Questions: Pythagorean Identity Use the Pythagorean identity to find sin x. cos x = 3 sqrt(13) / 19 sin x = [?] sqrt(square) /

Solution Steps

The Pythagorean identity states that: \[ \sin^2 x + \cos^2 x = 1 \] We are given \(\cos x = \frac{3 \sqrt{13}}{19}\), and we need to find \(\sin x\).

Substitute \(\cos x = \frac{3 \sqrt{13}}{19}\) into the identity: \[ \sin^2 x + \left(\frac{3 \sqrt{13}}{19}\right)^2 = 1 \]

First, calculate \(\left(\frac{3 \sqrt{13}}{19}\right)^2\): \[ \left(\frac{3 \sqrt{13}}{19}\right)^2 = \frac{9 \cdot 13}{361} = \frac{117}{361} \] Now, substitute this back into the equation: \[ \sin^2 x + \frac{117}{361} = 1 \]

Subtract \(\frac{117}{361}\) from both sides to isolate \(\sin^2 x\): \[ \sin^2 x = 1 - \frac{117}{361} = \frac{361 - 117}{361} = \frac{244}{361} \]

Take the square root of both sides to solve for \(\sin x\): \[ \sin x = \sqrt{\frac{244}{361}} = \frac{\sqrt{244}}{19} \] Simplify \(\sqrt{244}\): \[ \sqrt{244} = \sqrt{4 \cdot 61} = 2\sqrt{61} \] Thus: \[ \sin x = \frac{2\sqrt{61}}{19} \]

Final Answer