Question

Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.7. a. sec^(-1)(-√2)

Accepted Answer

Recall that the function $$ \sec^{-1}(x)  is $$the inverse secant function, which gives an angle $$ 	heta $$ such that $$ \sec(	heta) = x . $$The range of $$ \sec^{-1}(x)  is $$usually taken as $$ [0, \pi] $$ excluding $$ \frac{\pi}{2} $$, meaning the angle lies in either the first or second quadrant. Given $$ \sec^{-1}(-\sqrt{2}) $$, we want to find an angle $$ 	heta $$ where $$ \sec(	heta) = -\sqrt{2} . $$Since secant is the reciprocal of cosine, this means $$ \cos(	heta) = -\frac{1}{\sqrt{2}} . $$Identify the reference angle by considering the positive value $$ \frac{1}{\sqrt{2}} . $$The reference angle $$ \alpha $$ satisfies $$ \cos(\alpha) = \frac{1}{\sqrt{2}} $$, which corresponds to $$ \alpha = \frac{\pi}{4}  (or 45 $$degrees). Since $$ \cos(	heta)  is $$negative and the angle $$ 	heta $$ must be in the range $$ [0, \pi] $$ excluding $$ \frac{\pi}{2} $$, $$ 	heta $$ lies in the second quadrant. Therefore, $$ 	heta = \pi - \alpha = \pi - \frac{\pi}{4} . $$Express the final answer for $$ \sec^{-1}(-\sqrt{2})  as$$ $$ 	heta = \pi - \frac{\pi}{4} $$, which is the angle in the appropriate quadrant corresponding to the given inverse secant value.

Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.
7. a. sec^(-1)(-√2)

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Key Concepts

Inverse Secant Function (sec⁻¹)

Reference Triangles and Quadrants

Sign of Trigonometric Functions in Quadrants