Question 5 S.1 Givea: \( \tan x=\frac{3}{4} \), where \( x \in\left[180^{\circ} ; 270^{\circ}\right] \). With the aid of a skotch, and without the ase of a ealenalater, culoulate: \( 5.1 .1 . \sin x \) \( 5.1 .2 .2-\sin 2 x \)

To solve the problem, we start with the given condition \( \tan x = \frac{3}{4} \) and the interval \( x \in [180^{\circ}, 270^{\circ}] \). ### Step 1: Determine the quadrant Since \( x \) is in the interval \( [180^{\circ}, 270^{\circ}] \), it is in the third quadrant. In this quadrant, both sine and cosine are negative. ### Step 2: Use the tangent to find sine and cosine We know that: \[ \tan x = \frac{\sin x}{\cos x} \] Given \( \tan x = \frac{3}{4} \), we can set: \[ \sin x = -3k \quad \text{and} \quad \cos x = -4k \] for some positive \( k \) (since sine and cosine are negative in the third quadrant). ### Step 3: Use the Pythagorean identity Using the identity \( \sin^2 x + \cos^2 x = 1 \): \[ (-3k)^2 + (-4k)^2 = 1 \] \[ 9k^2 + 16k^2 = 1 \] \[ 25k^2 = 1 \] \[ k^2 = \frac{1}{25} \quad \Rightarrow \quad k = \frac{1}{5} \] ### Step 4: Calculate sine and cosine Now we can find \( \sin x \) and \( \cos x \): \[ \sin x = -3k = -3 \cdot \frac{1}{5} = -\frac{3}{5} \] \[ \cos x = -4k = -4 \cdot \frac{1}{5} = -\frac{4}{5} \] ### Step 5: Calculate \( \sin 2x \) Using the double angle formula for sine: \[ \sin 2x = 2 \sin x \cos x \] Substituting the values we found: \[ \sin 2x = 2 \left(-\frac{3}{5}\right) \left(-\frac{4}{5}\right) = 2 \cdot \frac{12}{25} = \frac{24}{25} \] ### Final Results Now we can summarize the results: - \( 5.1.1. \sin x = -\frac{3}{5} \) - \( 5.1.2. 2 - \sin 2x = 2 - \frac{24}{25} = \frac{50}{25} - \frac{24}{25} = \frac{26}{25} \) Thus, the final answers are: - \( \sin x = -\frac{3}{5} \) - \( 2 - \sin 2x = \frac{26}{25} \)