91Ó°ÊÓ

Identify the given angles which are \(\sin ^{-1} \frac{3}{5}\) and \(\cos ^{-1}(- \frac{4}{5})\). These are the inverse sin and inverse cos respectively of \(\frac{3}{5}\) and \(-\frac{4}{5}\) respectively. They represent angles whose sin and cos values are provided.

To calculate the sin of subtracted angles, use identity: \( \sin(A - B) = \sin A * \cos B - \cos A * \sin B \). In this case, A is \( sin^{-1} \frac{3}{5} \) and B is \( cos^{-1} (-\frac{4}{5}) \). Applying the rule, you get \( \sin (\sin ^{-1} \frac{3}{5} - \cos ^{-1} (-\frac{4}{5})) = \sin (sin^{-1} \frac{3}{5}) * \cos (cos^{-1} (-\frac{4}{5})) - \cos (sin^{-1} \frac{3}{5}) * \sin (cos^{-1} (-\frac{4}{5})). \)

Simplify the expression by noting that \( sin (sin^{-1} x) = x \) and \( cos (cos^{-1} x) = x \), therefore we get \( \sin (\sin ^{-1} \frac{3}{5} - \cos ^{-1} (-\frac{4}{5})) = \frac{3}{5} * (-\frac{4}{5}) - \cos (sin^{-1} \frac{3}{5}) * \sin (cos^{-1} (-\frac{4}{5})). \) Now identify \( \cos (sin^{-1} \frac{3}{5}) \) and \(\sin (cos^{-1} (-\frac{4}{5})) \) using the Pythagorean trigonometric identity \( \cos^2\theta = 1 - \sin^2\theta \) and \( \sin^2\theta = 1 - \cos^2\theta \) respectively, be careful with the signs. Because the negative sign only applies to \( sin^{-1} and \cos^{-1} \) in second and third quadrants. Hence, \( \cos (sin^{-1} \frac{3}{5}) = -\frac{4}{5} \) and \( \sin (cos^{-1} (-\frac{4}{5})) = \frac{3}{5} \)

Substitute \( \cos (sin^{-1} \frac{3}{5}) = -\frac{4}{5} \) and \( \sin (cos^{-1} (-\frac{4}{5})) = \frac{3}{5} \) into the evaluated expression from step 3. Therefore the expression becomes \(- \frac{3}{5} * \frac{4}{5} - (- \frac{4}{5}) * \frac{3}{5} = - \frac{12}{25} + \frac{12}{25} = 0 \)