How do you find the maclaurin series expansion of #f(x) =sinxcosx#?

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Answer 1 · 2017-01-21T14:43:33

#sinxcosx =sum_(n=0)^oo (-1)^n 2^(2n)x^(2n+1)/((2n+1)!)#

You can use the trigonometric identity:

#sin2x = 2sinx cosx#

then the MacLaurin expansion of #sint#:

#sint = sum_(n=0)^oo (-1)^n t^(2n+1)/((2n+1)!)#

substitute #t=2x#:

#sin2x = sum_(n=0)^oo (-1)^n (2x)^(2n+1)/((2n+1)!)= sum_(n=0)^oo (-1)^n 2^(2n+1)x^(2n+1)/((2n+1)!)#

and finally:

#sinxcosx = 1/2sin2x = 1/2sum_(n=0)^oo (-1)^n 2^(2n+1)x^(2n+1)/((2n+1)!) = sum_(n=0)^oo (-1)^n 2^(2n)x^(2n+1)/((2n+1)!)#