Question #5fced

1 Answer
mason m
Mar 1, 2017

sum_(n=0)^oo(-1)^n/((2n+1)!)(pi/6)^(2n+1)=sin(pi/6)=1/2∞∑n=0(−1)n(2n+1)!(π6)2n+1=sin(π6)=12

Explanation:

The Maclaurin series for sin(x)sin(x) is given by

sin(x)=sum_(n=0)^oo(-1)^n/((2n+1)!)x^(2n+1)sin(x)=∞∑n=0(−1)n(2n+1)!x2n+1

Thus,

sum_(n=0)^oo(-1)^n/((2n+1)!)(pi/6)^(2n+1)=sin(pi/6)=1/2∞∑n=0(−1)n(2n+1)!(π6)2n+1=sin(π6)=12

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