Question #1f828

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Answer 1 · 2017-07-31T14:24:21

#sinx = sum_(n=0)^oo (-1)^n (x-pi/2)^(2n)/((2n)!)#

Note that:

#cos(x-pi/2) = cosx cos(pi/2) + sinx sin(pi/2) = sinx#

Then:

#sinx = cos(x-pi/2)#

and expanding #cos(x-pi/2)# in MacLaurin series:

#cos(x-pi/2) = sum_(n=0)^oo (-1)^n (x-pi/2)^(2n)/(2n!)#

and:

#sinx = sum_(n=0)^oo (-1)^n (x-pi/2)^(2n)/((2n)!)#