iTutor Math Updates
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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 :?:
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