Questions: If tan θ = 1/2, -π/2 < θ < π/2, then sin θ =. (Simplify your answer, including any radicals. Use integers or fractions for any numbers.)

Solution Steps

To find \(\sin \theta\) given \(\tan \theta = \frac{1}{2}\) and \(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\), we can use the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). We can solve for \(\sin \theta\) and \(\cos \theta\) using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\).

1. Use the given \(\tan \theta = \frac{1}{2}\) to express \(\sin \theta\) and \(\cos \theta\) in terms of each other.
2. Use the Pythagorean identity to solve for \(\sin \theta\).

We are given that \( \tan \theta = \frac{1}{2} \) and the range of \( \theta \) is \( -\frac{\pi}{2} < \theta < \frac{\pi}{2} \).

Using the definition of tangent, we have: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{1}{2} \] This implies: \[ \sin \theta = \frac{1}{2} \cos \theta \]

Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we substitute \( \sin \theta \): \[ \left(\frac{1}{2} \cos \theta\right)^2 + \cos^2 \theta = 1 \] This simplifies to: \[ \frac{1}{4} \cos^2 \theta + \cos^2 \theta = 1 \] Combining the terms gives: \[ \frac{5}{4} \cos^2 \theta = 1 \]

Rearranging the equation, we find: \[ \cos^2 \theta = \frac{4}{5} \] Taking the square root, we have: \[ \cos \theta = \pm \sqrt{\frac{4}{5}} = \pm \frac{2}{\sqrt{5}} \] Since \( -\frac{\pi}{2} < \theta < \frac{\pi}{2} \), \( \cos \theta \) is positive: \[ \cos \theta = \frac{2}{\sqrt{5}} \]

Now substituting back to find \( \sin \theta \): \[ \sin \theta = \frac{1}{2} \cos \theta = \frac{1}{2} \cdot \frac{2}{\sqrt{5}} = \frac{1}{\sqrt{5}} \]

Final Answer