Question

In Exercises 39–54, find the exact value of each expression, if possible. Do not use a calculator. tan⁻¹ [tan(− π/6)]

Accepted Answer

Recall that the function $$ 	an^{-1}(x) $$, also known as arctangent, returns an angle $$ 	heta $$ such that $$ -\frac{\pi}{2} \u003c 	heta \u003c \frac{\pi}{2} $$ and $$ 	an(	heta) = x . $$This means the output of $$ 	an^{-1}  is $$always in the principal range $$ (-\frac{\pi}{2}, \frac{\pi}{2}) . $$Evaluate the inner function first: $$ 	an(-\frac{\pi}{6}) . $$Since tangent is an odd function, $$ 	an(-x) = -	an(x) $$, so $$ 	an(-\frac{\pi}{6}) = -	an(\frac{\pi}{6}) . $$Recall the exact value $$ 	an(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} $$, so $$ 	an(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}} . $$Now substitute back into the original expression: $$ 	an^{-1} \left( -\frac{1}{\sqrt{3}} ight) . We $$need to find the angle $$ 	heta  in $$the principal range $$ (-\frac{\pi}{2}, \frac{\pi}{2}) $$ such that $$ 	an(	heta) = -\frac{1}{\sqrt{3}} . $$Since $$ 	an(	heta) = -\frac{1}{\sqrt{3}} $$ and $$ 	an(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}} $$, the angle $$ 	heta = -\frac{\pi}{6} $$ lies within the principal range, so $$ 	an^{-1} [ 	an(-\frac{\pi}{6}) ] = -\frac{\pi}{6} .$$

In Exercises 39–54, find the exact value of each expression, if possible. Do not use a calculator. tan⁻¹ [tan(− π/6)]

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Key Concepts

Inverse Trigonometric Functions

Periodicity of the Tangent Function

Evaluating Composite Functions Involving Inverse Trigonometric Functions