Question

step 1 find the quadrant in which p(a, b) lies: p(a, b) is in quadrant iv. step 2 use the point and the pythagorean theorem to determine the value of r: r = ±√((a²)+(b²)), but since r must be positive, r = √(a² + b²). step 3 determine cos θ. cos θ = -a/√(a² + b²)= -a√(a² + b²)/(a² + b²), where a and b are positive. which of the following explains whether the student is correct? the student made an error in step 3 because a is positive in quadrant iv, therefore, cos θ = a/√(a² + b²)= a√(a² + b²)/(a² + b²). the student made an error in step 3 because cos θ = -b/√(a² + b²)= -b√(a² + b²)/(a² + b²).

Explanation:

Step1: Recall cosine definition in polar - rectangular conversion

In the polar - rectangular coordinate conversion, for a point \(P(a,b)\) and \(r = \sqrt{a^{2}+b^{2}}\), \(\cos\theta=\frac{x}{r}\), where \(x\) is the \(x\) - coordinate of the point in rectangular coordinates.

Step2: Analyze the sign of \(a\) in Quadrant IV

In Quadrant IV, the \(x\) - coordinate (\(a\)) of the point \(P(a,b)\) is positive and the \(y\) - coordinate (\(b\)) is negative.

Step3: Determine the correct formula for \(\cos\theta\)

Since \(\cos\theta=\frac{a}{r}\) and \(r = \sqrt{a^{2}+b^{2}}\), then \(\cos\theta=\frac{a}{\sqrt{a^{2}+b^{2}}}=\frac{a\sqrt{a^{2}+b^{2}}}{a^{2}+b^{2}}\) (by rationalizing the denominator). The student made an error in step 3 because \(a\) is positive in Quadrant IV.

Answer:

The student made an error in step 3 because \(a\) is positive in Quadrant IV; therefore, \(\cos\theta=\frac{a}{\sqrt{a^{2}+b^{2}}}=\frac{a\sqrt{a^{2}+b^{2}}}{a^{2}+b^{2}}\)