Questions: Let sin A = -3/5 with A in QIII and find sin (2A) =

Solution Steps

To find \(\sin(2A)\) given \(\sin A = -\frac{3}{5}\) and \(A\) is in the third quadrant, we can use the double angle identity for sine: \(\sin(2A) = 2 \sin A \cos A\). First, we need to determine \(\cos A\). Since \(A\) is in the third quadrant, both sine and cosine are negative. We use the Pythagorean identity \(\sin^2 A + \cos^2 A = 1\) to find \(\cos A\). Once we have \(\cos A\), we can substitute \(\sin A\) and \(\cos A\) into the double angle formula to find \(\sin(2A)\).

We are given that \(\sin A = -\frac{3}{5}\) and that angle \(A\) is in the third quadrant.

Using the Pythagorean identity: \[ \sin^2 A + \cos^2 A = 1 \] we can substitute \(\sin A\): \[ \left(-\frac{3}{5}\right)^2 + \cos^2 A = 1 \] This simplifies to: \[ \frac{9}{25} + \cos^2 A = 1 \] Subtracting \(\frac{9}{25}\) from both sides gives: \[ \cos^2 A = 1 - \frac{9}{25} = \frac{16}{25} \] Taking the square root, and noting that \(\cos A\) is negative in the third quadrant: \[ \cos A = -\frac{4}{5} \]

Using the double angle formula: \[ \sin(2A) = 2 \sin A \cos A \] Substituting the values we found: \[ \sin(2A) = 2 \left(-\frac{3}{5}\right) \left(-\frac{4}{5}\right) \] This simplifies to: \[ \sin(2A) = 2 \cdot \frac{12}{25} = \frac{24}{25} \]

Final Answer