Absolute Extrema of Trigonometric Functions on Closed Intervals

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SUMMARY

The discussion focuses on finding the absolute maximum and minimum values of the function f(t) = 2cos(t) + sin(2t) over the interval [0, π/2]. The critical point is identified by setting the derivative f'(t) = 0, leading to the equation 2sin(t) = 2cos(2t). Participants suggest using the double angle formula to convert the cosine term into a sine term for easier resolution of the equation.

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  • Understanding of trigonometric functions and their properties
  • Knowledge of calculus, specifically differentiation and critical points
  • Familiarity with the double angle formulas in trigonometry
  • Ability to analyze functions on closed intervals
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  • Study the application of double angle formulas in trigonometric equations
  • Learn how to find critical points and evaluate functions on closed intervals
  • Explore the concepts of absolute extrema in calculus
  • Practice solving similar problems involving trigonometric functions and their derivatives
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Students studying calculus, particularly those focusing on optimization problems involving trigonometric functions, as well as educators looking for examples of absolute extrema in real-world applications.

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Homework Statement



Find the absolute max and absolute min values of function on the given interval:
f(t) = 2cos(t) + sin(2t), [0,pi/2]

Homework Equations





The Attempt at a Solution



f '(t) = 0
0 = -2sin(t) + 2cos(2t)
2sin(t) = 2cos(2t)
stuck...
 
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Use a double angle formula to change the cosine term into a sine term. Then you'll have solvable equation.
 

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