6 please show work **Problem Statement:**Given that \(\sin A = -\frac{7}{25}\) with \(A\) in Quadrant III (QIII), find the following:**Task:**\[ \sin 2A \]**Solution Approach:**To find \(\sin 2A\), we use the double angle identity for sine:\[\sin 2A = 2 \sin A \cos A\]Given that \(\sin A = -\frac{7}{25}\), we need to determine \(\cos A\). In Quadrant III, both sine and cosine are negative.Using the Pythagorean identity:\[\sin^2 A + \cos^2 A = 1\]Substitute \(\sin A\):\[\left(-\frac{7}{25}\right)^2 + \cos^2 A = 1\]Calculate:\[\frac{49}{625} + \cos^2 A = 1\]\[\cos^2 A = 1 - \frac{49}{625} = \frac{576}{625}\]Thus, \(\cos A = -\frac{24}{25}\) (negative in Quadrant III).Now substitute back into the double angle identity:\[\sin 2A = 2 \times \left(-\frac{7}{25}\right) \times \left(-\frac{24}{25} \right)\]Calculate:\[\sin 2A = \frac{336}{625}\]So, \(\sin 2A = \frac{336}{625}\).

Name: 6 please show work **Problem Statement:**Given that \(\sin A = -\frac{
Uploaded: 2024-03-20T06:46:14.000Z
Channel: Bartleby
Description: 6 please show work **Problem Statement:**Given that \(\sin A = -\frac{7}{25}\) with \(A\) in Quadrant III (QIII), find the following:**Task:**\[ \sin 2A \]**Solution Approach:**To find \(\sin 2A\), we use the double angle identity for sine:\[\sin 2A