Question

How do you find maclaurin series for `sinx cosx` using maclaurin for `sinx`?

Answer 1

`sinx cosx = sum_{k=0}^{oo}2^{2k}(-1)^k(x^{2k+1})/((2k+1)!)`

Explanation:

We know
`sin(a+b)=sin(a)cos(b)+sin(b)cos(a)` if `a = b`
`sin(2a)=2sin(a)cos(a)` then
`sinx cosx = sin(2x)/2`

Supposing that

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

then

`sinx cosx = sum_{k=0}^{oo}2^{2k}(-1)^k(x^{2k+1})/((2k+1)!)`