Question

Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.6. b. arccsc(-2/√3)

Accepted Answer

Recall that $$ \arccsc(x)  is $$the inverse cosecant function, which gives an angle $$ 	heta $$ such that $$ \csc(	heta) = x . $$Here, we want to find $$ 	heta = \arccsc\left(-\frac{2}{\sqrt{3}}ight) . $$Rewrite the cosecant in terms of sine: since $$ \csc(	heta) = \frac{1}{\sin(	heta)} $$, we have $$ \sin(	heta) = \frac{1}{\csc(	heta)} = \frac{1}{-\frac{2}{\sqrt{3}}} = -\frac{\sqrt{3}}{2} . $$Determine the reference angle by considering the positive value of sine: $$ \sin(	heta_{ref}) = \frac{\sqrt{3}}{2} . $$From the unit circle, this corresponds to an angle of $$ \frac{\pi}{3} $$ radians (or 60 degrees). Since the sine value is negative and cosecant is negative, identify the quadrant where sine is negative. Sine is negative in Quadrants III and IV. The principal range of $$ \arccsc(x)  is$$ $$ [-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}] $$ excluding zero, so the angle must be in Quadrant IV (negative angle between 0 and $$ -\frac{\pi}{2} ). $$Therefore, the angle $$ 	heta  is $$the negative of the reference angle, so $$ 	heta = -\frac{\pi}{3} . $$This is the angle in the appropriate quadrant corresponding to $$ \arccsc\left(-\frac{2}{\sqrt{3}}ight) .$$

Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.
6. b. arccsc(-2/√3)

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Key Concepts

Inverse Trigonometric Functions

Reference Triangles

Quadrant Sign Rules