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

Let’s break it down! We know that if \( \tan x = \frac{3}{4} \), we can find \( \sin x \) and \( \cos x \) using the following relationships: 1. From \( \tan x = \frac{\sin x}{\cos x} \), we can let \( \sin x = 3k \) and \( \cos x = 4k \) for some \( k \). Using the Pythagorean theorem: \[ (3k)^2 + (4k)^2 = 1 \Rightarrow 9k^2 + 16k^2 = 1 \Rightarrow 25k^2 = 1 \Rightarrow k= \frac{1}{5} \] Thus, \( \sin x = \frac{3}{5} \) and \( \cos x = \frac{4}{5} \). 2. Now we can calculate \( \sin 2x \) and \( \cos 2x \): \[ \sin 2x = 2 \sin x \cos x = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25} \] \[ \cos 2x = \cos^2 x - \sin^2 x = \left( \frac{4}{5} \right)^2 - \left( \frac{3}{5} \right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25} \] So, \( \sin 2x = \frac{24}{25} \) and \( \cos 2x = \frac{7}{25} \) are indeed correct!