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.
 A: If you think about these as parametrized curves $(t,\sin t)$ and $(t,\cos t)$, then the angle between them at a point $t_0$ is the angle between $(1,\cos t_0)$ and $(1,-\sin t_0)$. If $t_0 = \frac{\pi}{4} + \pi k$, then we get
$$\cos\theta = \frac{(1,\pm\frac{1}{\sqrt{2}})\cdot(1,\mp\frac{1}{\sqrt{2}})}{\frac{3}{2}} = \frac{1}{2}\cdot\frac{2}{3} = \frac{1}{3}.$$
This, of course, yields an angle of $\arccos\frac{1}{3}$, which is the same as $\arctan 2\sqrt{2}$.
A: Hint:
If $u$ is the angle $$\tan u=\left|\dfrac{\cos x-(-\sin x)}{1+\cos x(-\sin x)}\right|$$
At the point of intersection,
$$\cos x+\sin x=\sqrt2\sin(x+\pi/4)=\cdots=\sqrt2(-1)^n$$
$$\sin x\cos x=\dfrac{\sin2(n\pi+\pi/4)}2=\dfrac12$$
A: 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 $\theta = \arctan 2\sqrt 2 \approx 70.53^\circ$
A: f(x): y=sin(x)
g(x): y=cos(x)
let m1 and m2 be the slopes of f(x) and g(x)
m1 = d(sin(x))/dx = cos(x)
m2 = d(cos(x))/dx = -sin(x)
We know that if k is the angle between two lines with slope ma and mb
Tan(k)=|(ma-mb)/(1+ma*mb)|
At the point of intersection, the angle between the curves is the same as angle between tangents of the curves at the point of intersection (say (x0,y0)) and these tangents have slope f'(x0) and g'(x0).
          [f'(x)= d(f(x))/dx and g'(x)=d(g(x))/d(x)]

we know x0=pi/4
Therefore at x=pi/4 :
m1=cos(pi/4)=1/sqrt(2)
m2=-sin(pi/4)=-1/sqrt(2)
If @(I don't have theta on my keyboard, sorry) is the angle between f(x) and g(x) at x=pi/4
tan(@)=|(m1-m2)/(1-m1*m2)|
=|{(1/sqrt(2))-[-(1/sqrt(2))]}/{1-(1/sqrt(2)^2)}|
=|{2/sqrt(2)}/{1-(1/2)}|
=|sqrt(2)/(1/2)|
=|2*sqrt(2)|
=2*sqrt(2)
Therefore @=Tan^-1(2sqrt(2)) or arctan(2sqrt(2))
