If \(\cos^2 x=\frac{9}{16}\) and x is in the fourth quadrant, what is the value of \(\sin x\)?

Correct answer: \(-\frac{\sqrt{7}}{4}\)

Begin with \(\sin^2x+\cos^2x=1\). Since \(\cos^2x=9/16\), we get \(\sin^2x=1-9/16=7/16\). Taking square roots gives \(|\sin x|=\sqrt7/4\). A square-root step gives two possible signs, so the quadrant information is essential. In the fourth quadrant, sine is negative; therefore \(\sin x=-\sqrt7/4\), which is option B. Option A has the correct magnitude but the wrong sign. Options C and D use \(3/4\), which is the magnitude of cosine, not sine. Thus the identity supplies the magnitude and the quadrant supplies the sign.