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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 7.6.8b

Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.
8. b. arccot(√3)

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1
Recall that \( \arccot(x) \) is the angle \( \theta \) such that \( \cot(\theta) = x \). Here, we want to find \( \theta = \arccot(\sqrt{3}) \).
Use the definition of cotangent in terms of sine and cosine: \( \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} \). So, we need to find an angle \( \theta \) where \( \frac{\cos(\theta)}{\sin(\theta)} = \sqrt{3} \).
Recognize that \( \cot(\theta) = \sqrt{3} \) corresponds to a reference triangle where the adjacent side is \( \sqrt{3} \) and the opposite side is 1. This is a 30°-60°-90° triangle, where \( \cot(30^\circ) = \sqrt{3} \).
Determine the quadrant for \( \arccot(\sqrt{3}) \). Since \( \sqrt{3} > 0 \), \( \theta \) lies in the first quadrant where cotangent is positive.
Conclude that \( \arccot(\sqrt{3}) \) is the angle in the first quadrant with reference angle 30°, so \( \theta = 30^\circ \) or in radians \( \theta = \frac{\pi}{6} \).

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