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Hi I know that differentiating sin = cos , and differentiating cos = -sin. Time to prove it.

**Q3:**

Prove that: [itex]\frac{d}{{dz}}\sin z = \cos z[/itex]

We know the McLaurin form for sin is:

[itex]

\sum\limits_{n = 0}^\infty {( - 1)^n } \frac{{z^{2n+1} }}{{(2n+1)!}} = z - z^3/3! + z^5/5! - z^7/7! +...[/itex]

differentiating sin term by term i get:

[itex]1 -3z^2/3! + 5z^4/5! - 7z^6/7! +...[/itex] =

[itex]1 -z^2/2! + z^4/4! - z^6/6! +...[/itex] =

cos z

would this be considered a "sufficient proof" . Even though I have only proved the first few terms, I am guessing we can just assume the rest is correct by the theory of induction.?