iTutor Math Updates

Here i = iota = square root(-1) - square root(a^2 + b^2) <= asinx + bcosx <= square root(a^2 + b^2) now , lets take (sinx + icosx) - square root(1^2 + i^2) <=(sinx + icosx)<= square root(1^2 + i^2) = - square root(1-1) <=(sinx + icosx)<= square root(1-1) So, sinx + icosx = 0 sinx = (-1)icosx squaring both sides, [sinx]^2 = [(-1)icosx]^2 now, we know [sinx]^2 = 1 - [cosx]^2 putting this in [sinx]^2 = [(-1)icosx]^2 1 - [cosx]^2 = [i]^2[cosx]^2 1 - [cosx]^2 = -[cosx]^2 1- [cosx]^2 + [cosx]^2 = 0 so, 1 = 0 :?: