Question

rewrite \\( \cos \left( \tan ^ { - 1 } 4 v \
ight) \\) as an algebraic expression in \\( v \\).
\\
\cos \left( \tan ^ { - 1 } 4 v \
ight) = \\

Explanation:

Step1: Let \(\theta=\tan^{- 1}(4v)\)

By the definition of inverse tangent function, \(\tan\theta = 4v=\frac{4v}{1}\), where \(\theta\in(-\frac{\pi}{2},\frac{\pi}{2})\). Consider a right - triangle where the opposite side \(y = 4v\) and the adjacent side \(x = 1\).

Step2: Find the hypotenuse \(r\)

Using the Pythagorean theorem \(r=\sqrt{x^{2}+y^{2}}\). Substitute \(x = 1\) and \(y = 4v\) into the formula, we get \(r=\sqrt{1+(4v)^{2}}=\sqrt{1 + 16v^{2}}\).

Step3: Find \(\cos\theta\)

By the definition of cosine function \(\cos\theta=\frac{x}{r}\). Since \(x = 1\) and \(r=\sqrt{1 + 16v^{2}}\), and \(\theta=\tan^{-1}(4v)\), then \(\cos(\tan^{-1}(4v))=\frac{1}{\sqrt{1 + 16v^{2}}}\).

Answer:

\(\frac{1}{\sqrt{1 + 16v^{2}}}\)