What is 2 cot \(\left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right)\) equal to ?
3 + √5
Evaluating the Trigonometric Expression: 2 cot(1/2 cos⁻¹(√5/3))
We are asked to find the value of the expression \(2 \cot \left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right)\). This problem involves inverse trigonometric functions and trigonometric identities, specifically half-angle formulas for cotangent.
Step-by-Step Solution for Evaluating the Expression
Let the expression inside the cotangent function be denoted by an angle. Let \( \theta = \cos ^{-1} \frac{\sqrt{5}}{3} \). This means that \( \cos \theta = \frac{\sqrt{5}}{3} \). Since the range of \( \cos^{-1}x \) is \( [0, \pi] \), and \( \frac{\sqrt{5}}{3} \) is positive, \( \theta \) must be in the first quadrant, specifically \( 0 < \theta < \frac{\pi}{2} \).
The expression we need to evaluate becomes \( 2 \cot \left(\frac{\theta}{2}\right) \).
Finding sin \(\theta\)
We know \( \cos \theta = \frac{\sqrt{5}}{3} \). To use half-angle identities for cotangent, we often need \( \sin \theta \). Since \( \theta \) is in the first quadrant, \( \sin \theta \) is positive.
\( \sin \theta = \sqrt{1 - \cos^2 \theta} \)
\( \sin \theta = \sqrt{1 - \left(\frac{\sqrt{5}}{3}\right)^2} \)
\( \sin \theta = \sqrt{1 - \frac{5}{9}} \)
\( \sin \theta = \sqrt{\frac{9 - 5}{9}} \)
\( \sin \theta = \sqrt{\frac{4}{9}} \)
\( \sin \theta = \frac{2}{3} \)
Using Half-Angle Identity for cot(\(\theta/2\))
We can use one of the half-angle identities for cotangent. A useful identity is:
\( \cot \left(\frac{\theta}{2}\right) = \frac{1 + \cos \theta}{\sin \theta} \)
Alternatively, we could use \( \cot \left(\frac{\theta}{2}\right) = \frac{\sin \theta}{1 - \cos \theta} \) or \( \cot \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}} \). Since \( 0 < \theta < \frac{\pi}{2} \), we have \( 0 < \frac{\theta}{2} < \frac{\pi}{4} \), which means \( \cot \left(\frac{\theta}{2}\right) \) is positive, so the positive square root is used.
Using \( \cot \left(\frac{\theta}{2}\right) = \frac{1 + \cos \theta}{\sin \theta} \):
Substitute the values of \( \cos \theta \) and \( \sin \theta \):
\( \cot \left(\frac{\theta}{2}\right) = \frac{1 + \frac{\sqrt{5}}{3}}{\frac{2}{3}} \)
\( \cot \left(\frac{\theta}{2}\right) = \frac{\frac{3 + \sqrt{5}}{3}}{\frac{2}{3}} \)
\( \cot \left(\frac{\theta}{2}\right) = \frac{3 + \sqrt{5}}{2} \)
Final Calculation
The original expression is \( 2 \cot \left(\frac{\theta}{2}\right) \). Substitute the value we found for \( \cot \left(\frac{\theta}{2}\right) \):
\( 2 \cot \left(\frac{\theta}{2}\right) = 2 \times \left(\frac{3 + \sqrt{5}}{2}\right) \)
\( 2 \cot \left(\frac{\theta}{2}\right) = 3 + \sqrt{5} \)
Thus, the value of \( 2 \cot \left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right) \) is \( 3 + \sqrt{5} \).
Revision Table: Key Concepts for Trigonometric Evaluations
| Concept | Description | Relevance to Problem |
|---|---|---|
| Inverse Cosine Function (\(\cos^{-1}x\)) | The angle \(\theta\) such that \(\cos \theta = x\). The range is typically \( [0, \pi] \). | Used to define the initial angle \( \theta \). |
| Pythagorean Identity | \( \sin^2 \theta + \cos^2 \theta = 1 \). Used to find \( \sin \theta \) if \( \cos \theta \) is known, or vice versa. | Used to calculate \( \sin \theta \) from \( \cos \theta \). |
| Cotangent Half-Angle Identity | Formulas relating \( \cot(\theta/2) \) to \( \sin \theta \) and \( \cos \theta \), e.g., \( \cot \left(\frac{\theta}{2}\right) = \frac{1 + \cos \theta}{\sin \theta} \). | Crucial for evaluating \( \cot(\theta/2) \). |
Additional Information: Understanding Inverse Trig and Half Angles
Inverse Trigonometric Functions: Inverse trigonometric functions, like \( \cos^{-1}x \), return the angle whose trigonometric value is \(x\). It's important to remember their defined ranges to determine the quadrant of the resulting angle, which affects the signs of other trigonometric functions of that angle.
Half-Angle Identities: Half-angle identities express the trigonometric functions of an angle \( \theta/2 \) in terms of trigonometric functions of the angle \( \theta \). They are derived from the double-angle or power-reduction formulas. For cotangent, the identities are particularly useful as they avoid square roots when expressed in terms of \( \sin \theta \) and \( \cos \theta \), provided \( \sin \theta \neq 0 \).
- \( \cot \left(\frac{\theta}{2}\right) = \frac{1 + \cos \theta}{\sin \theta} \)
- \( \cot \left(\frac{\theta}{2}\right) = \frac{\sin \theta}{1 - \cos \theta} \)
- \( \cot \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}} \) (The sign depends on the quadrant of \( \theta/2 \))
In this specific problem, since \( 0 < \theta < \frac{\pi}{2} \), the angle \( \theta/2 \) is in the range \( 0 < \frac{\theta}{2} < \frac{\pi}{4} \). In this range, all trigonometric functions, including cotangent, are positive.
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