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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.6.7c

Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.
7. c. arcsec(-2)

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1
Recall that the function \( \arcsec(x) \) is the inverse of the secant function, so \( \arcsec(-2) \) is the angle \( \theta \) such that \( \sec(\theta) = -2 \).
Since \( \sec(\theta) = \frac{1}{\cos(\theta)} \), rewrite the equation as \( \frac{1}{\cos(\theta)} = -2 \), which implies \( \cos(\theta) = -\frac{1}{2} \).
Identify the quadrants where \( \cos(\theta) \) is negative. Cosine is negative in Quadrants II and III, so the angle \( \theta \) must lie in one of these quadrants.
Use a reference triangle to find the reference angle \( \alpha \) where \( \cos(\alpha) = \frac{1}{2} \). The reference angle \( \alpha \) corresponds to \( \frac{\pi}{3} \) (or 60 degrees).
Determine the actual angle \( \theta \) in the appropriate quadrant(s) by adjusting the reference angle: in Quadrant II, \( \theta = \pi - \alpha \); in Quadrant III, \( \theta = \pi + \alpha \). Since the principal value of \( \arcsec(x) \) is usually taken in \([0, \pi]\) excluding \( \frac{\pi}{2} \), select the angle in Quadrant II.

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