Explain why a circle tangent to y=sin(x) at x=π/2 has its center on the vertical line x=π/2. Find the equation of the osculating circle at x=π/2. That is, the circle that is tangent to y=sin(x) and x=π/2 and has the same radius of curvature there.

Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.
Question

Explain why a circle tangent to y=sin(x) at x=π/2 has its center on the vertical line x=π/2. Find the equation of the osculating circle at x=π/2. That is, the circle that is tangent to y=sin(x) and x=π/2 and has the same radius of curvature there.

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