Absolute Extrema of Trigonometric Functions on Closed Intervals

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To find the absolute maximum and minimum values of the function f(t) = 2cos(t) + sin(2t) on the interval [0, pi/2], the first derivative f '(t) is set to zero, leading to the equation 0 = -2sin(t) + 2cos(2t). This simplifies to 2sin(t) = 2cos(2t), which can be further manipulated using double angle formulas to express cosine in terms of sine. The discussion highlights the need for a trigonometric identity to solve the resulting equation. Ultimately, applying these methods will help determine the extrema of the function within the specified interval.
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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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