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Question: c) i) Show that $sin(e^x-1) = x^1 +\frac{x^2}{2} - \frac{(5x^4)}{24} + ......$
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Subject:- Applied Mathematics

Marks:- 3

Mumbai Unversity>FE>Sem1>Applied Maths1

m1(81) • 9 views
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modified 4 days ago  • written 4 days ago by gravatar for Mayank Aggarwal Mayank Aggarwal0
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We have $sin(e^x-1) = sin (1+x+\frac{x^2}{2}+\frac{x^3}{6} + \frac{x^4}{24}$ ……………..-1)

$∴sin(e^x-1) = sin (x +\frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} …………)$

But $sinθ = θ–θ^3/3!+ θ^5/5! - …..$

$∴sin(e^x-1) = x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} +……. –1/6 (x+\frac{x^2}{2}+⋯…..)^3+……..$

$= x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} +……. –\frac{x^3}{6}–\frac{x^4}{24} + …….$

$= x + \frac{x^2}{2} - 5\frac{x^4}{24} + ……….$

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written 4 days ago by gravatar for Mayank Aggarwal Mayank Aggarwal0
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