Question

Find the values in Exercises 9–12.11. tan(arcsin(-1/2))

Accepted Answer

Recognize that the expression is $$ 	an(\arcsin(-\frac{1}{2})) . $$Here, $$ \arcsin(-\frac{1}{2}) $$ represents an angle $$ 	heta $$ such that $$ \sin(	heta) = -\frac{1}{2} . $$Let $$ 	heta = \arcsin(-\frac{1}{2}) . $$Since $$ \sin(	heta) = -\frac{1}{2} $$, we want to find $$ 	an(	heta) . $$Recall that $$ 	an(	heta) = \frac{\sin(	heta)}{\cos(	heta)} . $$Use the Pythagorean identity to find $$ \cos(	heta) $$: $$ \cos(	heta) = \pm \sqrt{1 - \sin^2(	heta)} = \pm \sqrt{1 - \left(-\frac{1}{2}ight)^2} = \pm \sqrt{1 - \frac{1}{4}} = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2} . $$Determine the correct sign of $$ \cos(	heta)  by $$considering the range of $$ \arcsin $$, which is $$ \left[-\frac{\pi}{2}, \frac{\pi}{2}ight] . In $$this interval, cosine is positive for angles where sine is negative, so $$ \cos(	heta) = \frac{\sqrt{3}}{2} . $$Finally, compute $$ 	an(	heta) = \frac{\sin(	heta)}{\cos(	heta)} = \frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} .$$

Find the values in Exercises 9–12.
11. tan(arcsin(-1/2))

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Key Concepts

Inverse Trigonometric Functions

Relationship Between Sine and Tangent

Pythagorean Identity