If $\tan \theta$ = $-\frac{3}{8}$, which of the following is the value of $\sin \theta$ if $\theta$ is in the fourth quadrant?

• A. $-\frac{3}{8}$
• B. $-\frac{3}{\sqrt{73}}$
• C. $\frac{3}{\sqrt{73}}$
• D. $-\frac{3}{\sqrt{73}}$

Answer: (D)

Verified step by step guidance

1. Recall the definition of tangent in terms of sine and cosine: \(\tan\theta = \frac{\sin\theta}{\cos\theta}\).
2. Given \(\tan\theta = -\frac{3}{8}\), let \(\sin\theta = y\) and \(\cos\theta = x\). Then \(\frac{y}{x} = -\frac{3}{8}\), which implies \(y = -\frac{3}{8}x\).
3. Use the Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\) to relate \(x\) and \(y\): substitute \(y = -\frac{3}{8}x\) into the identity to get \(\left(-\frac{3}{8}x\right)^2 + x^2 = 1\).
4. Simplify the equation: \(\frac{9}{64}x^2 + x^2 = 1\), which combines to \(\frac{73}{64}x^2 = 1\). Solve for \(x^2\) to find \(x^2 = \frac{64}{73}\), then take the square root to find \(x = \pm \frac{8}{\sqrt{73}}\).
5. Determine the correct sign of \(\cos\theta\) and \(\sin\theta\) based on the quadrant: since \(\theta\) is in the fourth quadrant, \(\cos\theta\) is positive and \(\sin\theta\) is negative. Use \(y = -\frac{3}{8}x\) with \(x = \frac{8}{\sqrt{73}}\) to find \(\sin\theta = -\frac{3}{\sqrt{73}}\).