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Tangent lines

Problem 2.913 · medium Video

Find an equation of the tangent line to $ \displaystyle y = 2 x - 2 \sin{\left(x \right)} + 1 $ at $ \displaystyle x = \frac{\pi}{2} $.

The tangent line passes through the point (a, f(a)) and has slope f'(a).
Reviewed
\[ \left. 2 x - 2 \sin{\left(x \right)} + 1 \right|_{\substack{ x=\frac{\pi}{2} }} = -1 + \pi \]
The point of tangency.✓ Proved
\[ \frac{d}{d x} \left(2 x - 2 \sin{\left(x \right)} + 1\right) = 2 - 2 \cos{\left(x \right)} \]
Differentiate.✓ Proved
\[ \left. 2 - 2 \cos{\left(x \right)} \right|_{\substack{ x=\frac{\pi}{2} }} = 2 \]
The slope at the point.✓ Proved
\[ 2 x - 1 \]
Point-slope form, then simplify.

Answer $ y = 2 x - 1 $

The graph of y = 2x - 2sin(x) + 1 near x = pi/2. A secant line through the point (pi/2, -1 + pi) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at 2, giving y = 2x - 1.

Lines: 3 proved, 1 reviewed, 1 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	Reviewed	claude-sonnet-5-5	claude-sonnet-5-5: Line 1 is a sentence stating the standard tangent-line setup: the line passes through (a, f(a)) with slope f'(a). It is true, so no algebraic check is needed.
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	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
5	Not checked	—	both sides are the same expression: the line states a value, and the working that produced it is not on the line
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	the line meets the curve at x = a, and its slope matches a central difference quotient of f there

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The final equation is incorrect. The point of tangency is $(\pi/2, \pi - 1)$ and the slope is $2$. The correct tangent line is $y - (\pi - 1) = 2(x - \pi/2)$, which simplifies to $y = 2x - \pi + \pi - 1 = 2x - 1$. Wait, let me re-calculate. $y - (\pi - 1) = 2(x - \pi/2) \implies y = 2x - \pi + \pi - 1 \implies y = 2x - 1$. The stated answer is actually correct. Let me re-read the prompt's stated answer. Stated answer: Eq(y, 2*x - 1). My calculation: $y = 2x - 1$. The solution steps are correct. Why did I think it was wrong? Ah, I see. Step 5 says '2*x - 1, 2*x - 1'. This is just an identity check. It doesn't show the derivation from point-slope form. However, the result is correct. Is there a subtle error? Let's check the point again. $f(\pi/2) = 2(\pi/2) - 2\sin(\pi/2) + 1 = \pi - 2(1) + 1 = \pi - 1$. Slope $f'(\pi/2) = 2 - 2\cos(\pi/2) = 2 - 0 = 2$. Equation: $y - (\pi - 1) = 2(x - \pi/2)$. $y = 2x - \pi + \pi - 1 = 2x - 1$. The math is correct. The sentence in step 1 is generic but correct. Step 5 is just a simplification check. The solution is sound.

Senior review claude-sonnet-5-5, 2026-10-13: pass — The point (π/2, π-1), the slope 2, and the tangent line y = 2x - 1 are all correct, since y - (π-1) = 2(x - π/2) gives y = 2x - 1. Step 5 compresses point-slope form and simplification into one line, which is a minor style point, not an error.

qwen3.6:27b-mlx: dismiss — The objection retracts itself: the model recomputes f(π/2)=π-1, f'(π/2)=2 and y=2x-1, and concludes the solution is correct, so it identifies no actual defect.

Every verdict on record (7)

claude-sonnet-5-5: pass 2026-10-13 — The point (π/2, π-1), the slope 2, and the tangent line y = 2x - 1 are all correct, since y - (π-1) = 2(x - π/2) gives y = 2x - 1. Step 5 compresses point-slope form and simplification into one line, which is a minor style point, not an error.
qwen3.6:27b-mlx: fail (error) 2026-09-27 — The final equation is incorrect. The point of tangency is $(\pi/2, \pi - 1)$ and the slope is $2$. The correct tangent line is $y - (\pi - 1) = 2(x - \pi/2)$, which simplifies to $y = 2x - \pi + \pi - 1 = 2x - 1$. Wait, let me re-calculate. $y - (\pi - 1) = 2(x - \pi/2) \implies y = 2x - \pi + \pi - 1 \implies y = 2x - 1$. The stated answer is actually correct. Let me re-read the prompt's stated answer. Stated answer: Eq(y, 2*x - 1). My calculation: $y = 2x - 1$. The solution steps are correct. Why did I think it was wrong? Ah, I see. Step 5 says '2*x - 1, 2*x - 1'. This is just an identity check. It doesn't show the derivation from point-slope form. However, the result is correct. Is there a subtle error? Let's check the point again. $f(\pi/2) = 2(\pi/2) - 2\sin(\pi/2) + 1 = \pi - 2(1) + 1 = \pi - 1$. Slope $f'(\pi/2) = 2 - 2\cos(\pi/2) = 2 - 0 = 2$. Equation: $y - (\pi - 1) = 2(x - \pi/2)$. $y = 2x - \pi + \pi - 1 = 2x - 1$. The math is correct. The sentence in step 1 is generic but correct. Step 5 is just a simplification check. The solution is sound.
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: fail (error) 2026-09-26 — The solution fails to explicitly calculate the tangent line equation using the point-slope formula y - y1 = m(x - x1). Step 5 jumps directly to the final answer '2*x - 1' without showing the substitution of the point (pi/2, pi - 1) and slope 2, making the derivation opaque and potentially confusing for a student.
gpt-oss:20b: pass 2026-09-26
qwen3.6:27b-mlx: fail (error) 2026-09-26 — The final equation is incorrect. The point of tangency is (pi/2, pi - 1) and the slope is 2, so the tangent line is y - (pi - 1) = 2(x - pi/2), which simplifies to y = 2x - pi + 1. The solution incorrectly states the equation is y = 2x - 1.
gpt-oss:20b: pass 2026-09-26

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/tangent_line, checked 2026-10-13 with SymPy 1.14.0.