Master Trigonometry with These Essential Practice Questions

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Try this:

##-1\le \cos x \le 1##

##-\frac 5 4 \le \cos x - \frac 1 4 \le \frac 3 4##

##0\le (\cos x - \frac 1 4)^2 \le \frac {25} {16}##

##0\ge -2(\cos x - \frac 1 4)^2\ge -\frac {25} 8##

##\frac 9 8 \ge \frac 9 8 -2(\cos x - \frac 1 4)^2\ge -\frac {16} 8 = -2##

Think about what that gives.
 
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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