If \( \tan x=\frac{3}{4} \), pind \( \sin 2 x, \cos 2 x \) \( 2 x=\frac{2 \tan x}{1+\tan ^{2} x}=\frac{24}{25} \) \( 2 x=\frac{1-\tan ^{2} x}{1+\tan ^{2} x}=\frac{7}{25} \)

Certainly! Let's walk through the problem step by step to find \( \sin 2x \) and \( \cos 2x \) given that \( \tan x = \frac{3}{4} \). ### Given: \[ \tan x = \frac{3}{4} \] ### Formulas to Use: 1. **Double Angle Formulas:** \[ \sin 2x = \frac{2 \tan x}{1 + \tan^2 x} \] \[ \cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x} \] ### Calculations: 1. **Calculate \( \sin 2x \):** \[ \sin 2x = \frac{2 \cdot \frac{3}{4}}{1 + \left(\frac{3}{4}\right)^2} = \frac{\frac{6}{4}}{1 + \frac{9}{16}} = \frac{\frac{6}{4}}{\frac{25}{16}} = \frac{6}{4} \times \frac{16}{25} = \frac{96}{100} = \frac{24}{25} \] 2. **Calculate \( \cos 2x \):** \[ \cos 2x = \frac{1 - \left(\frac{3}{4}\right)^2}{1 + \left(\frac{3}{4}\right)^2} = \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}} = \frac{\frac{7}{16}}{\frac{25}{16}} = \frac{7}{16} \times \frac{16}{25} = \frac{7}{25} \] ### Final Results: \[ \sin 2x = \frac{24}{25} \] \[ \cos 2x = \frac{7}{25} \] ### Summary: Given \( \tan x = \frac{3}{4} \), using the double angle formulas, we've determined that: - \( \sin 2x = \frac{24}{25} \) - \( \cos 2x = \frac{7}{25} \) If you have any further questions or need additional clarification, feel free to ask!