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Grade 12Trigonometry

how to solve or find sum of series of inversae trigonometric function

Profile image of sneha
12 Years agoGrade 12
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1 Answer

Profile image of Sher Mohammad
12 Years ago
y = sin-1(x) then x = sin(y)

\displaystyle \frac{dx}{dy} = \cos(y) = \sqrt{1-\sin^2(y)} = \sqrt{1-x^2}.
\displaystyle \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{\sqrt{1-x^2}}
\displaystyle \frac{d}{dx} \cos^{-1}(x) = -\frac{1}{\sqrt{1-x^2}}.
\displaystyle \frac{dx}{dy} = \sec^2(y) = 1+\tan^2(y) = 1+x^2.

\displaystyle \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{1+x^2}
\displaystyle \frac{dx}{dy} = \sec(y)\tan(y) = x \sqrt{x^2-1}.

\displaystyle \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{x \sqrt{x^2-1}}
\displaystyle \frac{1}{\sqrt{1-x^2}} = 1 + \frac{x^2}{2} + \frac{3x^4}{8} + \frac{5x^6}{16} + \frac{35x^8}{128} + ...

\displaystyle \frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + ...

Use above series and integrate accordingly to get complete series expansion of inverse trigonometric functions.


Sher Mohammad
B.Tech, IIT Delhi.







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