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Amplitude, period and phase shift

Problem 13.473 · medium

For \( \displaystyle y = - 2 \cos{\left(\frac{x}{2} + \frac{\pi}{6} \right)} - 2 \), find the amplitude, the period, the phase shift (with its direction) and the vertical shift.
  1. Write it as A·cos(B(x − h)) + D: A = -2, B = 1/2, h = -pi/3, D = -2.
  2. \[ \frac{2 \pi}{1 \cdot \frac{1}{2}} = 4 \pi \]
    Period = 2π/B.✓ Proved
  3. \[ \left(2 \cos{\left(\frac{x}{2} + \frac{\pi}{6} \right)} + 2\right) + \left. - 2 \cos{\left(\frac{x}{2} + \frac{\pi}{6} \right)} - 2 \right|_{\substack{ x=x + 4 \pi }} = 0 \]
    Shifting x by one period gives the same function.✓ Proved
  4. \[ \frac{\left(-1\right) \pi}{6 \cdot 1 \cdot \frac{1}{2}} = - \frac{\pi}{3} \]
    Phase shift h = C/B.✓ Proved
  5. \[ \left. - 2 \cos{\left(\frac{x}{2} + \frac{\pi}{6} \right)} - 2 \right|_{\substack{ x=\frac{\left(-1\right) \pi}{3} }} = -4 \]
    At x = -pi/3 the curve is 2 from its midline y = -2: that is the amplitude.✓ Proved
Answer \( \text{amplitude } 2,\ \text{period } 4 \pi,\ \text{phase shift } \frac{\pi}{3}\text{ to the left},\ \text{vertical shift } -2 \)

Lines: 4 proved, 1 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
LineStatusChecked byDetail
1Not checked—a sentence; read, not computed
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
5✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0max 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] The solution incorrectly identifies the phase shift as pi/3 to the left. The argument is x/2 + pi/6 = 1/2(x + pi/3), so the shift is -pi/3, which is pi/3 to the left. Wait, let me re-evaluate. h = -pi/3 means shift is to the left by pi/3. The stated answer says 'pi/3 to the left'. The solution says h = -pi/3. This is consistent. Let me check the amplitude calculation. Step 5 says at x = -pi/3, the value is -4. The midline is -2. The distance is |-4 - (-2)| = 2. This is correct. Is there an error? Let's check the phase shift direction again. y = A cos(B(x-h)) + D. Here we have -2 cos(1/2(x + pi/3)) - 2. So h = -pi/3. The shift is h. A shift of h means the graph is shifted by h. If h is negative, it's to the left. So 'pi/3 to the left' is correct. Why did I think it was wrong? Let me re-read carefully. Step 1: A=-2, B=1/2, h=-pi/3, D=-2. Correct. Step 4: h = C/B = (pi/6)/(1/2) = pi/3? No, the term is +pi/6. So B(x-h) = x/2 - h/2? No. B(x-h) = Bx - Bh. We have x/2 + pi/6. So -Bh = pi/6 => -1/2 h = pi/6 => h = -pi/3. Step 4 calculates (-pi/6)/(1/2) = -pi/3. Wait, where did -pi/6 come from? The term is +pi/6. To get h, we solve B(x-h) = Bx - Bh. We want Bx - Bh = x/2 + pi/6. So -Bh = pi/6. h = -pi/6 / B = -pi/6 / (1/2) = -pi/3. Step 4 writes (-pi/6)/(1/2). This implies the numerator is -pi/6. But the constant term in the argument is +pi/6. The formula for phase shift from cos(Bx + C) is -C/B. So - (pi/6) / (1/2) = -pi/3. The step shows (-pi/6)/(1/2). This is numerically correct for the value of h, but the derivation logic is slightly obscured if one assumes the numerator is just C. However, -C/B is the standard formula. If C = pi/6, then -C/B = -pi/3. The step shows -pi/6 in the numerator. This is correct because h = -C/B. So the calculation is correct. Is there any other error? Let's check the amplitude. Amplitude is |A| = |-2| = 2. Step 5 checks a point. At x = -pi/3, y = -2 cos(0) - 2 = -4. Midline is -2. Distance is 2. Correct. Period 4pi. Correct. Vertical shift -2. Correct. Phase shift pi/3 to the left. Correct. Why would this fail? Maybe the 'direction' is ambiguous? 'pi/3 to the left' is clear. Let me look closer at Step 1. 'Write it as A·cos(B(x − h)) + D'. A = -2. Usually, amplitude is defined as |A|. The problem asks for amplitude, which is positive. The solution identifies A=-2. This is fine as an intermediate step. Is there a subtle error in Step 5? 'At x = -pi/3 the curve is 2 from its midline y = -2: that is the amplitude.' This is a valid way to verify amplitude. It seems correct. Let me re-read the prompt. 'find the amplitude, the period, the phase shift (with its direction) and the vertical shift.' Stated answer: amplitude 2, period 4pi, phase shift pi/3 to the left, vertical shift -2. Solution steps support this. Is there a convention issue? Some define phase shift as just the value h. Others as the vector. 'pi/3 to the left' is unambiguous. Let me check the sign of the phase shift again. y = -2 cos(x/2 + pi/6) - 2. Let x/2 + pi/6 = 0 => x/2 = -pi/6 => x = -pi/3. The 'start' of the cosine cycle (if it were positive) is at -pi/3. Since it's negative cosine, it starts at a minimum. The shift is indeed -pi/3. Left by pi/3. Everything seems correct. Wait, look at Step 4 again. `(-pi/6)/(1/2) = -pi/3`. The comment says `Phase shift h = C/B`. This is the error. The formula is h = -C/B. The step calculates -pi/6 / 1/2, which is -C/B if C=pi/6. But the comment says `h = C/B`. If a student reads `h = C/B` and plugs in C=pi/6, they get pi/3, which is wrong. The calculation uses -pi/6, implying the user applied -C/B but wrote the formula as C/B. This is a misleading statement of the theorem/formula. The comment claims the formula is h = C/B, but the numbers used are -C/B. This teaches the student the wrong formula.
Every verdict on record (6)
  • qwen3.6:27b-mlx: fail (style) 2026-11-08 — [domain objection, downgraded to style] The solution incorrectly identifies the phase shift as pi/3 to the left. The argument is x/2 + pi/6 = 1/2(x + pi/3), so the shift is -pi/3, which is pi/3 to the left. Wait, let me re-evaluate. h = -pi/3 means shift is to the left by pi/3. The stated answer says 'pi/3 to the left'. The solution says h = -pi/3. This is consistent. Let me check the amplitude calculation. Step 5 says at x = -pi/3, the value is -4. The midline is -2. The distance is |-4 - (-2)| = 2. This is correct. Is there an error? Let's check the phase shift direction again. y = A cos(B(x-h)) + D. Here we have -2 cos(1/2(x + pi/3)) - 2. So h = -pi/3. The shift is h. A shift of h means the graph is shifted by h. If h is negative, it's to the left. So 'pi/3 to the left' is correct. Why did I think it was wrong? Let me re-read carefully. Step 1: A=-2, B=1/2, h=-pi/3, D=-2. Correct. Step 4: h = C/B = (pi/6)/(1/2) = pi/3? No, the term is +pi/6. So B(x-h) = x/2 - h/2? No. B(x-h) = Bx - Bh. We have x/2 + pi/6. So -Bh = pi/6 => -1/2 h = pi/6 => h = -pi/3. Step 4 calculates (-pi/6)/(1/2) = -pi/3. Wait, where did -pi/6 come from? The term is +pi/6. To get h, we solve B(x-h) = Bx - Bh. We want Bx - Bh = x/2 + pi/6. So -Bh = pi/6. h = -pi/6 / B = -pi/6 / (1/2) = -pi/3. Step 4 writes (-pi/6)/(1/2). This implies the numerator is -pi/6. But the constant term in the argument is +pi/6. The formula for phase shift from cos(Bx + C) is -C/B. So - (pi/6) / (1/2) = -pi/3. The step shows (-pi/6)/(1/2). This is numerically correct for the value of h, but the derivation logic is slightly obscured if one assumes the numerator is just C. However, -C/B is the standard formula. If C = pi/6, then -C/B = -pi/3. The step shows -pi/6 in the numerator. This is correct because h = -C/B. So the calculation is correct. Is there any other error? Let's check the amplitude. Amplitude is |A| = |-2| = 2. Step 5 checks a point. At x = -pi/3, y = -2 cos(0) - 2 = -4. Midline is -2. Distance is 2. Correct. Period 4pi. Correct. Vertical shift -2. Correct. Phase shift pi/3 to the left. Correct. Why would this fail? Maybe the 'direction' is ambiguous? 'pi/3 to the left' is clear. Let me look closer at Step 1. 'Write it as A·cos(B(x − h)) + D'. A = -2. Usually, amplitude is defined as |A|. The problem asks for amplitude, which is positive. The solution identifies A=-2. This is fine as an intermediate step. Is there a subtle error in Step 5? 'At x = -pi/3 the curve is 2 from its midline y = -2: that is the amplitude.' This is a valid way to verify amplitude. It seems correct. Let me re-read the prompt. 'find the amplitude, the period, the phase shift (with its direction) and the vertical shift.' Stated answer: amplitude 2, period 4pi, phase shift pi/3 to the left, vertical shift -2. Solution steps support this. Is there a convention issue? Some define phase shift as just the value h. Others as the vector. 'pi/3 to the left' is unambiguous. Let me check the sign of the phase shift again. y = -2 cos(x/2 + pi/6) - 2. Let x/2 + pi/6 = 0 => x/2 = -pi/6 => x = -pi/3. The 'start' of the cosine cycle (if it were positive) is at -pi/3. Since it's negative cosine, it starts at a minimum. The shift is indeed -pi/3. Left by pi/3. Everything seems correct. Wait, look at Step 4 again. `(-pi/6)/(1/2) = -pi/3`. The comment says `Phase shift h = C/B`. This is the error. The formula is h = -C/B. The step calculates -pi/6 / 1/2, which is -C/B if C=pi/6. But the comment says `h = C/B`. If a student reads `h = C/B` and plugs in C=pi/6, they get pi/3, which is wrong. The calculation uses -pi/6, implying the user applied -C/B but wrote the formula as C/B. This is a misleading statement of the theorem/formula. The comment claims the formula is h = C/B, but the numbers used are -C/B. This teaches the student the wrong formula.
  • gpt-oss:20b: pass 2026-11-08
  • qwen3.6:27b-mlx: fail (style) 2026-11-07 — [domain objection, downgraded to style] The solution incorrectly identifies the phase shift as \pi/3 to the left. The term inside the cosine is \frac{x}{2} + \frac{\pi}{6} = \frac{1}{2}(x + \frac{\pi}{3}), so the phase shift is \frac{\pi}{3} to the left. However, the stated answer claims \frac{\pi}{3} to the left, which matches the calculation h = -\pi/3 (shift is -h? No, standard form A cos(B(x-h)) + D implies shift is h. If h = -\pi/3, the shift is \pi/3 to the left. Wait, let's re-read carefully. Standard form: y = A cos(B(x - h)) + D. Here, \frac{x}{2} + \frac{\pi}{6} = \frac{1}{2}(x + \frac{\pi}{3}) = \frac{1}{2}(x - (-\frac{\pi}{3})). So h = -\frac{\pi}{3}. A shift of h means the graph is shifted by h. A negative h means a shift to the left. So \frac{\pi}{3} to the left is correct. Why did I think it was wrong? Let's check the sign convention. Usually, phase shift is defined as -C/B. Here C = \pi/6, B = 1/2. -C/B = -\pi/3. This represents a shift of -\pi/3, i.e., \pi/3 to the left. The stated answer says \pi/3 to the left. The solution says h = -\pi/3. This is consistent. Is there an error? Let's check the amplitude. A = -2. Amplitude is |A| = 2. Correct. Period 4\pi. Correct. Vertical shift -2. Correct. Phase shift \pi/3 left. Correct. Why would this fail? Let's look at step 5. It says 'At x = -pi/3 the curve is 2 from its midline... that is the amplitude.' This is a weird way to define amplitude, but numerically correct. Is there a subtle error? The problem asks for 'phase shift (with its direction)'. The solution calculates h = -\pi/3. The stated answer says \pi/3 to the left. These are equivalent. Is the 'direction' part handled correctly in the text? Step 1 says h = -pi/3. It does not explicitly state the direction in the sentence, but the final answer does. The prompt asks to review the *solution*. The solution steps are: 1. Identify parameters. 2. Calculate period. 3. Verify period. 4. Calculate phase shift value. 5. Verify amplitude. The final answer is provided separately. The solution steps themselves are mathematically sound. Wait, look at step 4: 'Phase shift h = C/B'. The formula for phase shift in y = A cos(Bx + C) + D is usually -C/B. The solution calculates (-\pi/6)/(1/2) = -\pi/3. But the term is +\pi/6. So C = \pi/6. The calculation shown is (-\pi/6)/(1/2). Where did the negative sign come from? It seems the solver used the formula -C/B but wrote the numerator as -\pi/6 explicitly, or perhaps they thought C = -\pi/6? If the form is B(x-h), then Bx - Bh = Bx + C => -Bh = C => h = -C/B. Here C = \pi/6. So h = -(\pi/6)/(1/2) = -\pi/3. The equation in step 4 is (-\pi/6)/(1/2) = -\pi/3. This is arithmetically correct. It implicitly uses the formula h = -C/B. The label 'Phase shift h = C/B' is technically incorrect if C is the coefficient in Bx+C, because it should be -C/B. However, if the solver defined the form as B(x+h), then h would be C/B. But step 1 says 'Write it as A·cos(B(x − h))'. In this form, the shift is h. And h = -C/B. The text in step 4 says 'Phase shift h = C/B'. This is a contradiction with the form in step 1 unless C is defined differently. If step 1 implies standard form, step 4's label is wrong (missing negative sign), even though the calculation includes the negative sign. This is misleading/wrong labeling. A student might learn that phase shift is C/B. It is actually -C/B. This is an error in the explanation.
  • gpt-oss:20b: fail (misleading) 2026-11-07 — The solution states A = -2 and later uses it as the amplitude, which can mislead students into thinking the amplitude is negative. The amplitude should be reported as the positive value 2.
  • qwen3.6:27b-mlx: inconclusive 2026-11-07 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The solution incorrectly identifies the phase shift as \\pi/3 to the left. The standard form y = A cos(B(x - h)) + D yields h = -\\pi/3, which corres
  • gpt-oss:20b: pass 2026-11-07

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 2026-11-08 with SymPy 1.14.0.

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