What is the sec(inverse cos (5^0.5/2))?

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What is the sec(inverse cos (5^0.5/2))?

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Answer 1 · 2015-12-04T21:38:02

$sec (arccos (\frac{\sqrt{5}}{2})) = \frac{2}{\sqrt{5}}$ $sec(arccos(sqrt(5)/2))=2/sqrt(5)$

Explanation:

For this problem, we just need two facts:

$sec (x) = \frac{1}{cos (x)}$ $sec(x) = 1/cos(x)$ (by definition of the secant function)
$cos (arccos (x)) = arccos (cos (x)) = x$ $cos(arccos(x)) = arccos(cos(x)) = x$ (by definition of the inverse of a function)

Applying those gives us

$sec (arccos (\frac{\sqrt{5}}{2})) = \frac{1}{cos (arccos (\frac{\sqrt{5}}{2}))} = \frac{1}{\frac{\sqrt{5}}{2}} = \frac{2}{\sqrt{5}}$ $sec(arccos(sqrt(5)/2)) = 1/cos(arccos(sqrt(5)/2)) = 1/(sqrt(5)/2) = 2/sqrt(5)$

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What is the sec(inverse cos (5^0.5/2))?

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