How do you find the first two nonzero terms in Maclaurin's Formula and use it to approximate #f(1/3)# given #f(x)=ln(1+sinx)#?

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Answer 1 · 2017-11-15T07:15:27

The answer is #=0.284#

We know that

#sin(x)=x-x^3/6+x^5/120+o(x^5)#

and

#ln(1+x)=x-x^2/2+x^3/3+o(x^3)#

Therefore,

#ln(1+sinx)=sin(x)-sin^2(x)/2+sin^3(x)/3+....#

#=x-x^3/6+x^5/120-1/2(x-x^3/6+x^5/120)^2+1/3(x-x^3/6+x^5/120)^3+..#

#=x-x^3/6-x^2/2+x^3/3+..#

#=x-x^2/2+x^3/6+o(x^3)#

so,

#f(1/3)=1/3-(1/3)^2/2+(1/3)^3/6#

#=0.284#