Question #1fcac

1 Answer
Nov 8, 2016

graph{(abs(x^2-1)-y)(abs(x^2-3)-y)((x-sqrt2)^2+(y-1)^2-0.002)=0 [-1, 4, -0.5, 2]}

alpha=pi-2arctan2sqrt(2)

Explanation:

The curves can intersect only when the arguments signs are opposite so

x^2-1=3-x^2\ \ \ \ =>\ \ \ \ x=+-sqrt2

x_0=sqrt(2)

f_(1,2)^'(x_0)=+-2sqrt2

So the obtuse angle is

beta=arctan2sqrt2-arctan(-2sqrt2)=2arctan2sqrt2

then

alpha=pi-2arctan2sqrt(2)

Now try to convert it into a single arctan value.

tan(alpha)=-tan(2arctan2sqrt(2))

Now remember duplication formula for tan.

tan(2x)=(2tanx)/(1-tan^2x)

tan(alpha)=-(2tan(arctan2sqrt(2)))/(1-tan^2(arctan2sqrt2))=-(2*2sqrt2)/(1-4*2)=(4sqrt2)/7

So

alpha=arctan((4sqrt2)/7)