TgA=1/2 sinA-? cosA-?

To find \(\sin A\) and \(\cos A\) given that \(\tan A = \frac{1}{2}\), we can use the definitions of sine, cosine, and tangent. 1. We start with the identity for tangent: \[ \tan A = \frac{\sin A}{\cos A} \] Given that \(\tan A = \frac{1}{2}\), we can set: \[ \sin A = k \quad \text{and} \quad \cos A = 2k \] for some \(k\). 2. Now using the Pythagorean identity: \[ \sin^2 A + \cos^2 A = 1 \] Substituting \(k\) and \(2k\) into the identity: \[ k^2 + (2k)^2 = 1 \] Simplifying this gives: \[ k^2 + 4k^2 = 1 \\ 5k^2 = 1 \\ k^2 = \frac{1}{5} \\ k = \frac{1}{\sqrt{5}} \] 3. Consequently, we can find \(\sin A\) and \(\cos A\): \[ \sin A = k = \frac{1}{\sqrt{5}} \\ \cos A = 2k = 2 \cdot \frac{1}{\sqrt{5}} = \frac{2}{\sqrt{5}} \] 4. For clarity, we can rationalize the denominators: \[ \sin A = \frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{5} \\ \cos A = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \] So, we have: \[ \sin A = \frac{\sqrt{5}}{5}, \quad \cos A = \frac{2\sqrt{5}}{5}. \]