Trignometric functions and identities

nil1996
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Homework Statement


How to quickly solve problems on maximum and minimum values of trig functions with help of calculus:
Ex. 10cos2x-6sinxcosx+2sin2x

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The Attempt at a Solution


I know the method of simplification. But i want to do it quickly with calculus. How to do that??
 
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If you "want to do it quickly with calculus" then take the derivative, set the derivative equal to 0, and solve for x. However, the derivative is fairly complicated and I'm not sure this is "quicker" than just completing the square in the original.

Setting f(x)= 10cos^2(x)- 6sin(x)cos(x)+ 2sin^2(x) then f'(x)= -20cos(x)sin(x)- 6cos^2(x)+ 6sin^2(x)+ 4sin(x)cos(x)= -16sin(x)cos(x)- 6(sin^2(x)- cos^2(x))= 0.
 
OK. but we get maximum values from that what about minimum values?
 
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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