AP Precalculus: Equivalent Trigonometric Expressions (Drill 1)

Question 1. Which of the following expressions is equivalent to \( 14\sin(x)\cos(x) \)?

• A) \( 7\sin(2x) \) ✓
• B) \( 14\sin(2x) \)
• C) \( 7\cos(2x) \)
• D) \( 14\cos(2x) \)

Explanation: Choice A is correct. The double-angle identity states \( \sin(2x) = 2\sin(x)\cos(x) \), so \( \sin(x)\cos(x) = \dfrac{\sin(2x)}{2} \). Therefore \( 14\sin(x)\cos(x) = 14 \cdot \dfrac{\sin(2x)}{2} = 7\sin(2x) \). Choice B is incorrect because the student applied \( \sin(2x) = 2\sin(x)\cos(x) \) without dividing by 2, leaving the coefficient at 14 instead of halving it to 7. Choice C is incorrect because the student used the cosine double-angle identity in place of the sine double-angle identity; \( \cos(2x) \neq 2\sin(x)\cos(x) \). Choice D is incorrect because the student made both errors simultaneously: wrong identity and wrong coefficient.

Question 2. Which of the following expressions is equivalent to \( \sin^2(x) + \sin^2(x)\tan^2(x) \)?

• A) \( \sin^2(x) \)
• B) \( \cos^2(x) \)
• C) \( \tan^2(x) \) ✓
• D) 1

Explanation: Choice C is correct. Factor: \( \sin^2(x)\bigl(1 + \tan^2(x)\bigr) \). The Pythagorean identity gives \( 1 + \tan^2(x) = \sec^2(x) \), so the expression becomes \( \sin^2(x)\sec^2(x) = \dfrac{\sin^2(x)}{\cos^2(x)} = \tan^2(x) \). Choice A is incorrect because the student dropped the \( \tan^2(x) \) term, as if \( \tan^2(x) = 0 \). Choice B is incorrect because the student confused this structure with \( \sin^2(x) + \cos^2(x) = 1 \) and then solved for \( \cos^2(x) \), which does not apply here. Choice D is incorrect because the result is a trigonometric function of \( x \), not a constant; the student likely confused this with the identity \( \sin^2(x) + \cos^2(x) = 1 \).

Question 3. A student claims that \( \cos(2x) \) and \( 1 - \sin^2(x) \) are equivalent expressions. The table below shows values of each expression at selected inputs.

x	cos(2x)	1 − sin²(x)
0	1	1
π/6	1/2	3/4
π/4	0	1/2
π/3	−1/2	1/4

Based on the table, which of the following best describes the student's claim?

• A) The claim is correct because both expressions equal 1 when \( x = 0 \).
• B) The claim is correct because the values in the table are close enough to confirm equivalence.
• C) The claim is incorrect; the correct equivalent form is \( 1 - 2\sin^2(x) \). ✓
• D) The claim is incorrect; the correct equivalent form is \( 2\cos^2(x) - 1 \).

Explanation: Choice C is correct. The table shows the expressions differ: at \( x = \pi/6 \), \( \cos(2 \cdot \pi/6) = \cos(\pi/3) = \dfrac{1}{2} \), but \( 1 - \sin^2(\pi/6) = 1 - \dfrac{1}{4} = \dfrac{3}{4} \). The student's expression is missing a factor of 2 on the \( \sin^2(x) \) term. The correct identity is \( \cos(2x) = 1 - 2\sin^2(x) \), derived by substituting \( \cos^2(x) = 1 - \sin^2(x) \) into \( \cos(2x) = \cos^2(x) - \sin^2(x) \). Choice A is incorrect because agreement at a single input does not establish equivalence; two different functions can share an equal value at an isolated point. Choice B is incorrect because the table values are not close: at \( x = \pi/3 \), one expression equals \( -\dfrac{1}{2} \) and the other equals \( \dfrac{1}{4} \). Choice D is incorrect because, although \( 2\cos^2(x) - 1 \) is also equivalent to \( \cos(2x) \), it does not directly address the student's incorrect expression \( 1 - \sin^2(x) \). Choice C specifically identifies the missing factor of 2 on \( \sin^2(x) \), making it the best correction of the student's claim.

Question 4. A student is solving the equation \( \cos(2\theta) + \cos\theta = 0 \) on \( [0, 2\pi) \). After substituting \( \cos(2\theta) = 2\cos^2\theta - 1 \), the equation becomes \( 2\cos^2\theta + \cos\theta - 1 = 0 \). Which of the following lists all solutions?

• A) \( \theta = \dfrac{\pi}{3}, \dfrac{5\pi}{3} \) only
• B) \( \theta = \dfrac{2\pi}{3}, \dfrac{4\pi}{3} \) only
• C) \( \theta = \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, \dfrac{5\pi}{3} \)
• D) \( \theta = \dfrac{\pi}{3}, \pi, \dfrac{5\pi}{3} \) ✓

Explanation: Choice D is correct. Factor \( 2\cos^2\theta + \cos\theta - 1 = (2\cos\theta - 1)(\cos\theta + 1) = 0 \). From \( 2\cos\theta - 1 = 0 \): \( \cos\theta = \dfrac{1}{2} \), giving \( \theta = \dfrac{\pi}{3} \) and \( \theta = \dfrac{5\pi}{3} \). From \( \cos\theta + 1 = 0 \): \( \cos\theta = -1 \), giving \( \theta = \pi \). All three lie in \( [0, 2\pi) \). Choice A is incorrect because the student found only the solutions from \( \cos\theta = \dfrac{1}{2} \) and omitted \( \theta = \pi \) from the second factor. Choice B is incorrect because \( \cos(\dfrac{2\pi}{3}) = \cos(\dfrac{4\pi}{3}) = -\dfrac{1}{2} \), which satisfies neither factor; the student applied the wrong reference angle, confusing \( \cos\theta = -\dfrac{1}{2} \) with \( \cos\theta = \dfrac{1}{2} \) in the wrong quadrants. Choice C is incorrect because it combines the errors of choice B with the correct values from \( \cos\theta = \dfrac{1}{2} \), overcounting with extraneous values and also omitting \( \theta = \pi \).

Question 5. Which of the following expressions is equivalent to \( \sin^4(x) - \cos^4(x) \), written only in terms of \( \sin(x) \)?

• A) \( \sin^2(x) - \cos^2(x) \)
• B) \( 2\sin^2(x) - 1 \) ✓
• C) \( -\cos(2x) \)
• D) \( \sin^2(x) + \cos^2(x) \)

Explanation: Choice B is correct. Factor as a difference of squares: \( \sin^4(x) - \cos^4(x) = (\sin^2(x) - \cos^2(x))(\sin^2(x) + \cos^2(x)) \). Since \( \sin^2(x) + \cos^2(x) = 1 \), this reduces to \( \sin^2(x) - \cos^2(x) \). Replace \( \cos^2(x) \) using the Pythagorean identity \( \cos^2(x) = 1 - \sin^2(x) \): \( \sin^2(x) - (1 - \sin^2(x)) = 2\sin^2(x) - 1 \). This is the only answer written entirely in terms of \( \sin(x) \). Choice A is incorrect because it represents only the first factoring step and still contains \( \cos^2(x) \), so it does not meet the requirement of being written only in terms of \( \sin(x) \). Choice C is incorrect because \( -\cos(2x) \) is mathematically equivalent to \( 2\sin^2(x) - 1 \), but it introduces a cosine function and therefore does not satisfy the requirement of being written only in terms of \( \sin(x) \). Choice D is incorrect because \( \sin^2(x) + \cos^2(x) = 1 \) by the Pythagorean identity; this is the sum factor that simplifies to 1 during the factoring step, not the final result.