What Mistake Am I Making When Finding the Tangent Plane for a Torus?

[margaret23]

I know I am making a stupid mistake but I am not sure what it is...

Find an equation for the plane tangent to the torus X(s,t)=((5+2cost)coss, (5+2cost)sins, 2sint) at the point ((5-(3)^1/2)/(2)^1/2, (5-(3)^1/2/(2)^1/2, 1).

First I have to find what s and t are in order to sub them in for dT/ds and dT/dt. So first I solved for t using 2sint=1 and get t=pi/6. However, when I attempt to sub it into one of the other equations to solve for s I get 2 different answers for s when using each of the different equations. What am I doing wrong?

 

[HallsofIvy]

sin(t)= 1/2 means either t= \pi/6, in which case cos(t)= \sqrt{3}/2 or t= 5\pi/6, in which case cos(t)= -\sqrt{3}/2. Obviously, since x= (5+ 2cos(t))cos(s)) and y= (5+ 2cos(t))sin(s), if x= y, as you have here, then sin(s)= cos(s)= \pm1/\sqrt{2}. Then 5+ 2cos(t)= 5-sqrt(3) so that cos(t)= -\frac{\sqrt{3}}{2}. From sin²(t)+ cos²(t)= 1 that gives immediately sin(t)= 1/2 as needed. The given point has t= 5\pi/6, s= \pi/4.