How do you find the Maclaurin Series for #(sinx)(cosx)#? Calculus Power Series Constructing a Maclaurin Series 1 Answer Cesareo R. Dec 7, 2016 #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)!)# Answer link Related questions How do you find the Maclaurin series of #f(x)=(1-x)^-2# ? How do you find the Maclaurin series of #f(x)=cos(x^2)# ? How do you find the Maclaurin series of #f(x)=cosh(x)# ? How do you find the Maclaurin series of #f(x)=cos(x)# ? How do you find the Maclaurin series of #f(x)=e^(-2x)# ? How do you find the Maclaurin series of #f(x)=e^x# ? How do you find the Maclaurin series of #f(x)=ln(1+x)# ? How do you find the Maclaurin series of #f(x)=ln(1+x^2)# ? How do you find the Maclaurin series of #f(x)=sin(x)# ? How do you use a Maclaurin series to find the derivative of a function? See all questions in Constructing a Maclaurin Series Impact of this question 22873 views around the world You can reuse this answer Creative Commons License