Use the Double Angle formula Given that \( \tan \theta=\frac{3}{2} \) and \( \cos \theta<0 \) Make a sketch of \( \theta \) in the appr quadrant. \( \sin (2 \theta) \) is \( \cos (2 \theta) \) is \( \tan (2 \theta) \) is

Certainly! Let's work through the problem step by step using the double angle formulas. ### **Given:** - \( \tan \theta = \frac{3}{2} \) - \( \cos \theta < 0 \) ### **1. Determining the Quadrant for \( \theta \)** - **Understanding the Signs:** - \( \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{2} \) implies that both sine and cosine have the same sign since tangent is positive. - \( \cos \theta < 0 \) indicates that cosine is negative. - **Quadrant Analysis:** - **Quadrant I:** All trigonometric functions are positive. (Not suitable since \( \cos \theta < 0 \)) - **Quadrant II:** Sine is positive, cosine is negative. (Tangent is negative here.) - **Quadrant III:** Both sine and cosine are negative, making tangent positive. (This fits both conditions.) - **Quadrant IV:** Sine is negative, cosine is positive. (Tangent is negative here.) - **Conclusion:** \( \theta \) is in **Quadrant III**. - **Sketch of \( \theta \) in Quadrant III:** Imagine the Cartesian plane divided into four quadrants. Quadrant III is the lower-left quadrant where both \( x \) (cosine) and \( y \) (sine) coordinates are negative. Here's a rough sketch: ``` | | II | I | ------------------- | III | IV | ``` \( \theta \) lies in the lower-left quadrant (Quadrant III). ### **2. Calculating \( \sin(2\theta) \), \( \cos(2\theta) \), and \( \tan(2\theta) \)** **First, determine \( \sin \theta \) and \( \cos \theta \):** Given \( \tan \theta = \frac{3}{2} \), we can consider a right triangle where: - Opposite side = 3 - Adjacent side = 2 - Hypotenuse = \( \sqrt{3^2 + 2^2} = \sqrt{13} \) However, since \( \theta \) is in Quadrant III, both \( \sin \theta \) and \( \cos \theta \) are negative. - \( \sin \theta = -\frac{3}{\sqrt{13}} \) - \( \cos \theta = -\frac{2}{\sqrt{13}} \) **Now, apply the Double Angle Formulas:** 1. **\( \sin(2\theta) \):** \[ \sin(2\theta) = 2 \sin \theta \cos \theta \] Substitute the values: \[ \sin(2\theta) = 2 \left(-\frac{3}{\sqrt{13}}\right) \left(-\frac{2}{\sqrt{13}}\right) = 2 \left(\frac{6}{13}\right) = \frac{12}{13} \] 2. **\( \cos(2\theta) \):** \[ \cos(2\theta) = \cos^2 \theta - \sin^2 \theta \] Substitute the values: \[ \cos(2\theta) = \left(-\frac{2}{\sqrt{13}}\right)^2 - \left(-\frac{3}{\sqrt{13}}\right)^2 = \frac{4}{13} - \frac{9}{13} = -\frac{5}{13} \] 3. **\( \tan(2\theta) \):** \[ \tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta} \] Substitute \( \tan \theta = \frac{3}{2} \): \[ \tan(2\theta) = \frac{2 \times \frac{3}{2}}{1 - \left(\frac{3}{2}\right)^2} = \frac{3}{1 - \frac{9}{4}} = \frac{3}{-\frac{5}{4}} = -\frac{12}{5} \] ### **Final Results:** - \( \sin(2\theta) = \frac{12}{13} \) - \( \cos(2\theta) = -\frac{5}{13} \) - \( \tan(2\theta) = -\frac{12}{5} \) ### **Summary:** | Function | Value | |----------------|--------------| | \( \sin(2\theta) \) | \( \frac{12}{13} \) | | \( \cos(2\theta) \) | \( -\frac{5}{13} \) | | \( \tan(2\theta) \) | \( -\frac{12}{5} \) | These results are derived using the double angle formulas tailored to the given values and the quadrant in which \( \theta \) lies.