AP Precalculus -3.2 Sine, Cosine, and Tangent- FRQ Exam Style Questions - Effective Fall 2023

(A)(i) Rewrite \(g(x)\)

The function is given by \(g(x) = 3 \ln x – \frac{1}{2} \ln x\).
Combine the like terms:
\(g(x) = \left(3 – \frac{1}{2}\right) \ln x\)
\(g(x) = \frac{5}{2} \ln x\)
Apply the power property of logarithms, \(a \ln b = \ln(b^a)\):
\(g(x) = \ln\left(x^{5/2}\right)\)

(A)(ii) Rewrite \(h(x)\)

The function is given by \(h(x) = \frac{\sin^2 x – 1}{\cos x}\).
Recall the Pythagorean identity: \(\sin^2 x + \cos^2 x = 1\).
Rearrange the identity to isolate the numerator expression: \(\sin^2 x – 1 = -\cos^2 x\).
Substitute this into the function:
\(h(x) = \frac{-\cos^2 x}{\cos x}\)
Simplify the expression by canceling one \(\cos x\) term:
\(h(x) = -\cos x\)

(B)(i) Solve \(j(x) = 0\)

Set the function equal to zero: \(2(\sin x)(\cos x) = 0\).
Divide both sides by 2:
\(\sin x \cos x = 0\)
By the zero product property, either \(\sin x = 0\) or \(\cos x = 0\).
Case 1: \(\sin x = 0\). In the interval \(\left[0, \frac{\pi}{2}\right]\), \(x = 0\).
Case 2: \(\cos x = 0\). In the interval \(\left[0, \frac{\pi}{2}\right]\), \(x = \frac{\pi}{2}\).
The solutions are \(x = 0\) and \(x = \frac{\pi}{2}\).

(B)(ii) Solve \(k(x) = 3e\)

Set the function equal to \(3e\):
\(8e^{(3x)} – e = 3e\)
Add \(e\) to both sides:
\(8e^{(3x)} = 4e\)
Divide both sides by 8:
\(e^{(3x)} = \frac{4e}{8}\)
\(e^{(3x)} = \frac{e}{2}\)
Take the natural logarithm (\(\ln\)) of both sides:
\(\ln\left(e^{3x}\right) = \ln\left(\frac{e}{2}\right)\)
Use logarithm properties to simplify (\(\ln(e^a) = a\) and \(\ln(a/b) = \ln a – \ln b\)):
\(3x = \ln e – \ln 2\)
Since \(\ln e = 1\):
\(3x = 1 – \ln 2\)
Divide by 3:
\(x = \frac{1 – \ln 2}{3}\)

(C) Find input values for \(m(x) = \frac{9}{2}\)

Set the function equal to \(\frac{9}{2}\):
\(\cos(2x) + 4 = \frac{9}{2}\)
Subtract 4 from both sides (note that \(4 = \frac{8}{2}\)):
\(\cos(2x) = \frac{9}{2} – \frac{8}{2}\)
\(\cos(2x) = \frac{1}{2}\)
The reference angle for cosine equal to \(\frac{1}{2}\) is \(\frac{\pi}{3}\).
The general solution for \(2x\) is:
\(2x = \frac{\pi}{3} + 2\pi n\) or \(2x = -\frac{\pi}{3} + 2\pi n\) (where \(n\) is an integer).
Solve for \(x\) by dividing by 2:
\(x = \frac{\pi}{6} + \pi n\) or \(x = -\frac{\pi}{6} + \pi n\)
Combining these, the input values are \(x = \pm \frac{\pi}{6} + \pi n\) for any integer \(n\).