Amplitude, period and phase shift
Problem 13.101 · medium
For \( \displaystyle y = 3 \sin{\left(3 x \right)} - 3 \), 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 = -3, B = 3, h = pi/2, D = -3.
- Period = 2π/B: 2*pi/3.
- \[ \left(3 - 3 \sin{\left(3 x \right)}\right) + \left. 3 \sin{\left(3 x \right)} - 3 \right|_{\substack{ x=x + \frac{2 \pi}{3} }} = 0 \]Shifting x by one period gives the same function.✓ Proved
- \[ \frac{3 \pi}{2 \cdot 3} = \frac{\pi}{2} \]Phase shift h = C/B.✓ Proved
- \[ \left. 3 \sin{\left(3 x \right)} - 3 \right|_{\substack{ x=\frac{\pi}{2} }} = -6 \]At x = pi/2 the curve is 3 from its midline y = -3: that is the amplitude.✓ Proved
Answer \( \text{amplitude } 3,\ \text{period } \frac{2 \pi}{3},\ \text{phase shift } \frac{\pi}{2}\text{ to the right},\ \text{vertical shift } -3 \)
Lines: 3 proved, 2 not checked. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see the receipt. 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: fail (error) — The solution incorrectly rewrites the function as A·cos(B(x−h))+D with A = –3 and h = π/2. The original function y = 3 sin(3x) – 3 has amplitude 3, period 2π/3, no phase shift (h = 0), and vertical shift –3. The statements about A, h, and the phase shift are therefore wrong.qwen3.6:27b-mlx: fail (error) — The solution incorrectly identifies the phase shift as \pi/2 to the right. The function y = 3\sin(3x) - 3 has no horizontal phase shift relative to the sine function (or a shift of 0). The value \pi/2 corresponds to a shift relative to a cosine function, but the solution fails to clarify this distinction, leading to an incorrect statement about the phase shift of the given sine function.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-17 — The solution incorrectly identifies the phase shift as \pi/2 to the right. The function y = 3\sin(3x) - 3 has no horizontal phase shift relative to the sine function (or a shift of 0). The value \pi/2 corresponds to a shift relative to a cosine function, but the solution fails to clarify this distinction, leading to an incorrect statement about the phase shift of the given sine function.gpt-oss:20b: fail (error) 2026-10-17 — The solution incorrectly rewrites the function as A·cos(B(x−h))+D with A = –3 and h = π/2. The original function y = 3 sin(3x) – 3 has amplitude 3, period 2π/3, no phase shift (h = 0), and vertical shift –3. The statements about A, h, and the phase shift are therefore wrong.qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution incorrectly identifies the phase shift as pi/2 to the right. The function y = 3sin(3x) - 3 has no horizontal phase shift relative to the standard sine function (phase shift is 0). The value pi/2 is a root of the function, not a phase shift. Additionally, the setup in step 1 claims A=-3 for a cosine form, which implies a specific transformation not justified or correctly linked to the sine form's phase shift.gpt-oss:20b: inconclusive 2026-10-14 — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"The solution incorrectly rewrites the function as \(A\cos(B(x-h))+D\) with \(A=-3\) and \(h=\pi/2\). The amplitude is the absolute value of the coefficienqwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution incorrectly identifies the phase shift as pi/2 to the right. The function y = 3sin(3x) - 3 has a phase shift of 0 (or multiples of the period) relative to the sine function, or -pi/6 relative to a cosine function. The value pi/2 corresponds to a point where the function is at its minimum, not the start of a standard cycle, and the derivation in step 4 is mathematically unjustified.gpt-oss:20b: fail (error) 2026-10-14 — The solution incorrectly rewrites the function as A·cos(B(x−h))+D with A = −3 and h = π/2. The amplitude is |A| = 3, not −3, and the sine function 3 sin(3x) has no horizontal shift, so h should be 0. Thus the setup and subsequent conclusions about phase shift and amplitude 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.