?
Answer: $\frac{3\pi}{4}$ When you see a trig equation like this, think about the unit circle and the signs of the trig functions in each quadrant. You need an angle in $(-\pi,\pi]$ where $\cos\theta$ is negative and $\tan\theta=-1$ . Since $\tan\theta=\frac{\sin\theta}{\cos\theta}$ , a tangent of $-1$ means sine and cosine have opposite signs. But cosine is negative, so sine must be positive. That places you in Quadrant II, where sine is positive and cosine is negative. The reference angle for $|\tan\theta|=1$ is $\frac{\pi}{4}$ , so the Quadrant II angle is $\theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4}.$ Indeed, $\cos\frac{3\pi}{4}=-\frac{\sqrt2}{2}$ and $\tan\frac{3\pi}{4}=-1$ . The other choices fail for different reasons. $-\frac{3\pi}{4}$ is in Quadrant III, where both sine and cosine are negative, so tangent is $+1$ , not $-1$ . $-\frac{\pi}{4}$ is in Quadrant IV, where cosine is positive, so it fails the given cosine condition even though tangent is $-1$ . $\frac{5\pi}{4}$ is outside the interval $(-\pi,\pi]$ , and its tangent is also $+1$ . A strong strategy: for angle problems, first decide the quadrant from the signs of cosine and tangent, then use the reference angle. That combination identifies the angle quickly and avoids sign mistakes.
The terminal side of an angle $\theta$ in standard position intersects the unit circle at $\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$ . If $0 \le \theta < 2\pi$ , what is $\theta$ ?
Answer: $\frac{5\pi}{3}$ Whenever you see a point on the unit circle, remember that its coordinates are directly $\cos\theta$ and $\sin\theta$ . Here, $\cos\theta=\frac12$ and $\sin\theta=-\frac{\sqrt3}{2}$ , so your job is to find the angle in $0 \le \theta < 2\pi$ with that cosine and sine pair. The positive cosine and negative sine tell you the terminal side is in quadrant IV. The reference angle is the angle above the positive x-axis whose cosine is $\frac12$ , which is $\frac{\pi}{3}$ . In quadrant IV, the angle is $2\pi-\frac{\pi}{3}=\frac{5\pi}{3}$ , so the choice $\frac{5\pi}{3}$ is correct. The other options all come from mixing up quadrants or reference angles. $\frac{2\pi}{3}$ is in quadrant II, where cosine is negative and sine is positive, so it would correspond to $\left(-\frac12,\frac{\sqrt3}{2}\right)$ . $\frac{11\pi}{6}$ is in quadrant IV, but its reference angle is $\frac{\pi}{6}$ , giving $\left(\frac{\sqrt3}{2},-\frac12\right)$ , not the given point. $\frac{7\pi}{4}$ is also quadrant IV, but its reference angle is $\frac{\pi}{4}$ , so its coordinates are $\left(\frac{\sqrt2}{2},-\frac{\sqrt2}{2}\right)$ . These are close but not exact matches. A reliable strategy: first read the signs of the coordinates to determine the quadrant, then use the absolute values to find the reference angle. Finally, convert that reference angle into the correct quadrant form: in quadrant IV, use $2\pi-\text{reference angle}$ .
Find the exact value of $\sin\frac{5\pi}{6} - \cos\frac{5\pi}{6}$ .
Answer: $\frac{1+\sqrt{3}}{2}$ $\sin\frac{5\pi}{6}=\frac{1}{2}$ and $\cos\frac{5\pi}{6}=-\frac{\sqrt{3}}{2}$ . Subtracting the cosine gives $\frac{1}{2}-\left(-\frac{\sqrt{3}}{2}\right)=\frac{1+\sqrt{3}}{2}$ . The choice $\frac{1-\sqrt{3}}{2}$ treats cosine as positive; $\frac{\sqrt{3}-1}{2}$ reverses the subtraction; the negative choice makes sine negative as well.
For which angle $\theta$ in $[0,2\pi)$ does $\csc\theta = 2$ and $\sin(2\theta) > 0$ ?
Answer: $\frac{\pi}{6}$ The condition $\csc\theta=2$ means $\sin\theta=\frac{1}{2}$ , so the possible angles are $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ . For $\theta=\frac{\pi}{6}$ , $\sin(2\theta)=\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}>0$ . For $\theta=\frac{5\pi}{6}$ , $\sin(2\theta)=\sin\frac{5\pi}{3}=-\frac{\sqrt{3}}{2}<0$ . Therefore the valid angle is $\frac{\pi}{6}$ . The angle $\frac{\pi}{3}$ has $\csc\theta=\frac{2}{\sqrt{3}}$ , and $\frac{7\pi}{6}$ has $\csc\theta=-2$ .
Find the exact value of $\csc\frac{5\pi}{6} + \cot\frac{5\pi}{6}$ .
Answer: $2-\sqrt{3}$ At $\frac{5\pi}{6}$ , $\sin\theta=\frac{1}{2}$ , so $\csc\theta=2$ . Also $\cos\theta=-\frac{\sqrt{3}}{2}$ , so $\cot\theta=\frac{\cos\theta}{\sin\theta}=-\sqrt{3}$ . The sum is $2-\sqrt{3}$ . The choice $-2-\sqrt{3}$ makes cosecant negative; $-2+\sqrt{3}$ makes both signs wrong; $2+\sqrt{3}$ takes cotangent positive.
For $\theta = \frac{5\pi}{6}$ , which point is the intersection of the terminal side with the unit circle?
Answer: $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ Whenever you see a question about the unit circle, remember that the terminal side of an angle $\theta$ intersects the circle at exactly $(\cos\theta,\sin\theta)$ . So for $\theta=\frac{5\pi}{6}=150^\circ$ , the reference angle is $30^\circ$ , and the coordinates are $\cos 150^\circ=-\frac{\sqrt3}{2}$ , $\sin 150^\circ=\frac12$ . That gives the point $\left(-\frac{\sqrt3}{2},\frac12\right)$ . Since $150^\circ$ is in Quadrant II, the x-coordinate must be negative and the y-coordinate positive—a sign check that immediately kills two choices. The point $\left(-\frac{1}{2},\frac{\sqrt3}{2}\right)$ would correspond to $120^\circ$ , not $150^\circ$ ; it's easy to confuse the shared reference-angle structure, but the cosine value for a $30^\circ$ reference angle is $\sqrt3/2$ , not $1/2$ . The point $\left(\frac{\sqrt3}{2},-\frac12\right)$ is the angle $330^\circ$ (or $-30^\circ$ ); it has the wrong signs for QII, because you forgot to make x negative. The point $\left(-\frac{\sqrt3}{2},-\frac12\right)$ is $210^\circ$ , where the sine is negative, not positive—you correctly made x negative but forgot y stays positive in Quadrant II. A powerful habit: draw or visualize the quadrant, then apply the signs $(x,y)=(-,+)$ for angles between $90^\circ$ and $180^\circ$ . That alone prevents most unit-circle sign errors.
Which angle $\theta$ in $[0, 2\pi)$ satisfies both $\sin\theta = -\frac{1}{2}$ and $\cos\theta = -\frac{\sqrt{3}}{2}$ ?
Answer: $\frac{7\pi}{6}$ When you're asked for the angle that satisfies both a sine and a cosine value, your first move is to identify the quadrant. Here, both $\sin\theta$ and $\cos\theta$ are negative, so $\theta$ must land in the third quadrant of the unit circle, between $\pi$ and $3\pi/2$ . The reference angle is the acute angle whose sine has magnitude $1/2$ and cosine has magnitude $\sqrt{3}/2$ — that's $\pi/6$ . In the third quadrant, you add that reference angle to $\pi$ , giving $\theta = \pi + \pi/6 = 7\pi/6$ . Check: $\sin(7\pi/6) = -1/2$ and $\cos(7\pi/6) = -\sqrt{3}/2$ , exactly what the problem demands. Now look at the other choices to see what each one traps. $5\pi/6$ is in the second quadrant: its sine is positive $1/2$ , not $-1/2$ , so it fails immediately. $4\pi/3$ is also in the third quadrant, but its sine is $-\sqrt{3}/2$ and its cosine is $-1/2$ — the two values are swapped relative to what's given. That's a classic sign/value mix-up. $11\pi/6$ is in the fourth quadrant: its sine is negative $-1/2$ , but its cosine is positive $\sqrt{3}/2$ , so it only gets half the story. Your study tip: always test quadrant first using the signs of sine and cosine. Then find the reference angle from the absolute values. That two-step process — quadrant then reference angle — eliminates most guesswork on unit-circle questions.
Which angle in $[0,2\pi)$ has the same sine as $\frac{5\pi}{6}$ and a positive cosine?
Answer: $\frac{\pi}{6}$ $\sin\frac{5\pi}{6}=\frac{1}{2}$ . The other angle in $[0,2\pi)$ with sine $\frac{1}{2}$ is $\frac{\pi}{6}$ , and its cosine is positive. The angle $\frac{5\pi}{6}$ has the same sine but negative cosine; $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$ have sine $-\frac{1}{2}$ .
If $\tan\theta = \frac{\sqrt{3}}{3}$ and $\sin\theta < 0$ , which angle $\theta$ in $[0,2\pi)$ satisfies both conditions?
Answer: $\frac{7\pi}{6}$ $\tan\theta=\frac{\sqrt{3}}{3}$ has reference angle $\frac{\pi}{6}$ . Tangent is positive in quadrants I and III, but $\sin\theta<0$ excludes quadrant I. In quadrant III, the angle is $\pi+\frac{\pi}{6}=\frac{7\pi}{6}$ . The angle $\frac{\pi}{6}$ has positive sine; $\frac{5\pi}{6}$ and $\frac{11\pi}{6}$ have tangent $-\frac{\sqrt{3}}{3}$ .
Find the exact value of $\sec\frac{7\pi}{4} + \csc\frac{7\pi}{4}$ .
Answer: $0$ At $\frac{7\pi}{4}$ , $\cos\theta=\frac{\sqrt{2}}{2}$ and $\sin\theta=-\frac{\sqrt{2}}{2}$ . Therefore $\sec\theta=\sqrt{2}$ and $\csc\theta=-\sqrt{2}$ , so their sum is $0$ . The choice $-\sqrt{2}$ treats secant as negative; $\sqrt{2}$ omits the cosecant term; $2\sqrt{2}$ treats both reciprocals as positive.
If $\sec\theta = -2$ and $\sin\theta > 0$ , what is $\cot\theta$ ?
Answer: $-\frac{\sqrt{3}}{3}$ When a trig question gives you one function value and a sign condition, start by identifying the quadrant — it controls the signs of every answer. Here $\sec\theta=-2$ means $\cos\theta=\frac{1}{\sec\theta}=-\frac12$ . Since $\sin\theta>0$ , $\theta$ is in Quadrant II, where cosine is negative and sine is positive. Now use the Pythagorean identity: $\sin^2\theta=1-\cos^2\theta=1-\left(-\frac12\right)^2=1-\frac14=\frac34$ so $\sin\theta=\frac{\sqrt3}{2}$ (positive, as required). Then $\cot\theta=\frac{\cos\theta}{\sin\theta} =\frac{-\frac12}{\frac{\sqrt3}{2}} =-\frac{1}{\sqrt3} =-\frac{\sqrt3}{3}.$ That makes $-\frac{\sqrt3}{3}$ correct. Why are the others traps? $-\sqrt3$ is actually the value of $\tan\theta$ , since $\tan\theta=\frac{\sin\theta}{\cos\theta}=-\sqrt3$ ; $\cot\theta$ is its reciprocal, not the tangent itself. $\frac{\sqrt3}{3}$ has the right magnitude but the wrong sign — it ignores that cosine is negative in Quadrant II. And $\sqrt3$ is the positive reciprocal of $\frac{\sqrt3}{3}$ , so it doubles the sign mistake and flips the magnitude. Study tip: when you see $\sec$ or $\csc$ , convert to $\cos$ or $\sin$ immediately. Then always use the quadrant to assign the final sign.
What is the exact value of $\csc\left(-\frac{\pi}{6}\right)$ ?
Answer: $-2$ When you see $\csc$ of an angle, your first move should be to rewrite it in terms of sine: $\csc\theta = \frac{1}{\sin\theta}$ . This question is testing both your special-angle values and your ability to track signs using the unit circle. The angle $-\frac{\pi}{6}$ is located in the fourth quadrant, where sine is negative. You know $\sin\left(\frac{\pi}{6}\right)=\frac12$ , so $\sin\left(-\frac{\pi}{6}\right)=-\frac12$ . Therefore $\csc\left(-\frac{\pi}{6}\right)=\frac{1}{-\frac12}=-2$ . The answer $2$ comes from forgetting that the negative angle keeps the sine value negative. The choices $-\frac12$ and $\frac12$ are traps: they give the value of sine itself, not the cosecant. In particular, $-\frac12$ correctly names $\sin\left(-\frac{\pi}{6}\right)$ , but the question asks for its reciprocal. A good habit is to pause and ask, "Is this a trig function or its reciprocal?" Then confirm the sign by picturing the quadrant. Memorize the special-angle sines like $\sin\frac{\pi}{6}=\frac12$ , and remember that cosecant is the reciprocal of sine, so it "flips" the fraction but keeps the sign.
What is the exact value of $\sin\left(\frac{5\pi}{4}\right)\cos\left(\frac{7\pi}{6}\right)$ ?
Answer: $\frac{\sqrt{6}}{4}$ Whenever you see exact trig values at angles like $\frac{5\pi}{4}$ or $\frac{7\pi}{6}$ , go to the unit circle. Identify the reference angle, then attach the correct sign based on the quadrant. For $\frac{5\pi}{4}$ , the reference angle is $\frac{\pi}{4}$ , and it lies in Quadrant III, where sine is negative. So $\sin\left(\frac{5\pi}{4}\right)=-\frac{\sqrt{2}}{2}$ . For $\frac{7\pi}{6}$ , the reference angle is $\frac{\pi}{6}$ , also in Quadrant III, where cosine is negative. So $\cos\left(\frac{7\pi}{6}\right)=-\frac{\sqrt{3}}{2}$ . Multiplying them: $\left(-\frac{\sqrt{2}}{2}\right)\left(-\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{6}}{4}$ That is the choice $\frac{\sqrt{6}}{4}$ . The choice $-\frac{\sqrt{6}}{4}$ is what you get if you forget that the product of two negatives is positive, or if you accidentally make one of the values positive. The choice $-\frac{\sqrt{2}}{4}$ would require $\cos\left(\frac{7\pi}{6}\right)$ to be $\frac{\sqrt{2}}{2}$ , but that is not its value. The choice $\frac{\sqrt{3}}{4}$ mixes up the special-angle values, perhaps using $\sin\frac{\pi}{3}$ and $\cos\frac{\pi}{6}$ , which are not the angles given. On this exam, always pause to check signs: both $5\pi/4$ and $7\pi/6$ are Quadrant III angles, and they share needing negative sine/cosine values. A quick quadrant check can turn an easy calculation into a confident correct answer.
Which of the following is equal to $\cos\frac{25\pi}{6}$ ?
Answer: $\frac{\sqrt{3}}{2}$ Subtract $4\pi$ from $\frac{25\pi}{6}$ : $\frac{25\pi}{6}-4\pi=\frac{\pi}{6}$ . Since cosine has period $2\pi$ , $\cos\frac{25\pi}{6}=\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$ . The negative choices come from quadrant-sign errors; $\frac{1}{2}$ is the sine value.
If $\sin\theta = \cos\theta = -\frac{\sqrt{2}}{2}$ , what is the value of $\tan\theta + \sec\theta$ ?
Answer: $1 - \sqrt{2}$ This question tests your ability to connect quadrant signs with reciprocal identities. Seeing $\sin\theta=\cos\theta=-\frac{\sqrt2}{2}$ should immediately signal an angle in the third quadrant, where both sine and cosine are negative. Since tangent is sine divided by cosine, equal numerators and denominators give $\tan\theta=1$ . Secant is the reciprocal of cosine, so $\sec\theta=\frac{1}{-\sqrt2/2}=-\frac{2}{\sqrt2}=-\sqrt2$ . Adding gives $1+(-\sqrt2)=1-\sqrt2$ . Each wrong answer comes from a sign error. Choosing $1+\sqrt2$ means you used $\sec\theta=+\sqrt2$ , forgetting that cosine is negative. Choosing $-1-\sqrt2$ means you made $\tan\theta=-1$ , perhaps confusing this with a situation where sine and cosine are opposites; but here they are equal, so tangent must be positive. Choosing $-1+\sqrt2$ combines both mistakes: tangent negative and secant positive. The key is that in the third quadrant, both sine and cosine are negative, so tangent is positive but secant is negative. On exam day, when you see equal sine and cosine values, immediately recall the $45^\circ$ reference angle and ask which quadrant makes both negative. Then assign signs to tangent and secant before adding. Sign awareness is the entire battle here.

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