Absolute Extrema of Trigonometric Functions on Closed Intervals

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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.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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