How do you find the exact value of $sec (\frac{5 π}{12})$ $sec((5pi)/12)$ ?

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How do you find the exact value of $sec (\frac{5 π}{12})$ $sec((5pi)/12)$ ?

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Answer 1 · 2016-02-29T17:43:33

$\frac{2}{\sqrt{2 - \sqrt{3}}}$ $2/(sqrt(2 - sqrt3)$

Explanation:

$sec (\frac{5 π}{12}) = \frac{1}{cos}$ $sec ((5pi)/12) = 1/cos$ . Find $cos (\frac{5 π}{12})$ $cos ((5pi)/12)$
On the trig unit circle,
$cos (\frac{5 π}{12}) = cos (\frac{6 π}{12} - \frac{π}{12}) = cos (\frac{π}{2} - \frac{π}{12})$ $cos ((5pi)/12) = cos ((6pi)/12 - pi/12) = cos (pi/2 - (pi)/12)$ =
$= sin (\frac{π}{12})$ $= sin (pi/12)$ . (complementary arcs)
Find $sin (\frac{π}{12})$ $sin (pi/12)$ by using trig identity:
$cos (\frac{π}{6}) = \frac{\sqrt{3}}{3} = 1 - 2 {sin}^{2} (\frac{π}{12})$ $cos (pi/6) = sqrt3/3 = 1 - 2sin^2 (pi/12)$
${sin}^{2} (\frac{π}{12}) = 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{4}$ $sin^2 (pi/12) = 1 -sqrt3/2 = (2 - sqrt3)/4$
$sin (\frac{π}{12}) = \frac{\sqrt{2 - \sqrt{3}}}{2}$ $sin (pi/12) = sqrt(2 - sqrt3)/2$ --> (sin (pi/12) is positive.
We have:
$cos (\frac{5 π}{12}) = sin (\frac{π}{12}) = \frac{\sqrt{2 - \sqrt{3}}}{2}$ $cos ((5pi)/12) = sin (pi/12) = sqrt(2 - sqrt3)/2$
Finally, flip the value of cos.
$sec (\frac{5 π}{12}) = \frac{2}{\sqrt{2 - \sqrt{3}}}$ $sec ((5pi)/12) = 2/(sqrt(2 - sqrt3)$

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How do you find the exact value of $sec (\frac{5 π}{12})$ $sec((5pi)/12)$ ?

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