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Question
Show that `sin(e^x-1)=x^1+x^2/2-(5x^4)/24+`...................
Solution
We have `sin(e^x-1)=sin(1+x+x^2/2+x^3/6+x^4/24...........-1)`
`therefore sin(e^x-1)=sin(x+x^2/2+x^3/6+x^4/24...........)`
But `sintheta=theta-theta^3/(3!)+theta^5/(5!)-......`
`therefore sin(e^x-1)=x+x^2/2+x^3/6+x^4/24+.......-1/6(x+x^2/2+........)^3+........`
`= x+x^2/2+x^3/6+x^4/24+......-x^3/6-x^4/4+......`
`= x+x^2/2-(5x^4)/24+.........`
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