Unit Circle: Sine & Cosine - Math 3

Point $R$ on the unit circle has coordinates $\left(\cos\frac{7\pi}{4}, \sin\frac{7\pi}{4}\right)$ . If point $S$ is obtained by rotating point $R$ counterclockwise by $\frac{\pi}{4}$ radians about the origin, what are the coordinates of point $S$ ?
Answer: $\left(1, 0\right)$ Point R corresponds to angle 7π/4. When rotated counterclockwise by π/4, point S corresponds to angle 7π/4 + π/4 = 8π/4 = 2π. Since 2π ≡ 0 (mod 2π), point S is at angle 0, giving coordinates (cos 0, sin 0) = (1, 0). Choice A corresponds to angle π/2. Choice C corresponds to angle π/4. Choice D corresponds to angle 3π/4.
On the unit circle, angle $\phi$ in standard position has $\cos\phi = \frac{\sqrt{3}}{2}$ . If $0 \leq \phi < 2\pi$ , what is the sum of all possible values of $\sin\phi$ ?
Answer: $0$ When you encounter unit circle problems involving finding multiple angle values, remember that trigonometric functions can have the same value at different angles within one complete rotation. Given $\cos\phi = \frac{\sqrt{3}}{2}$ on the unit circle, you need to identify all angles where cosine equals this value. From your knowledge of special angles, cosine equals $\frac{\sqrt{3}}{2}$ at $\phi = \frac{\pi}{6}$ (30°) and $\phi = \frac{11\pi}{6}$ (330°). These are the only two solutions in the interval $[0, 2\pi)$ . Now find the corresponding sine values. At $\phi = \frac{\pi}{6}$ , we have $\sin\phi = \frac{1}{2}$ . At $\phi = \frac{11\pi}{6}$ , we have $\sin\phi = -\frac{1}{2}$ . The sum of all possible sine values is $\frac{1}{2} + (-\frac{1}{2}) = 0$ . Looking at the wrong answers: Choice A ( $\sqrt{3}$ ) incorrectly uses the tangent values at these angles. Choice B ( $1$ ) might result from mistakenly thinking there's only one angle or confusing sine and cosine relationships. Choice D ( $\frac{1}{2}$ ) represents only one of the sine values, missing the negative value in the fourth quadrant. The key insight is recognizing the symmetry: when cosine is positive, the angle can be in either the first or fourth quadrant, where sine values are equal in magnitude but opposite in sign. Always check both quadrants where your given trigonometric function is positive, and remember that opposite sine values will sum to zero.
The terminal side of angle $\beta$ passes through point $(-8, 15)$ . Which of the following represents the exact value of $\cos\beta$ ?
Answer: $-\frac{8}{17}$ To find cos β, we need the x-coordinate divided by the radius. First, find r = √((-8)² + 15²) = √(64 + 225) = √289 = 17. Therefore, cos β = x/r = -8/17. Choice B gives sin β instead of cos β. Choice C incorrectly uses the y-coordinate in the denominator. Choice D omits the negative sign from the x-coordinate.
If $\sin\gamma = -\frac{2}{3}$ and $\gamma$ is in the third quadrant, what is the value of $\frac{\cos\gamma}{\sin\gamma}$ ?
Answer: $\frac{\sqrt{5}}{2}$ First find cos γ using sin²γ + cos²γ = 1. With sin γ = -2/3: (-2/3)² + cos²γ = 1, so 4/9 + cos²γ = 1, giving cos²γ = 5/9. Therefore cos γ = ±√5/3. Since γ is in quadrant III, cos γ = -√5/3. Now: cos γ/sin γ = (-√5/3)/(-2/3) = (-√5/3) × (-3/2) = √5/2. Choice B incorrectly keeps a negative sign. Choice C uses incorrect algebraic manipulation. Choice D uses the wrong denominator value.
If $\cos\delta = \frac{4}{5}$ and $\sin\delta < 0$ , what is the value of $\sin\delta \cdot \cos\delta$ ?
Answer: $-\frac{12}{25}$ This problem tests your understanding of the Pythagorean identity and how to determine the sign of trigonometric functions based on quadrant location. Since $\cos\delta = \frac{4}{5}$ is positive and $\sin\delta < 0$ , angle $\delta$ must be in Quadrant IV, where cosine is positive and sine is negative. To find $\sin\delta$ , use the Pythagorean identity: $\sin^2\delta + \cos^2\delta = 1$ . Substituting the known value: $\sin^2\delta + \left(\frac{4}{5}\right)^2 = 1$ , so $\sin^2\delta + \frac{16}{25} = 1$ . This gives us $\sin^2\delta = \frac{9}{25}$ , meaning $\sin\delta = \pm\frac{3}{5}$ . Since we're told $\sin\delta < 0$ , we have $\sin\delta = -\frac{3}{5}$ . Therefore: $\sin\delta \cdot \cos\delta = -\frac{3}{5} \cdot \frac{4}{5} = -\frac{12}{25}$ , which is answer choice C. Let's examine the wrong answers: Choice A gives $-\frac{3}{5}$ , which is just the value of $\sin\delta$ alone, not the product. Choice B gives $\frac{12}{25}$ , which would be correct if you forgot that sine is negative in Quadrant IV—a sign error. Choice D gives $-\frac{4}{5}$ , which is the negative of the cosine value, suggesting confusion about which function is negative. Study tip: Always determine the quadrant first when given one trig function value and a sign condition. This tells you the signs of all other trig functions and prevents costly sign errors.
On the unit circle, point $P$ has coordinates $\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$ . If angle $\theta$ in standard position has its terminal side passing through point $P$ , and $0 \leq \theta < 2\pi$ , what is the value of $\sin\theta + \cos\theta$ ?
Answer: $0$ On the unit circle, if point P has coordinates (x,y), then cos θ = x and sin θ = y. Therefore, cos θ = √2/2 and sin θ = -√2/2. Adding these: sin θ + cos θ = -√2/2 + √2/2 = 0. Choice B would be the value of cos θ alone. Choice C would be the value of sin θ alone. Choice D incorrectly adds the absolute values.
If $\sin\xi = \frac{\sqrt{3}}{2}$ and $\xi$ is in the second quadrant, what is the value of $2\cos^2\xi - 1$ ?
Answer: $-\frac{1}{2}$ This problem tests your understanding of trigonometric identities and the signs of trig functions in different quadrants. When you see an expression like $2\cos^2\xi - 1$ , recognize it as the double angle formula for cosine: $\cos(2\xi) = 2\cos^2\xi - 1$ . Since $\sin\xi = \frac{\sqrt{3}}{2}$ and $\xi$ is in the second quadrant, you first need to find $\cos\xi$ . Using the Pythagorean identity $\sin^2\xi + \cos^2\xi = 1$ : $\cos^2\xi = 1 - \sin^2\xi = 1 - \left(\frac{\sqrt{3}}{2}\right)^2 = 1 - \frac{3}{4} = \frac{1}{4}$ So $\cos\xi = \pm\frac{1}{2}$ . Since $\xi$ is in the second quadrant where cosine is negative, $\cos\xi = -\frac{1}{2}$ . Now you can calculate: $2\cos^2\xi - 1 = 2\left(\frac{1}{4}\right) - 1 = \frac{1}{2} - 1 = -\frac{1}{2}$ Choice A ( $\frac{3}{4}$ ) likely comes from incorrectly using $\sin^2\xi$ instead of $\cos^2\xi$ in the calculation. Choice B ( $\frac{1}{2}$ ) is what you'd get if you forgot the "-1" in the expression. Choice C ( $-\frac{3}{4}$ ) results from using $\sin^2\xi$ and remembering the negative sign from the second quadrant incorrectly. Remember: always check which quadrant you're in to determine the correct signs, and watch for double angle identities that can simplify your work.
Point P moves counterclockwise around the unit circle starting from $(1, 0)$ . After rotating through an angle of $\frac{5\pi}{3}$ radians, what are the coordinates of point P?
Answer: $(\frac{1}{2}, -\frac{\sqrt{3}}{2})$ The coordinates of a point on the unit circle after rotating through angle θ are (cos θ, sin θ). For θ = 5π/3, we have cos(5π/3) = cos(2π - π/3) = cos(-π/3) = cos(π/3) = 1/2, and sin(5π/3) = sin(2π - π/3) = sin(-π/3) = -sin(π/3) = -√3/2. Therefore, the coordinates are (1/2, -√3/2). Choice B represents the coordinates at 2π/3. Choice C represents coordinates at π/6. Choice D represents coordinates at 7π/6.
If an angle $\theta$ in standard position has $\sin(\theta) = \frac{2}{3}$ , what are the possible values for $\cos(\theta)$ ?
Answer: $\pm\frac{\sqrt{5}}{3}$ Using the Pythagorean identity sin²(θ) + cos²(θ) = 1, we get cos²(θ) = 1 - sin²(θ) = 1 - (2/3)² = 1 - 4/9 = 5/9. Therefore, cos(θ) = ±√5/3. Since sin(θ) = 2/3 > 0, the angle could be in quadrant I (where cos > 0) or quadrant II (where cos < 0). Choice B incorrectly assumes cosine must be positive. Choice C uses an incorrect calculation: (2√2/3)² = 8/9 ≠ 5/9. Choice D incorrectly assumes sine and cosine must have opposite signs.
Point $A$ is located at $\left(\cos\frac{5\pi}{6}, \sin\frac{5\pi}{6}\right)$ on the unit circle. What are the coordinates of point $B$ , which is the reflection of point $A$ across the x-axis?
Answer: $\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$ First, find point A's coordinates. Since 5π/6 is in quadrant II: cos(5π/6) = -√3/2 and sin(5π/6) = 1/2. So A = (-√3/2, 1/2). When reflecting across the x-axis, the x-coordinate stays the same and the y-coordinate changes sign. Therefore, B = (-√3/2, -1/2). Choice A gives the original point A. Choice B reflects across the y-axis instead. Choice D confuses the sine and cosine values and reflects incorrectly.
If $\cos\eta = -\frac{\sqrt{2}}{2}$ and $\sin\eta = \frac{\sqrt{2}}{2}$ , what is the value of $\frac{1 + \sin\eta}{1 + \cos\eta}$ ?
Answer: $\frac{2 + \sqrt{2}}{2 - \sqrt{2}}$ This question tests your ability to work with trigonometric expressions and rationalize complex fractions. When you see specific trigonometric values like these, recognize that they correspond to special angles on the unit circle. To find $\frac{1 + \sin\eta}{1 + \cos\eta}$ , substitute the given values: $\cos\eta = -\frac{\sqrt{2}}{2}$ and $\sin\eta = \frac{\sqrt{2}}{2}$ . This gives us: $\frac{1 + \frac{\sqrt{2}}{2}}{1 + \left(-\frac{\sqrt{2}}{2}\right)} = \frac{1 + \frac{\sqrt{2}}{2}}{1 - \frac{\sqrt{2}}{2}}$ Converting to a common denominator: $\frac{\frac{2 + \sqrt{2}}{2}}{\frac{2 - \sqrt{2}}{2}} = \frac{2 + \sqrt{2}}{2 - \sqrt{2}}$ This matches answer choice D. Answer A gives only the numerator portion of our final answer, missing the denominator entirely. Answer B ( $2 + \sqrt{2}$ ) would be the result if you incorrectly assumed the denominator equaled 1, ignoring the $1 + \cos\eta$ term completely. Answer C ( $\sqrt{2} + 1$ ) represents a common error where students might try to simplify the fraction incorrectly or confuse the order of terms. Remember that when working with trigonometric expressions involving special angle values, substitute carefully and work through fraction operations step-by-step. Don't try to take shortcuts with complex fractions—convert everything to common denominators first, then simplify systematically.
The coordinates of point $T$ on the unit circle are $\left(a, \frac{3}{5}\right)$ where $a < 0$ . What is the value of $a^2 + 6a$ ?
Answer: $\frac{16}{25} - \frac{24}{5}$ When you see coordinates on the unit circle, remember that any point $(x,y)$ on the unit circle must satisfy the equation $x^2 + y^2 = 1$ . This is your key constraint for solving these problems. Since point $T$ has coordinates $\left(a, \frac{3}{5}\right)$ and lies on the unit circle, we can substitute into the unit circle equation: $a^2 + \left(\frac{3}{5}\right)^2 = 1$ . This gives us $a^2 + \frac{9}{25} = 1$ , so $a^2 = 1 - \frac{9}{25} = \frac{16}{25}$ . Since $a < 0$ , we have $a = -\frac{4}{5}$ . Now we can calculate $a^2 + 6a = \frac{16}{25} + 6\left(-\frac{4}{5}\right) = \frac{16}{25} - \frac{24}{5}$ . Looking at the wrong answers: Choice A gives $\frac{16}{25} + \frac{24}{5}$ , which results from incorrectly using $a = +\frac{4}{5}$ instead of recognizing that $a < 0$ . Choice B gives $-\frac{24}{5}$ , which is just the $6a$ term alone—you'd get this if you forgot to include the $a^2$ term entirely. Choice C gives $\frac{16}{25}$ , which is only the $a^2$ term without adding $6a$ . The correct answer is D: $\frac{16}{25} - \frac{24}{5}$ . Study tip: Always use the constraint $x^2 + y^2 = 1$ first to find unknown coordinates on the unit circle, and pay careful attention to sign conditions—they determine which root to choose when solving quadratic equations.
An angle $\phi$ has its terminal side passing through the point $(-5, 12)$ on a circle centered at the origin. If this point were projected onto the unit circle, what would be the value of $\cos(\phi)$ ?
Answer: $-\frac{5}{13}$ To find cos(φ), we need the x-coordinate of the corresponding point on the unit circle. First, find the distance from origin to (-5, 12): r = √((-5)² + 12²) = √(25 + 144) = √169 = 13. The unit circle point is (-5/13, 12/13), so cos(φ) = -5/13. Choice B incorrectly uses 12 as the denominator. Choice C incorrectly makes the result positive. Choice D confuses the x and y coordinates.
On the unit circle, if point $Q$ has coordinates $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ , what is the value of $\left(\sin\theta\right)\left(\cos\theta\right)$ where $\theta$ is the angle in standard position whose terminal side passes through $Q$ ?
Answer: $\frac{1}{2}$ When you encounter a point on the unit circle, remember that the coordinates directly give you the cosine and sine values. For any point $(x, y)$ on the unit circle, $x = \cos\theta$ and $y = \sin\theta$ , where $\theta$ is the angle in standard position. Since point $Q$ has coordinates $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ , we know that $\cos\theta = \frac{\sqrt{2}}{2}$ and $\sin\theta = \frac{\sqrt{2}}{2}$ . To find $(\sin\theta)(\cos\theta)$ , we simply multiply these values: $(\sin\theta)(\cos\theta) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{(\sqrt{2})^2}{4} = \frac{2}{4} = \frac{1}{2}$ This confirms that **(B) $\frac{1}{2}$ ** is correct. Let's examine why the other choices are wrong: **(A) $\frac{\sqrt{2}}{2}$ ** is the individual value of both $\sin\theta$ and $\cos\theta$ , but not their product. Students often confuse the individual trigonometric values with their product. **(C) $1$ ** would be the result if you incorrectly thought $(\sqrt{2})^2 = 4$ instead of $2$ , or if you confused this with $\sin^2\theta + \cos^2\theta = 1$ . **(D) $\frac{1}{4}$ ** might result from forgetting to square the $\sqrt{2}$ terms and just multiplying the denominators. Study tip: Always remember that coordinates on the unit circle are $(x,y) = (\cos\theta, \sin\theta)$ . When multiplying expressions with square roots, be careful to square them properly: $(\sqrt{a})^2 = a$ .
On the unit circle, if $\sin\theta = \frac{1}{2}$ and $\cos\theta < 0$ , what is the value of $4\cos\theta - 3\sin\theta$ ?
Answer: $-2\sqrt{3} - \frac{3}{2}$ Since sin θ = 1/2 and cos θ < 0, θ is in quadrant II. Using sin²θ + cos²θ = 1: (1/2)² + cos²θ = 1, so cos²θ = 3/4, giving cos θ = ±√3/2. Since cos θ < 0, cos θ = -√3/2. Therefore: 4cos θ - 3sin θ = 4(-√3/2) - 3(1/2) = -2√3 - 3/2. Choice B uses the positive value of cos θ. Choice C has an error in the sign of the second term. Choice D uses cos θ = √3/4 instead of √3/2.
If $\cos\alpha = -\frac{3}{5}$ and $\alpha$ terminates in the second quadrant, what is the value of $\sin^2\alpha + \cos^2\alpha + \sin\alpha$ ?
Answer: $\frac{9}{5}$ Since sin²α + cos²α = 1 (Pythagorean identity), we need to find sin α. Using sin²α + cos²α = 1 with cos α = -3/5: sin²α + 9/25 = 1, so sin²α = 16/25, giving sin α = ±4/5. Since α is in quadrant II, sin α = 4/5. Therefore: sin²α + cos²α + sin α = 1 + 4/5 = 9/5. Choice B incorrectly uses negative sin α. Choice C adds terms incorrectly. Choice D makes computational errors.
The terminal side of angle $\omega$ passes through point $(3k, -4k)$ where $k > 0$ . What is $\cos\omega + \sin\omega$ ?
Answer: $-\frac{1}{5}$ When you see an angle whose terminal side passes through a point, you're working with trigonometric functions in standard position. The key is finding the cosine and sine values using the coordinates and distance from the origin. Given the point $(3k, -4k)$ where $k > 0$ , you first need the distance from the origin: $r = \sqrt{(3k)^2 + (-4k)^2} = \sqrt{9k^2 + 16k^2} = \sqrt{25k^2} = 5k$ . Since $k > 0$ , we have $r = 5k$ . Now you can find the trigonometric values: $\cos\omega = \frac{x}{r} = \frac{3k}{5k} = \frac{3}{5}$ and $\sin\omega = \frac{y}{r} = \frac{-4k}{5k} = -\frac{4}{5}$ . Therefore: $\cos\omega + \sin\omega = \frac{3}{5} + \left(-\frac{4}{5}\right) = \frac{3}{5} - \frac{4}{5} = -\frac{1}{5}$ Looking at the wrong answers: Choice A gives $\frac{7}{5}$ , which would result from incorrectly adding the absolute values $\frac{3}{5} + \frac{4}{5}$ , ignoring the negative sign of sine. Choice C gives $\frac{1}{5}$ , which happens if you forget that sine is negative and calculate $\frac{3}{5} - \frac{4}{5}$ as positive. Choice D gives $-\frac{7}{5}$ , which occurs from adding the magnitudes but making both terms negative. The correct answer is B: $-\frac{1}{5}$ . Study tip: When a point has coordinates involving a parameter like $k$ , the parameter cancels out when computing trigonometric ratios. Always pay careful attention to signs—the quadrant determines whether sine and cosine are positive or negative.

Start this quiz