**Title: Solving Double-Angle Trigonometric Functions****Problem Statement:**Find \( \sin 2x \), \( \cos 2x \), and \( \tan 2x \) if \( \tan x = -\frac{5}{12} \) and \( x \) terminates in quadrant II.**Required Calculations:**1. **For \( \sin 2x \):** \[ \sin 2x = \Box \]2. **For \( \cos 2x \):** \[ \cos 2x = \Box \]3. **For \( \tan 2x \):** \[ \tan 2x = \Box \]**Instructions:**Given that \( \tan x = -\frac{5}{12} \) in quadrant II, recall the trigonometric identities and relationships between the functions to solve for \( \sin 2x \), \( \cos 2x \), and \( \tan 2x \). - Use the double-angle identity for sine: \[ \sin 2x = 2 \sin x \cos x \]- Use the double-angle identity for cosine: \[ \cos 2x = \cos^2 x - \sin^2 x \] or \[ \cos 2x = 2 \cos^2 x - 1 \] or \[ \cos 2x = 1 - 2 \sin^2 x \]- Use the double-angle identity for tangent: \[ \tan 2x = \frac{2 \tan x}{1 - \tan^2 x} \]First, determine \( \sin x \) and \( \cos x \) using the Pythagorean identity:\[\tan x = \frac{\sin x}{\cos x}\]Given that \( \tan x = -\frac{5}{12} \) and knowing that sine is positive and cosine is negative in quadrant II, you can find \( \sin x \) and \( \cos x \) before applying the double-angle formulas.**Graphical Representation:**On the right side of the problem, there's a

Name: **Title: Solving Double-Angle Trigonometric Functions****Problem Stat
Uploaded: 2024-03-20T06:46:55.000Z
Channel: Bartleby
Description: **Title: Solving Double-Angle Trigonometric Functions****Problem Statement:**Find \( \sin 2x \), \( \cos 2x \), and \( \tan 2x \) if \( \tan x = -\frac{5}{12} \) and \( x \) terminates in quadrant II.**Required Calculations:**1. **For \( \sin 2x \):