Home›Calculus 1›Amplitude, period and phase shift›Problem 13.392

Amplitude, period and phase shift

Problem 13.392 · medium

For \( \displaystyle y = 3 - 4 \cos{\left(3 x - 3 \right)} \), find the amplitude, the period, the phase shift (with its direction) and the vertical shift.

Write it as A·cos(B(x − h)) + D: A = -4, B = 3, h = 1, D = 3.
Period = 2π/B: 2*pi/3.
\[ \left(4 \cos{\left(3 x - 3 \right)} - 3\right) + \left. 3 - 4 \cos{\left(3 x - 3 \right)} \right|_{\substack{ x=x + \frac{2 \pi}{3} }} = 0 \]
Shifting x by one period gives the same function.✓ Proved
\[ \frac{3}{3} = 1 \]
Phase shift h = C/B.✓ Proved
\[ \left. 3 - 4 \cos{\left(3 x - 3 \right)} \right|_{\substack{ x=1 }} = -1 \]
At x = 1 the curve is 4 from its midline y = 3: that is the amplitude.✓ Proved

Answer \( \text{amplitude } 4,\ \text{period } \frac{2 \pi}{3},\ \text{phase shift } 1\text{ to the right},\ \text{vertical shift } 3 \)

Lines: 3 proved, 2 not checked. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify

The full receipt

Line	Status	Checked by	Detail
1	Not checked	—	a sentence; read, not computed
2	Not checked	—	a sentence; read, not computed
3	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
4	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
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: pass
qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] Step 1 incorrectly identifies A as -4; for the form A·cos(B(x-h)) + D, A is 4 (since 3-4cos(...) = 3 + 4cos(... - pi) or simply by factoring out 4 from the cosine term's coefficient magnitude for amplitude, but strictly A=4 if we write 4cos(3(x-1)+pi) or A=-4 if we write -4cos(3(x-1)). However, standard convention usually takes A>0 for amplitude definition, but the bigger issue is Step 5 claims the value -1 proves the amplitude is 4 because it is '4 from its midline'. While the calculation 3-4cos(0) = -1 is correct and |-1-3|=4 is correct, the logic in Step 1 is ambiguous about the sign of A relative to the phase shift definition. More critically, Step 1 says A=-4. If A=-4, the function is -4cos(3(x-1))+3. This is equivalent to 4cos(3(x-1)+pi)+3. The phase shift h=1 is correct for the form -4cos(3(x-1)). But standard textbooks often define phase shift for A cos(Bx - C) as C/B. Here 3x-3 = 3(x-1), so h=1. The solution is technically consistent if A=-4 is allowed, but Step 5's justification 'At x=1 the curve is 4 from its midline' is a valid check for amplitude, not a derivation. The main error is that Step 1 sets A=-4, which is a valid parameterization, but usually amplitude is defined as |A|. The solution lists amplitude 4, which is correct. The phase shift is 1 to the right. This is correct. Wait, let's look closer. y = 3 - 4cos(3x-3). Rewrite as 3 + 4cos(3x-3 + pi). Or 3 - 4cos(3(x-1)). If we use the form A cos(B(x-h)) + D, we can have A=4, h=1+pi/3? No. -4cos(3(x-1)) = 4cos(3(x-1) - pi) = 4cos(3(x - 1 - pi/3)). So if A=4, h = 1 + pi/3. If A=-4, h=1. The solution chooses A=-4, h=1. This is a valid representation. Is it misleading? Students are often taught A>0. If A must be positive, the phase shift is different. However, the problem asks for 'the' phase shift. Usually, phase shift is defined for the standard form y = A cos(B(x - C)) + D. If A is negative, it's often rewritten. But many calculators/texts accept negative A. The bigger issue is Step 5: 'At x=1 the curve is 4 from its midline... that is the amplitude.' This is a verification, not a derivation, but acceptable. The real issue is Step 1: 'A=-4'. Amplitude is |A|=4. The solution states A=-4. This is a parameter, not the amplitude itself. The final answer lists amplitude 4, so it's consistent. Is there an error? Let's check the period. 2pi/3. Correct. Vertical shift 3. Correct. Phase shift 1 right. Correct. The solution seems correct in its results. Why would it fail? Step 1 says 'Write it as A·cos(B(x − h)) + D: A = -4...'. This is a valid algebraic manipulation. Step 5 checks the amplitude. It seems sound. Let me re-read the prompt. 'Find... amplitude... phase shift'. The solution provides them. Is there a subtle error? 'Phase shift 1 to the right'. For y = 3 - 4cos(3(x-1)), the graph is a cosine wave reflected and shifted. The 'standard' phase shift often assumes A>0. If A>0, y = 3 + 4cos(3x - 3 + pi) = 3 + 4cos(3(x - 1 + pi/3)). Phase shift would be 1 - pi/3. But 'phase shift' is ambiguous for negative A. Most contexts accept the shift of the argument. 3x-3=0 => x=1. This is the start of the cycle for cos(3x-3) if we consider the standard cosine starting at max. But here we have -cos, so it starts at min. Does 'phase shift' refer to the shift of the parent function cos(x)? If so, cos(3(x-1)) is shifted right by 1. The negative sign is a reflection, not a shift. So h=1 is correct for the argument. The solution is likely 'pass'. Wait, look at Step 5 again. 'Subs(..., x, 1) = -1'. 3 - 4cos(0) = -1. Correct. '4 from its midline y=3'. |-1-3|=4. Correct. 'That is the amplitude'. Correct. I see no error. I will mark pass.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (style) 2026-10-20 — [domain objection, downgraded to style] Step 1 incorrectly identifies A as -4; for the form A·cos(B(x-h)) + D, A is 4 (since 3-4cos(...) = 3 + 4cos(... - pi) or simply by factoring out 4 from the cosine term's coefficient magnitude for amplitude, but strictly A=4 if we write 4cos(3(x-1)+pi) or A=-4 if we write -4cos(3(x-1)). However, standard convention usually takes A>0 for amplitude definition, but the bigger issue is Step 5 claims the value -1 proves the amplitude is 4 because it is '4 from its midline'. While the calculation 3-4cos(0) = -1 is correct and |-1-3|=4 is correct, the logic in Step 1 is ambiguous about the sign of A relative to the phase shift definition. More critically, Step 1 says A=-4. If A=-4, the function is -4cos(3(x-1))+3. This is equivalent to 4cos(3(x-1)+pi)+3. The phase shift h=1 is correct for the form -4cos(3(x-1)). But standard textbooks often define phase shift for A cos(Bx - C) as C/B. Here 3x-3 = 3(x-1), so h=1. The solution is technically consistent if A=-4 is allowed, but Step 5's justification 'At x=1 the curve is 4 from its midline' is a valid check for amplitude, not a derivation. The main error is that Step 1 sets A=-4, which is a valid parameterization, but usually amplitude is defined as |A|. The solution lists amplitude 4, which is correct. The phase shift is 1 to the right. This is correct. Wait, let's look closer. y = 3 - 4cos(3x-3). Rewrite as 3 + 4cos(3x-3 + pi). Or 3 - 4cos(3(x-1)). If we use the form A cos(B(x-h)) + D, we can have A=4, h=1+pi/3? No. -4cos(3(x-1)) = 4cos(3(x-1) - pi) = 4cos(3(x - 1 - pi/3)). So if A=4, h = 1 + pi/3. If A=-4, h=1. The solution chooses A=-4, h=1. This is a valid representation. Is it misleading? Students are often taught A>0. If A must be positive, the phase shift is different. However, the problem asks for 'the' phase shift. Usually, phase shift is defined for the standard form y = A cos(B(x - C)) + D. If A is negative, it's often rewritten. But many calculators/texts accept negative A. The bigger issue is Step 5: 'At x=1 the curve is 4 from its midline... that is the amplitude.' This is a verification, not a derivation, but acceptable. The real issue is Step 1: 'A=-4'. Amplitude is |A|=4. The solution states A=-4. This is a parameter, not the amplitude itself. The final answer lists amplitude 4, so it's consistent. Is there an error? Let's check the period. 2pi/3. Correct. Vertical shift 3. Correct. Phase shift 1 right. Correct. The solution seems correct in its results. Why would it fail? Step 1 says 'Write it as A·cos(B(x − h)) + D: A = -4...'. This is a valid algebraic manipulation. Step 5 checks the amplitude. It seems sound. Let me re-read the prompt. 'Find... amplitude... phase shift'. The solution provides them. Is there a subtle error? 'Phase shift 1 to the right'. For y = 3 - 4cos(3(x-1)), the graph is a cosine wave reflected and shifted. The 'standard' phase shift often assumes A>0. If A>0, y = 3 + 4cos(3x - 3 + pi) = 3 + 4cos(3(x - 1 + pi/3)). Phase shift would be 1 - pi/3. But 'phase shift' is ambiguous for negative A. Most contexts accept the shift of the argument. 3x-3=0 => x=1. This is the start of the cycle for cos(3x-3) if we consider the standard cosine starting at max. But here we have -cos, so it starts at min. Does 'phase shift' refer to the shift of the parent function cos(x)? If so, cos(3(x-1)) is shifted right by 1. The negative sign is a reflection, not a shift. So h=1 is correct for the argument. The solution is likely 'pass'. Wait, look at Step 5 again. 'Subs(..., x, 1) = -1'. 3 - 4cos(0) = -1. Correct. '4 from its midline y=3'. |-1-3|=4. Correct. 'That is the amplitude'. Correct. I see no error. I will mark pass.
gpt-oss:20b: pass 2026-10-20
qwen3.6:27b-mlx: pass 2026-10-19 — The solution correctly identifies the parameters A, B, h, and D. The derived amplitude, period, phase shift, and vertical shift match the stated answer. The supporting equations, while slightly idiosyncratic in their justification (e.g., checking periodicity rather than deriving it), are mathematically valid checks that confirm the results.
gpt-oss:20b: pass 2026-10-19
qwen3.6:27b-mlx: fail (style) 2026-10-19 — [domain objection, downgraded to style] The solution incorrectly identifies the amplitude as -4. Amplitude is defined as the absolute value |A|, so it must be positive (4). While the final stated answer lists 4, the derivation in step 1 is mathematically incorrect and misleading.
gpt-oss:20b: pass 2026-10-19

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 2026-10-20 with SymPy 1.14.0.