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Difficulty: Very hardTrigonometric Ratios and Identities

In the xyxy-plane, the terminal ray of an angle θ\theta in standard position intersects the unit circle at a point PP. If the line passing through PP and the point (0,1)(0, 1) has a slope of −12-\frac{1}{2}, what is the value of sin⁡(θ)\sin(\theta)?

  1. A
    45\frac{4}{5}
  2. B
    −35-\frac{3}{5}
  3. 35\frac{3}{5}Answer
  4. D
    11

Answer

three-fifths
The correct answer is three-fifths. Since the point PP lies on the unit circle, its coordinates are (cos⁡(θ),sin⁡(θ))(\cos(\theta), \sin(\theta)). Using the slope formula between PP and the point (0,1)(0, 1), we get the equation sin⁡(θ)−1cos⁡(θ)=−12\frac{\sin(\theta) - 1}{\cos(\theta)} = -\frac{1}{2}. Cross-multiplying and rearranging gives cos⁡(θ)=2(1−sin⁡(θ))\cos(\theta) = 2(1 - \sin(\theta)). Substituting this into the Pythagorean identity cos⁡2(θ)+sin⁡2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1 yields the quadratic equation 5sin⁡2(θ)−8sin⁡(θ)+3=05\sin^2(\theta) - 8\sin(\theta) + 3 = 0. Factoring this equation gives (5sin⁡(θ)−3)(sin⁡(θ)−1)=0(5\sin(\theta) - 3)(\sin(\theta) - 1) = 0, which gives the solutions sin⁡(θ)=35\sin(\theta) = \frac{3}{5} or sin⁡(θ)=1\sin(\theta) = 1. Since the line is defined by two distinct points, PP cannot be (0,1)(0, 1), which means sin⁡(θ)≠1\sin(\theta) \neq 1. Thus, sin⁡(θ)=35\sin(\theta) = \frac{3}{5}.

Step-by-Step Solution

1
Represent the point PP on the unit circle using trigonometric functions and set up the slope equation with the point (0,1)(0,1).
The slope equation is sin⁡(θ)−1cos⁡(θ)=−12\frac{\sin(\theta) - 1}{\cos(\theta)} = -\frac{1}{2}.
Since PP lies on the unit circle, its coordinates are (cos⁡(θ),sin⁡(θ))(\cos(\theta), \sin(\theta)). The slope of a line through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is y2−y1x2−x1\frac{y_2 - y_1}{x_2 - x_1}.
2
Express cos⁡(θ)\cos(\theta) in terms of sin⁡(θ)\sin(\theta) by cross-multiplying and isolating cos⁡(θ)\cos(\theta) in the equation.
cos⁡(θ)=2(1−sin⁡(θ))\cos(\theta) = 2(1 - \sin(\theta)).
This substitution variable will allow us to rewrite the Pythagorean identity in terms of a single variable, sin⁡(θ)\sin(\theta).
3
Substitute the expression for cos⁡(θ)\cos(\theta) into the Pythagorean trigonometric identity cos⁡2(θ)+sin⁡2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1 and simplify.
4(1−sin⁡(θ))2+sin⁡2(θ)=1  ⟹  5sin⁡2(θ)−8sin⁡(θ)+3=04(1 - \sin(\theta))^2 + \sin^2(\theta) = 1 \implies 5\sin^2(\theta) - 8\sin(\theta) + 3 = 0.
The Pythagorean identity is a fundamental relationship between the sine and cosine of any angle.
4
Factor the quadratic equation to find the possible values of sin⁡(θ)\sin(\theta), and reject any extraneous solutions.
(5sin⁡(θ)−3)(sin⁡(θ)−1)=0(5\sin(\theta) - 3)(\sin(\theta) - 1) = 0, giving sin⁡(θ)=35\sin(\theta) = \frac{3}{5} or sin⁡(θ)=1\sin(\theta) = 1. The value sin⁡(θ)=1\sin(\theta) = 1 is rejected because it makes the point PP identical to (0,1)(0, 1), meaning a line cannot be defined.
A line requires two distinct points to be defined with a specific slope.

Key Concept

Unit circle definitions and the Pythagorean trigonometric identity
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Topics

Trigonometric Ratios and Identities

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