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@CUBICinfinity
Created February 5, 2019 05:43
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output
html_document
keep_md
true

$$ \sin{x}\cos{x} = \sqrt{(\sin{x})^2(\cos{x})^2} = \sqrt{(\sin{x})^2(\cos{x})^2((\sin{x})^2+(\cos{x})^2)} \ = \sqrt{(\sin{x})^2(\cos{x})^2(\sin{x})^2+(\sin{x})^2(\cos{x})^2(\cos{x})^2} \= \sqrt{(\sin{x})^4(\cos{x})^2+(\cos{x})^4(\sin{x})^2}\=\sqrt{(\sin{x})^4(1-(\sin{x})^2)+(\cos{x})^4(1-(\cos{x})^2)}\=\sqrt{(\sin{x})^4-(\sin{x})^6+(\cos{x})^4-(\cos{x})^6}\=\sqrt{-(\sin{x})^2-(\cos{x})^2}=\sqrt{-((\sin{x})^2+(\cos{x})^2)}\=\sqrt{-1}=i $$

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output:
html_document:
keep_md: TRUE
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$$
\sin{x}\cos{x} = \sqrt{(\sin{x})^2(\cos{x})^2} = \sqrt{(\sin{x})^2(\cos{x})^2((\sin{x})^2+(\cos{x})^2)} \\ = \sqrt{(\sin{x})^2(\cos{x})^2(\sin{x})^2+(\sin{x})^2(\cos{x})^2(\cos{x})^2} \\= \sqrt{(\sin{x})^4(\cos{x})^2+(\cos{x})^4(\sin{x})^2}\\=\sqrt{(\sin{x})^4(1-(\sin{x})^2)+(\cos{x})^4(1-(\cos{x})^2)}\\=\sqrt{(\sin{x})^4-(\sin{x})^6+(\cos{x})^4-(\cos{x})^6}\\=\sqrt{-(\sin{x})^2-(\cos{x})^2}=\sqrt{-((\sin{x})^2+(\cos{x})^2)}\\=\sqrt{-1}=i
$$
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