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What is the exact value of cos (tan ¹(-1)) 2 -? sin cos 08-¹(- +/-))

What is the exact value of cos (tan ¹(-1)) 2 -? sin cos 08-¹(- +/-))

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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### Problem Statement **What is the exact value of** \[ \frac{\cos\left(\tan^{-1}(-1)\right)}{\sin\left(\cos^{-1}\left( -\frac{\sqrt{3}}{2} \right)\right)}? \] Enter your answer as a simplified fraction in the box. \[ \frac{\cos\left(\tan^{-1}(-1)\right)}{\sin\left(\cos^{-1}\left( -\frac{\sqrt{3}}{2} \right)\right)} = \boxed{} \] ### Explanation This problem involves inverse trigonometric functions. Let's break down the problem step-by-step to find the exact value. 1. **Identify the angles:** - For \(\tan^{-1}(-1)\), we need to find the angle \(\theta\) such that \(\tan(\theta) = -1\). - For \(\cos^{-1}\left( -\frac{\sqrt{3}}{2} \right)\), we need to find the angle \(\phi\) such that \(\cos(\phi) = -\frac{\sqrt{3}}{2}\). 2. **Solve for \(\theta\) and \(\phi\):** - \(\tan^{-1}(-1) = \theta\) where \(\theta = -\frac{\pi}{4}\) (since \(\tan(-\frac{\pi}{4}) = -1\)). - \(\cos^{-1}\left( -\frac{\sqrt{3}}{2} \right) = \phi\) where \(\phi = \frac{5\pi}{6}\) (since \(\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}\)). 3. **Apply the trigonometric functions:** - \(\cos\left(-\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\). - \(\sin\left(\frac{5\pi}{6}\right) = \sin\left(\pi - \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{
Transcribed Image Text:### Problem Statement **What is the exact value of** \[ \frac{\cos\left(\tan^{-1}(-1)\right)}{\sin\left(\cos^{-1}\left( -\frac{\sqrt{3}}{2} \right)\right)}? \] Enter your answer as a simplified fraction in the box. \[ \frac{\cos\left(\tan^{-1}(-1)\right)}{\sin\left(\cos^{-1}\left( -\frac{\sqrt{3}}{2} \right)\right)} = \boxed{} \] ### Explanation This problem involves inverse trigonometric functions. Let's break down the problem step-by-step to find the exact value. 1. **Identify the angles:** - For \(\tan^{-1}(-1)\), we need to find the angle \(\theta\) such that \(\tan(\theta) = -1\). - For \(\cos^{-1}\left( -\frac{\sqrt{3}}{2} \right)\), we need to find the angle \(\phi\) such that \(\cos(\phi) = -\frac{\sqrt{3}}{2}\). 2. **Solve for \(\theta\) and \(\phi\):** - \(\tan^{-1}(-1) = \theta\) where \(\theta = -\frac{\pi}{4}\) (since \(\tan(-\frac{\pi}{4}) = -1\)). - \(\cos^{-1}\left( -\frac{\sqrt{3}}{2} \right) = \phi\) where \(\phi = \frac{5\pi}{6}\) (since \(\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}\)). 3. **Apply the trigonometric functions:** - \(\cos\left(-\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\). - \(\sin\left(\frac{5\pi}{6}\right) = \sin\left(\pi - \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{
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