"역삼각함수"의 두 판 사이의 차이

2010년 8월 21일 (토) 15:40 판

사인/아크사인함수 덧셈정리의 적분표현
\(\sin \left(x+y\right)=\sin x \cos y + \cos x \sin y\\)
\(\arcsin x+\arcsin y=\arcsin (x\sqrt{1-y^2}+y\sqrt{1-x^2})\)
\(\int_0^x{\frac{1}{\sqrt{1-x^2}}}dx+\int_0^y{\frac{1}{\sqrt{1-x^2}}}dx = \int_0^{x\sqrt{1-y^2}+y\sqrt{1-x^2}}{\frac{1}{\sqrt{1-x^2}}}dx \)
탄젠트/아크탄젠트 함수 덧셈정리의 적분표현
\(\tan(\theta_1+\theta_2)=\frac{\tan\theta_1+\tan\theta_2}{1-\tan\theta_1\tan\theta_2}\)
\(\arctan x+\arctan y = \arctan{\frac{x+y}{1-xy}}\)
\(\int_0^x \frac{dx}{1+x^2} + \int_0^y \frac{dx}{1+x^2} = \int_0^{\frac{x+y}{1-xy}} \frac{dx}{1+x^2}\)

\(2(\arcsin x)^2=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2\binom{2n}{n}}\)

\(\frac{2x \arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n\binom{2n}{n}}\)

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 * http://www.google.com/search?hl=en&tbs=tl:1&q=inverse+tangent
-* http://www.google.com/search?hl=en&tbs=tl:1&q=
+* http://www.google.com/search?hl=en&tbs=tl:1&q=arctanget
+* http://www.google.com/search?hl=en&tbs=tl:1&q=arctangent<br> http://www.google.com/search?hl=en&tbs=tl:1&q=
 * [[수학사연표 (역사)|수학사연표]]
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