Find the exact value of arcsin\left(-\frac{\sqrt{3}}{2}\right). a.) \text{ }arcsin\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3} b.) \text{ }arcsin\left(-\frac{\sqrt{3}}{2}\right) = \frac{4\pi}{3} c.) \text{ }arcsin\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{6} d.) \text{ }arcsin\left(-\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}
The Correct Answer and Explanation is:
The correct answer is a.) arcsin(-√3/2) = -π/3.
Here is a detailed explanation:
The expression arcsin(-√3/2) asks for the angle, let’s call it θ, whose sine is equal to -√3/2. This can be written as the equation sin(θ) = -√3/2.
A critical aspect of the inverse sine function, arcsin(x), is its restricted range. To ensure that the function gives a single, unique output for each input, its range is defined as [-π/2, π/2]. In degrees, this is equivalent to [-90°, 90°]. This means our final answer must be an angle in the first or fourth quadrant, with fourth quadrant angles represented as negative values.
To find the value of θ, we can first identify the reference angle. We do this by considering the positive version of the value: sin(θ) = √3/2. From our knowledge of the unit circle and standard trigonometric values, we know that the angle in the first quadrant whose sine is √3/2 is π/3 (or 60°). This is our reference angle.
Now, we must determine the correct quadrant for our original problem, sin(θ) = -√3/2. The sine function is negative in the third and fourth quadrants. However, we must choose the angle that falls within the [-π/2, π/2] range of the arcsin function. The third quadrant is outside this range. The fourth quadrant is included. Using our reference angle of π/3, the angle in the fourth quadrant is -π/3.
This angle, -π/3, is within the required range of [-π/2, π/2]. We can verify our answer by checking if sin(-π/3) equals -√3/2. Since sine is an odd function, sin(-x) = -sin(x). Therefore, sin(-π/3) = -sin(π/3) = -√3/2. This confirms our result.
Comparing this with the given options, option (a) is -π/3. Option (b), 4π/3, is incorrect because while sin(4π/3) is indeed -√3/2, this angle is not in the principal value range of the arcsin function. Options (c) and (d) are incorrect as they correspond to different sine values.thumb_upthumb_down
