Question

derive series expansion of sinx by tayler theorem
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Answer 1

Answer:

Taylor’s Series of sin x

In order to use Taylor’s formula to find the power series expansion of sin x we

Based on this power series expansion of #sin(x)#:

#sin(x) = x-x^3/(3!)+x^5/(5!)-x^7/(7!)+...#

#= (-1)^0[x^(2*0+1)/((2*0+1)!)]+(-1)^1[x^(2*1+1)/((2*1+1)!)]+(-1)^2[x^(2*2+1)/((2*2+1)!)]+(-1)^3[x^(2*3+1)/((2*3+1)!)]+...#

#= sum_(n=0)^(\infty)((-1)^nx^(2n+1))/((2n+1)!)#

we can derive it for #sin(sin(x))#.

What we can do is treat the inner #sin(x)# as the argument for the outer #sin(x)#, similar to how #x# is the argument for #ln(x)#. Doing that, we can just replace #x# with #sin(x)#:

#sin(sin(x)) = (sinx)/(1!)-(sinx)^3/(3!)+(sinx)^5/(5!)-(sinx)^7/(7!)+...#

#= sum_(n=0)^(\infty)((-1)^n(sinx)^(2n+1))/((2n+1)!)#

Answer 2

Answer:

Hope it helps

thanxx