Find the angles at which the following pairs of curves
Find the angles at which the following pairs of curves intersect: 1° $y=\sin x$ and $y=\cos x$; 2° $y=\arcsin x$ and $y=\arccos x$.
Explanation of Method
A function pair and its inverse pair, sharing the same crossing angle.
Step 1: Part 1°: $\sin x=\cos x$
$\tan x=1 \Rightarrow x=\dfrac\pi4+\pi n$; by periodicity work at $x_0=\dfrac\pi4$, where $y_0=\dfrac{\sqrt2}{2}$.
\[ k_{1}=\cos\frac\pi4=\frac{\sqrt2}2, \qquad k_{2}=-\sin\frac\pi4=-\frac{\sqrt2}2. \]
\[ \tan\theta_{1} = \left|\frac{\dfrac{\sqrt2}2-\!\left(-\dfrac{\sqrt2}2\right)}{1+\dfrac{\sqrt2}2\!\left(-\dfrac{\sqrt2}2\right)}\right| = \left|\frac{\sqrt2}{1/2}\right| = 2\sqrt2. \tag{1} \]
Step 2: Part 2°: $\arcsin x=\arccos x$
Using $\arcsin x+\arccos x=\dfrac\pi2$, equality forces $\arcsin x=\dfrac\pi4$, i.e.\ $x_0=\dfrac1{\sqrt2}$, $y_0=\dfrac\pi4$.
\[ k_{1}=\frac{1}{\sqrt{1-x_0^{2}}}=\sqrt2, \qquad k_{2}=-\frac{1}{\sqrt{1-x_0^{2}}}=-\sqrt2. \]
\[ \tan\theta_{2} = \left|\frac{\sqrt2-(-\sqrt2)}{1+\sqrt2(-\sqrt2)}\right| = \left|\frac{2\sqrt2}{-1}\right| = 2\sqrt2. \tag{2} \]
Result
Final answer:
\[\boxed{\,\theta_{1} = \theta_{2} = \arctan(2\sqrt2) \approx 70.5°\,}\]
Remark
Remark. The two angles coincide because $\arcsin$ and $\arccos$ are the reflections of $\sin$ and $\cos$ across $y=x$ (on the appropriate domains), and reflection across $y=x$ preserves the angle between two curves.