Increasing and Decreasing Functions (max/min)

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Homework Statement



Find a > 0 so that the curves y = sin ax and y = cos ax intersect at right angles (let them intersect at (x0, y0)).


2. The attempt at a solution

Thinking about the unit circle, if theta equals pi/4, then sin theta and cos theta would intersect at right angles at the point (1/sqrt2, 0). Does this imply that sin ax0 = cos ax0? I don't know where to go from here. This problem is at the end of a section that concerned what I can learn about the graph of a function from the first derivative of the function. However, I don't see how the derivatives of the above functions can help me here. I would greatly appreciate a walk-through.
 
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you know one thing if they intersect at right angle, the multiplication of their slope should be negative 1.
say
m1=3
m2=-1/3
that means that they are perpendicular

Now, what do derivatives give you ?

I'm not sure if this works, but you should try
 
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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