Question

How do you find the Maclaurin Series for `(sinx)(cosx)`?

Answer 1

`sin(x)cos(x)=1/2sum_(k=0)^oo(-1)^k(2x)^(2k+1)/((2k+1)!)`

Explanation:

We know that `sin(2x)=2sin(x)cos(x)` so

`sin(x)cos(x)=1/2sin(2x)` or

`sin(x)cos(x)=1/2sum_(k=0)^oo(-1)^k(2x)^(2k+1)/((2k+1)!)`