Home›Calculus 1›Amplitude, period and phase shift›Problem 13.100

Amplitude, period and phase shift

Problem 13.100 · medium

For \( \displaystyle y = - 4 \sin{\left(2 x \right)} \), find the amplitude, the period, the phase shift (with its direction) and the vertical shift.

Write it as A·sin(B(x − h)) + D: A = 4, B = 2, h = pi/2, D = 0.
\[ \frac{2 \pi}{2} = \pi \]
Period = 2π/B.✓ Proved
\[ 4 \sin{\left(2 x \right)} + \left. - 4 \sin{\left(2 x \right)} \right|_{\substack{ x=x + \pi }} = 0 \]
Shifting x by one period gives the same function.✓ Proved
Phase shift h = C/B: pi/2.
\[ \left. - 4 \sin{\left(2 x \right)} \right|_{\substack{ x=\frac{3 \pi}{4} }} = 4 \]
At x = 3*pi/4 the curve is 4 from its midline y = 0: that is the amplitude.✓ Proved

Answer \( \text{amplitude } 4,\ \text{period } \pi,\ \text{phase shift } \frac{\pi}{2}\text{ to the right},\ \text{vertical shift } 0 \)

Lines: 3 proved, 2 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify

The full receipt

Line	Status	Checked by	Detail
1	Not checked	—	a sentence; read, not computed
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
4	Not checked	—	a sentence; read, not computed
5	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	max and min over 16000 samples give the amplitude and midline; f(x + p) = f(x) for the period and not half of it

Reviewers

gpt-oss:20b: fail (style) — [domain objection, downgraded to style] Sentence 4 incorrectly refers to a parameter C that was never defined; it should state that the phase shift is h=π/2 to the right, not use the formula C/B. This could mislead a student about the correct form of the phase‑shift formula.
qwen3.6:27b-mlx: inconclusive — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly identifies the phase shift as pi/2 to the right. The function y = -4sin(2x) can be written as 4sin(2x - pi), which correspo

Every verdict on record (6)

qwen3.6:27b-mlx: inconclusive 2026-10-17 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly identifies the phase shift as pi/2 to the right. The function y = -4sin(2x) can be written as 4sin(2x - pi), which correspo
gpt-oss:20b: fail (style) 2026-10-17 — [domain objection, downgraded to style] Sentence 4 incorrectly refers to a parameter C that was never defined; it should state that the phase shift is h=π/2 to the right, not use the formula C/B. This could mislead a student about the correct form of the phase‑shift formula.
qwen3.6:27b-mlx: fail (style) 2026-10-14 — [domain objection, downgraded to style] The solution incorrectly identifies the phase shift as π/2 to the right. The function y = -4 sin(2x) is equivalent to 4 sin(2x - π), which corresponds to a phase shift of π/2 to the right only if the form is A sin(B(x-h)). However, -sin(θ) = sin(θ - π), so -4 sin(2x) = 4 sin(2x - π) = 4 sin(2(x - π/2)). Wait, let's re-evaluate. -sin(u) = sin(u + π) or sin(u - π). If we use sin(u - π), then -4 sin(2x) = 4 sin(2x - π) = 4 sin(2(x - π/2)). This implies a shift of π/2 to the right. But standard convention often prefers the smallest positive shift or specific forms. Let's check the graph. At x=0, y=0. At x=π/4, y=-4. A standard sin(2x) has peak at π/4. -sin(2x) has trough at π/4. The function 4 sin(2(x-π/2)) = 4 sin(2x - π). At x=π/2, 4 sin(0)=0. At x=3π/4, 4 sin(π/2)=4. This matches. So π/2 right is a valid phase shift representation. However, usually phase shift is defined for A sin(Bx - C) + D as C/B. Here we have -4 sin(2x). To write as A sin(B(x-h)), we need A>0 usually for amplitude definition, but A can be negative. If A=-4, B=2, C=0, h=0. If we force A=4, we get h=π/2. The solution claims h=π/2. Is this wrong? The problem asks for "the" phase shift. Phase shift is not unique. But typically, for y = A sin(Bx + C), shift is -C/B. Here C=0, so shift is 0. The negative sign is part of the amplitude/reflection. Identifying the shift as π/2 relies on converting -sin to sin with a phase change. While mathematically valid, it's not the standard "phase shift" of the given expression -4 sin(2x) which is usually considered 0 (with a reflection). More critically, step 1 says A=4, B=2, h=π/2. This implies the form 4 sin(2(x-π/2)). This is correct. But is it the *intended* answer? Often, phase shift is 0 for sin(2x) and the negative is handled by amplitude. If the question implies the standard form y = A sin(B(x-C)) + D, then C is the phase shift. For -4 sin(2x), we can write it as 4 sin(2x - π) = 4 sin(2(x - π/2)). So π/2 is a valid phase shift. However, many textbooks define phase shift for y = A sin(Bx + C) as -C/B. Here C=0, so shift is 0. The ambiguity is the issue. But let's look at Step 4: "Phase shift h = C/B". In the form A sin(B(x-h)), h is the shift. In the form A sin(Bx + C), shift is -C/B. The solution mixes notations or assumes a specific conversion. The bigger error is likely that for y = -4 sin(2x), the most direct reading is amplitude 4, period π, phase shift 0, vertical shift 0. The negative sign indicates a reflection, not a shift. Converting it to a shift of π/2 is an extra step that changes the "simplest" form. If the provided answer is considered correct, then the solution is fine. But if the standard answer is 0, then it's an error. Given the adversarial nature, claiming a non-zero phase shift for a function with no horizontal translation in its basic form is misleading/wrong in standard contexts. The phase shift of sin(x) is 0. The phase shift of -sin(x) is often cited as 0 (with reflection) or π. π/2 is for cos(x) = sin(x+π/2). -sin(x) = sin(x-π). So shift is π. For 2x, shift is π/2. So π/2 is correct for the form 4sin(2(x-h)). But is it the *only* answer? No. Is it the *best* answer? Debatable. However, looking at Step 1, it asserts A=4. The original A is -4. Changing the sign of A requires changing the phase. This is a valid transformation. But Step 4 says "Phase shift h = C/B". This formula applies to y = A sin(Bx + C). If we use that form on -4 sin(2x), C=0, so h=0. The solution implicitly converts to a different form without stating it clearly, leading to a different h. This is confusing and potentially misleading for a student learning to identify parameters directly from the given equation.
gpt-oss:20b: fail (style) 2026-10-14 — [domain objection, downgraded to style] The setup incorrectly identifies the phase shift. The function y=−4sin(2x) can be written as 4sin(2(x+π/2)), so the phase shift is π/2 to the left, not to the right. Moreover, the sentence “Phase shift h = C/B” refers to an undefined constant C, violating the proper application of the formula.
qwen3.6:27b-mlx: fail (style) 2026-10-14 — [domain objection, downgraded to style] The solution incorrectly identifies the phase shift as pi/2 to the right. The function y = -4sin(2x) can be written as 4sin(2x - pi), which corresponds to a phase shift of pi/2 to the right relative to sin(2x), but standard convention usually defines phase shift for -sin(x) as a shift of pi (or -pi) or simply notes the reflection. More critically, the standard form A sin(B(x-h)) + D requires factoring out B: -4sin(2x) = 4sin(2x - pi) = 4sin(2(x - pi/2)). So h = pi/2 is technically correct for the form sin(B(x-h)). However, the direction is often ambiguous without specifying the base function. Let's look closer. -sin(theta) = sin(theta - pi). So -4sin(2x) = 4sin(2x - pi) = 4sin(2(x - pi/2)). The phase shift is indeed pi/2 to the right. Wait, is it? sin(2(x - pi/2)) = sin(2x - pi) = -sin(2x). Yes. So the math is correct. Why did I think it was wrong? Let's re-read the prompt's stated answer. Stated answer: phase shift pi/2 to the right. Solution says h = pi/2. This matches. Is there an error? Let's check the amplitude. A=4. Correct. Period pi. Correct. Vertical shift 0. Correct. Phase shift pi/2 right. Correct. Is there a subtle error? The solution says 'Write it as A·sin(B(x − h)) + D: A = 4...'. It implicitly converts -4sin(2x) to 4sin(2x - pi). This is a valid transformation. However, usually, if A is negative, we might say amplitude is |A|=4 and phase shift is 0, or keep A negative and shift is 0. But the form A sin(B(x-h)) usually assumes A>0. If A>0, then -4sin(2x) = 4sin(2x - pi) = 4sin(2(x - pi/2)). So h=pi/2. This is correct. Let's look for a different error. Step 4 says 'Phase shift h = C/B'. Where did C come from? The standard form is often A sin(Bx - C) + D, where phase shift is C/B. The solution jumps to h = C/B without defining C in the context of the previous line which used h. It's a bit sloppy but not necessarily an 'error' in result. Let's look at Step 5. 'At x = 3*pi/4 the curve is 4 from its midline... that is the amplitude.' This is a verification, not a derivation. It's fine. Is there a mistake in the phase shift direction? sin(2(x - pi/2)) = sin(2x - pi). Graph of sin(2x) starts at 0, goes up. Graph of -sin(2x) starts at 0, goes down. Graph of sin(2(x - pi/2)) is sin(2x) shifted right by pi/2. sin(2(x - pi/2)) at x=pi/2 is sin(0)=0. At x=3pi/4, 2(3pi/4 - pi/2) = 2(pi/4) = pi/2, sin(pi/2)=1. So 4(1)=4. The point (3pi/4, 4) is on the curve. The peak of sin(2x) is at pi/4. The peak of -sin(2x) is at 3pi/4 (since -sin(2*3pi/4) = -sin(3pi/2) = -(-1) = 1). The peak of sin(2(x-h)) is at h + pi/4. If h=pi/2, peak is at 3pi/4. This matches. So the solution is actually correct. Why would it fail? Maybe the 'standard' phase shift for -sin(x) is considered 0 with a negative amplitude? Or pi? If we write -4sin(2x) = 4sin(2x + pi), then h = -pi/2 (left). Both pi/2 right and pi/2 left are valid phase shifts for sine waves due to periodicity. However, usually we pick the smallest magnitude. pi/2 is smaller than 3pi/2. So pi/2 is fine. Let's look at the 'unchecked' equations. Step 3 checks periodicity. Step 5 checks amplitude. These are just checks. The main issue might be the definition of Phase Shift. Some textbooks define phase shift only for the form sin(Bx - C). If the function is -4sin(2x), one could argue the phase shift is 0 and the amplitude is -4 (or 4 with a reflection). But the problem asks for 'the' phase shift, implying a specific convention. The convention A>0 is standard for 'amplitude' being positive. So converting to positive A is the right move. Is there a calculation error? 2x - pi = 2(x - pi/2). Correct. I suspect the 'error' is that the solution claims h=pi/2 is *the* phase shift, but phase shift is not unique. However, in multiple choice or standard answers, pi/2 is the standard principal value. Let's look really closely at Step 1. 'A=4, B=2, h=pi/2, D=0'. It asserts these values. It doesn't show the conversion from -4 to 4. It just states it. This is a 'sentence'. Is it misleading? It skips the step of showing -4sin(2x) = 4sin(2x - pi). A student might not know how to get from -4 to 4 and h=pi/2. They might think you just flip the sign and add pi/2 arbitrarily. This is 'misleading' or 'style'. But is it an 'error'? The prompt asks for 'error' if a step is wrong. The step is mathematically correct, just abbreviated. Let's look for a harder error. What if the period calculation is wrong? 2pi/2 = pi. Correct. What if the amplitude is wrong? |-4| = 4. Correct. What if the vertical shift is wrong? 0. Correct. I will bet on the lack of justification for the phase shift conversion being 'misleading' or 'style'. But wait, look at Step 4: 'Phase shift h = C/B: pi/2.' This implies C = pi. From where? The previous line didn't define C. It defined h. This is inconsistent notation. Step 1 uses h. Step 4 uses C. This is confusing. Also, Step 1 says 'Write it as... A=4... h=pi/2'. It presents the result of the conversion as a given. This is poor pedagogy. I will mark it as 'misleading' because it doesn't explain how the negative sign was absorbed into the phase shift, which is the core difficulty of the problem. A student might guess h=pi/2 or h=0. Without the identity -sin(u) = sin(u - pi), the jump is unjustified. Actually, looking at the options, 'error' is for wrong steps. 'Misleading' is for false learning. Teaching that you can just pick h=pi/2 for -sin is bad. You must show the identity. I'll go with misleading. Or perhaps there is a simpler error. Is the phase shift pi/2 to the RIGHT? sin(2(x - pi/2)) = sin(2x - pi). Yes. Is it possible the answer key expects 0? No, usually amplitude is positive. I will stick with misleading due to the unjustified jump in Step 1 and inconsistent notation in Step 4.
gpt-oss:20b: fail (error) 2026-10-14 — The solution misidentifies the amplitude and phase shift. The function y = –4 sin(2x) has amplitude |A| = 4, B = 2, no horizontal shift (h = 0), and vertical shift D = 0. The stated h = π/2 and the use of C/B are incorrect.

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/sinusoid_features, checked 2026-10-17 with SymPy 1.14.0.