### Problem StatementRewrite \(\sin\left(2 \cos^{-1} \frac{v}{5}\right)\) as an algebraic expression in \(v\).#### Expression to Rewrite\[ \sin\left(2 \cos^{-1} \frac{v}{5}\right) = \]### Step-by-Step Explanation1. **Understanding the Function Composition**: - The expression involves a sine function applied to an angle that is twice the inverse cosine of \(\frac{v}{5}\). 2. **Using Trigonometric Identities**: - Recall the double-angle identity for sine: \[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \] - Let \(\theta = \cos^{-1} \frac{v}{5}\). This implies: \[ \cos(\theta) = \frac{v}{5} \] - Using the Pythagorean identity: \[ \sin^2(\theta) + \cos^2(\theta) = 1 \] Since \(\cos(\theta) = \frac{v}{5}\), we can find \(\sin(\theta)\): \[ \sin(\theta) = \sqrt{1 - \left(\frac{v}{5}\right)^2} \] \[ \sin(\theta) = \sqrt{1 - \frac{v^2}{25}} \] \[ \sin(\theta) = \frac{\sqrt{25 - v^2}}{5} \]3. **Applying the Identity to Rewrite the Expression**: - Substitute \(\theta\) back into the double-angle identity: \[ \sin\left(2 \cos^{-1} \frac{v}{5}\right) = 2 \sin(\theta) \cos(\theta) \] - From earlier, we have: \[ \sin(\theta) = \frac{\sqrt{25 - v^2}}{5} \] \[ \cos(\theta) = \frac{v}{5} \] - Therefore: \[ \sin\left(2 \cos^{-1} \frac{v}{5}\right) = 2 \left(\frac{\sqrt{25 - v^2}}{5}\

Name: ### Problem StatementRewrite \(\sin\left(2 \cos^{-1} \frac{v}{5}\righ
Uploaded: 2024-03-20T06:46:55.000Z
Channel: Bartleby
Description: ### Problem StatementRewrite \(\sin\left(2 \cos^{-1} \frac{v}{5}\right)\) as an algebraic expression in \(v\).#### Expression to Rewrite\[ \sin\left(2 \cos^{-1} \frac{v}{5}\right) = \]### Step-by-Step Explanation1. **Understanding the Function Compo