Signs of trigonometric functions in different quadrants

Correct answer: A. Fourth quadrant

The governing concept is locating a radian angle by comparing it with the quadrant boundaries. From θ/π = 1.65, multiply by π to obtain θ = 1.65π. The four quadrants are bounded by 0, π/2, π, 3π/2, and 2π. Since 3π/2 = 1.5π and 1.5π < 1.65π < 2π, the angle lies between 3π/2 and 2π. This interval is the fourth quadrant, so option A is correct. It is not in the third quadrant because that quadrant extends only from π to 3π/2. The given condition 0 < θ < 2π confirms that the angle makes one complete revolution range without reaching either endpoint.

Correct answer: C. \(-\frac{8}{17}\)

Use the fundamental identity \(\sin^2x+\cos^2x=1\). Substituting \(\cos x=15/17\), we obtain \(\sin^2x=1-(15/17)^2=1-225/289=64/289\). Therefore, \(|\sin x|=8/17\). The quadrant determines the sign: in the fourth quadrant, cosine is positive but sine is negative. Hence \(\sin x=-8/17\), so option C is correct. Option A has the right magnitude but the wrong sign; options B and D confuse the sine value with the given cosine value or use an incorrect magnitude. The answer follows from both the identity and the quadrant sign rule.

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

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.