Question

Expansion of y = Sin^-1(x) is?

(a) \(x+\frac{x^3}{6}+\frac{3}{40} x^5+\frac{5}{112} x^7+…..\)

(b) \(x-\frac{x^3}{6}+\frac{3}{40} x^5-\frac{5}{112} x^7+…..\)

(c) \(\frac{x^3}{6}-\frac{3}{40} x^5+\frac{5}{112} x^7…..\)

(d) \(x+\frac{x^2}{6}+\frac{3}{40} x^2+\frac{5}{112} x^2+…..\)

I have been asked this question at a job interview.

This key question is from Taylor Mclaurin Series in portion Differential Calculus of Engineering Mathematics

Answer 1

The correct option is (a) \(x+\frac{x^3}{6}+\frac{3}{40} x^5+\frac{5}{112} x^7+…..\)

To explain: Given, y = Sin^-1(x), hence at x = 0, y = 0

Now, differentiating it, we get

\(\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}=(1-x^2)^{-1/2}\)

On expanding the R.H.S. by Binomial Theorem we get,

\(\frac{dy}{dx}=1+\frac{x^2}{2}+\frac{3}{8} x^4+\frac{5}{16} x^6+…\)

On integrating we get,

\(y=x+\frac{x^3}{6}+\frac{3}{40} x^5+\frac{5}{112} x^7+….+C\)

By putting x=0 hence we get,

y=c=0

Hence,

\(y=x+\frac{x^3}{6}+\frac{3}{40} x^5+\frac{5}{112} x^7+….\)