Sum

Show that `sin(e^x-1)=x^1+x^2/2-(5x^4)/24+`...................

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#### 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+.........`

Concept: Expansion of 𝑒^𝑥 , sin(x), cos(x), tan(x), sinh(x), cosh(x), tanh(x), log(1+x), 𝑠𝑖𝑛−1 (𝑥),𝑐𝑜𝑠−1 (𝑥),𝑡𝑎𝑛−1 (𝑥)

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