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Ch. 3 - Trigonometric Identities and Equations
Blitzer - Trigonometry 3rd Edition
Blitzer3rd EditionTrigonometryISBN: 9780137316601Non è quello che usi tu?Cambia libro di testo
Capitolo 3, Problema 11b

Use the given information to find the exact value of each of the following: cos 2θ
cot θ = 2, θ lies in quadrant III.

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Recall the double-angle identity for cosine: \(\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}\) or alternatively \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\). We will use the identity involving cotangent to find \(\cos 2\theta\).
Given \(\cot \theta = 2\), express \(\tan \theta\) as the reciprocal: \(\tan \theta = \frac{1}{2}\).
Since \(\theta\) lies in quadrant III, both sine and cosine are negative, but tangent (and cotangent) is positive, which matches \(\tan \theta = \frac{1}{2}\). Use this to find \(\sin \theta\) and \(\cos \theta\) by considering a right triangle or using the Pythagorean identity.
Set \(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{1}{2}\). Let \(\sin \theta = k\) and \(\cos \theta = 2k\). Use the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\) to solve for \(k\).
Once \(\sin \theta\) and \(\cos \theta\) are found (with correct signs for quadrant III), substitute them into the double-angle formula \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\) to find the exact value of \(\cos 2\theta\).

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