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Question #5fced

1 Answer

Mar 1, 2017

$sum_(n=0)^oo(-1)^n/((2n+1)!)(pi/6)^(2n+1)=sin(pi/6)=1/2$

Explanation:

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

$sin(x)=sum_(n=0)^oo(-1)^n/((2n+1)!)x^(2n+1)$

Thus,

$sum_(n=0)^oo(-1)^n/((2n+1)!)(pi/6)^(2n+1)=sin(pi/6)=1/2$

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