Find angle between $y=\sin x$ and $y=\cos x$ at their intersection point. Find angle between $y=\sin x$ and $y=\cos x$ at their intersection point.
Intersection points are $\frac{\pi}{4}+\pi k$ and to find angle between them we need to compute derivatives at intersection points but then I can't combine them to get an answer which is $\arctan2\sqrt2$. Will be thankful for your help.

The slope of the tangent line to $y=\sin x$ at $x = \frac{\pi}{4}+\pi k$ is
$$m_s = \cos\left(\frac{\pi}{4}+\pi k \right)
      = \frac{\cos \pi k - \sin \pi k}{\sqrt 2} = \frac{(-1)^k}{\sqrt 2} $$
The slope of the tangent line to $y=\cos x$ at $x = \frac{\pi}{4}+\pi k$ is
$$m_c = \cos\left(\frac{\pi}{4}+\pi k \right)
      = -\frac{\cos \pi k + \sin \pi k}{\sqrt 2} = -\frac{(-1)^k}{\sqrt 2} $$
The angle between the two lines is described by
$$ \tan \theta = 
   \left| \dfrac{m_s - m_c}{1 + m_s m_c}\right | 
   = \dfrac{\sqrt 2}{\left(\frac 12 \right)} = 2\sqrt 2$$
So $\theta = \arctan 2\sqrt 2 \approx 70.53^\circ$
