\( \tan \theta=3 \) and \( \cos \theta<0 \), use a sketch in the correct quadrant to determine the we of: \( \frac{\sin \theta+\cos \theta}{2 \sin \theta} \) without using a calculator.

To solve for \( \frac{\sin \theta + \cos \theta}{2 \sin \theta} \) given \( \tan \theta = 3 \) and \( \cos \theta < 0 \), we first note that since \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \cos \theta < 0 \), this implies that \( \theta \) is in the second quadrant. Now, using the definition of tangent, we can express sine and cosine in terms of a reference triangle. If we imagine a triangle where the opposite side (for sine) is 3 and the adjacent side (for cosine) is -1 (since cosine is negative), we can find the hypotenuse: \[ r = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} \] Now we can find \( \sin \theta \) and \( \cos \theta \): \[ \sin \theta = \frac{3}{\sqrt{10}}, \quad \cos \theta = \frac{-1}{\sqrt{10}} \] Next, substitute these values into the expression \( \frac{\sin \theta + \cos \theta}{2 \sin \theta} \): \[ \sin \theta + \cos \theta = \frac{3}{\sqrt{10}} + \frac{-1}{\sqrt{10}} = \frac{2}{\sqrt{10}} \] Now, note that \( 2 \sin \theta = 2 \cdot \frac{3}{\sqrt{10}} = \frac{6}{\sqrt{10}} \). So we plug these back into our expression: \[ \frac{\sin \theta + \cos \theta}{2 \sin \theta} = \frac{\frac{2}{\sqrt{10}}}{\frac{6}{\sqrt{10}}} = \frac{2}{6} = \frac{1}{3} \] Therefore, the value is: \[ \frac{\sin \theta + \cos \theta}{2 \sin \theta} = \frac{1}{3} \]