91Ó°ÊÓ

We are given \(x = 4 \tan \theta\). Since \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we can write \(x = 4 \frac{\sin \theta}{\cos \theta}\). From this, we get \(\cos \theta = \frac{4 \sin \theta}{x}\).

Recall the Pythagorean identity for sine and cosine: \(\sin^2 \theta + \cos^2 \theta = 1\). Since we have an expression for \(\cos \theta\) in terms of \(\sin \theta\) and \(x\), we can substitute it into the Pythagorean identity. So, we get \(\sin^2 \theta + \left(\frac{4 \sin \theta}{x}\right)^2 = 1\).

Start by simplifying the equation obtained in step 2: \(\sin^2 \theta + \frac{16 \sin^2 \theta}{x^2} = 1\). Now we have a quadratic equation in terms of \(\sin^2 \theta\). Factor out \(\sin^2 \theta\): \(\sin^2 \theta\left(1 + \frac{16}{x^2}\right) = 1\). To find \(\sin \theta\), we need to find \(\sin^2 \theta\) first. Divide both sides by \(\left(1 + \frac{16}{x^2}\right)\): \(\sin^2 \theta = \frac{1}{1 + \frac{16}{x^2}}\). Now, we can take the square root of both sides. We have two solutions: \(\sin \theta = \pm\sqrt{\frac{1}{1 + \frac{16}{x^2}}}\), but we are only interested in the positive solution as \(\sin \theta\) represents the ratio of the side opposite to the angle \(\theta\) and the hypotenuse in a right-angled triangle, and these sides must be positive. Therefore, the final answer is:

The final expression for \(\sin \theta\) in terms of \(x\) is \(\sin \theta = \sqrt{\frac{1}{1 + \frac{16}{x^2}}}\) or, equivalently, \(\sin \theta = \frac{x}{\sqrt{x^2 + 16}}\).