Q:Find $a>0$ so that the curves $y=\sin(ax)$ and $y=\cos(ax)$ intersect at right angles.
I took derivative of both ,than $a^2\cos(ax)(-\sin(ax))=-1$ i multiplied both side by 2 so, $sin(2ax)=$$\frac{2}{a^2}$
And since, $-1<sin<1$
$-1<$$\frac{2}{a^2}$$<1$ and $\frac{2}{a^2 }$ already pozitive so $0<$$\frac{2}{a^2}$$<1$ My answer is $a>$$\sqrt{2}$
Is it true?
The curves cross when $x=\frac\pi{4a}+k\frac\pi a$, with $k\in\mathbb Z$. At such point, the slope of the tangent of $\sin(ax)$ is $(-1)^k\frac a{\sqrt2}$ and the slope of the tangent of $\cos(ax)$ is $(-1)^{k+1}\frac a{\sqrt2}$. The product of this two numbers is $-1$ (that is, the lines intersect at a right angle) when (and only when) $a=\sqrt2$.