Question

Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.3. c. sin^(-1)(-√3/2)

Accepted Answer

Recognize that the problem asks for the angle $$ 	heta $$ such that $$ \sin(	heta) = -\frac{\sqrt{3}}{2} . $$This means we are looking for the inverse sine (arcsin) of $$ -\frac{\sqrt{3}}{2} . $$Recall that the sine function is negative in the third and fourth quadrants. Since the range of $$ \sin^{-1}  ($$arcsin) is restricted to $$ \left[-\frac{\pi}{2}, \frac{\pi}{2}ight] $$, the angle must lie in either the fourth quadrant (negative angles) or the first quadrant (positive angles, but sine positive there). So here, the angle will be in the fourth quadrant (between $$ -\frac{\pi}{2} $$ and 0). Identify the reference angle whose sine is $$ \frac{\sqrt{3}}{2} . $$From common special angles, $$ \sin\left(\frac{\pi}{3}ight) = \frac{\sqrt{3}}{2} . So $$the reference angle is $$ \frac{\pi}{3} . $$Since the sine value is negative and the angle must be in the fourth quadrant (due to the range of arcsin), the angle is $$ -\frac{\pi}{3} . $$This is the angle whose sine is $$ -\frac{\sqrt{3}}{2} $$ within the principal range of arcsin. Express the final answer as $$ 	heta = \sin^{-1}\left(-\frac{\sqrt{3}}{2}ight) = -\frac{\pi}{3} .$$

Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.
3. c. sin^(-1)(-√3/2)

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Key Concepts

Inverse Sine Function (sin⁻¹ or arcsin)

Reference Triangles and Quadrants

Sign of Sine in Different Quadrants