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pokergod
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#1
Report Thread starter 9 years ago
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How does 1/{8cos[arcsin(x/4]} = (+ or -) 1/{2[square root (16-x^2)]} ??
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nuodai
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Report 9 years ago
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\cos^2 A + \sin^2 A = 1 \Rightarrow \cos A = \sqrt{1 - \sin^2 A}, so if you let y = \cos \big( \arcsin{\dfrac{x}{4}} \big) then:
\newline \displaystyle

y = \sqrt{1 - \big(\sin (\arcsin x/4)\big)^2}\newline

 = \sqrt{1 - \dfrac{x^2}{16}}\newline

= \sqrt{\dfrac{16 - x^2}{16}}
... which simplifies to what you have.
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pokergod
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#3
Report Thread starter 9 years ago
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(Original post by nuodai)
\cos^2 A + \sin^2 A = 1 \Rightarrow \cos A = \sqrt{1 - \sin^2 A}, so if you let y = \cos \big( \arcsin{\dfrac{x}{4}} \big) then:
\newline \displaystyle

y = \sqrt{1 - \big(\sin (\arcsin x/4)\big)^2}\newline

 = \sqrt{1 - \dfrac{x^2}{16}}\newline

= \sqrt{\dfrac{16 - x^2}{16}}
... which simplifies to what you have.
Clever Clever lol. Thanks for the help
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