**Problem:**Find \(\cos(2A)\) if \(\sin(A) = \frac{15}{17}\), and \(A\) is in quadrant 2.**Solution:**First, we need to determine \(\cos(A)\). Since \(A\) is in the second quadrant, and sine is positive while cosine is negative in the second quadrant, we proceed as follows:Given \(\sin(A) = \frac{15}{17}\), we use the Pythagorean identity:\[\sin^2(A) + \cos^2(A) = 1.\]Substitute \(\sin(A) = \frac{15}{17}\) into the identity:\[\left(\frac{15}{17}\right)^2 + \cos^2(A) = 1.\]Calculate \(\left(\frac{15}{17}\right)^2\):\[\left(\frac{15}{17}\right)^2 = \frac{225}{289}.\]So,\[\frac{225}{289} + \cos^2(A) = 1.\]Solving for \(\cos^2(A)\):\[\cos^2(A) = 1 - \frac{225}{289} = \frac{289}{289} - \frac{225}{289} = \frac{64}{289}.\]Therefore, \(\cos(A) = \pm \sqrt{\frac{64}{289}} = \pm \frac{8}{17}\).Since \(A\) is in the second quadrant, \(\cos(A)\) is negative:\[\cos(A) = -\frac{8}{17}.\]Now, we use the double-angle identity for cosine:\[\cos(2A) = 2\cos^2(A) - 1.\]Substitute \(\cos(A) = -\frac{8}{17}\):\[\cos^2(A) = \left(-\frac{8}{17}\right)^2 = \frac{64}{289}.\]Therefore,\[\cos(2A) = 2\left(\frac{64}{289}\right) - 1.\]Simplify:\[\cos(2A) = \frac{128}{289} - 1 = \

Name: **Problem:**Find \(\cos(2A)\) if \(\sin(A) = \frac{15}{17}\), and \(A
Uploaded: 2024-03-20T06:46:55.000Z
Channel: Bartleby
Description: **Problem:**Find \(\cos(2A)\) if \(\sin(A) = \frac{15}{17}\), and \(A\) is in quadrant 2.**Solution:**First, we need to determine \(\cos(A)\). Since \(A\) is in the second quadrant, and sine is positive while cosine is negative in the second quadran