Tangent Plane in 4-dimensional Euclidean space.

Hello,

My question is, given a mapping $\displaystyle F: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ defined by $\displaystyle F(x,y) = (sin(x-y),cos(x-y))$,

the problem is to find the linear equations of the tangent plane in $\displaystyle \mathbb{R}^4$ to the graph of F at the point $\displaystyle (\pi/2, \pi/2,0,0)$

I am confused because the image of F has two components, so how could it have a four-dimensional tangent plane? Wouldn't there be infinetly tangent planes in $\displaystyle \mathbb{R}^4$? We can't even evaluate F at the given point...

Sorry I am so hopelessly confused - any clarity or guidance much appriciated.