How do you find the exact values of tan(5pi/12) & sin(5pi/12) using the half angle formula?

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How do you find the exact values of tan(5pi/12) & sin(5pi/12) using the half angle formula?

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Answer 1 · 2015-07-16T04:31:06

Find $tan (\frac{5 π}{12})$ $tan ((5pi)/12)$ and sin ((5pi)/12)
Answer: $\pm (2 \pm \sqrt{3}) and \pm \frac{\sqrt{2 + \sqrt{3}}}{2}$ $+- (2 +- sqrt3) and +- sqrt(2 + sqrt3)/2$

Explanation:

Call tan ((5pi/12) = t.
Use trig identity: $tan 2 a = \frac{2 tan a}{1 - {tan}^{2} a}$ $tan 2a = (2tan a)/(1 - tan^2 a)$
$tan (\frac{10 π}{12}) = tan (\frac{5 π}{6}) = - \frac{1}{\sqrt{3}} = \frac{2 t}{1 - t^{2}}$ $tan ((10pi)/12) = tan ((5pi)/6) = - 1/(sqrt3) = (2t)/(1 - t^2)$
$t^{2} - 2 \sqrt{3} t - 1 = 0$ $t^2 - 2sqrt3t - 1 = 0$

$D = d^{2} = b^{2} - 4 a c = 12 + 4 = 16$ $D = d^2 = b^2 - 4ac = 12 + 4 = 16$ --> $d = \pm 4$ $d = +- 4$

$t = tan (\frac{5 π}{12}) = \frac{2 \sqrt{3}}{2} \pm \frac{4}{2} = 2 \pm \sqrt{3}$ $t = tan ((5pi)/12) = (2sqrt3)/2 +- 4/2 = 2 +- sqrt3$

Call $sin (\frac{5 π}{12}) = sin y$ $sin((5pi)/12) = sin y$
Use trig identity: $cos 2 a = 1 - 2 {sin}^{2} a$ $cos 2a = 1 - 2sin^2 a$
$cos (\frac{10 π}{12}) = cos (\frac{5 π}{6}) = \frac{- \sqrt{3}}{2} = 1 - 2 {sin}^{2} y$ $cos ((10pi)/12) = cos ((5pi)/6) = (-sqrt3)/2 = 1 - 2sin^2 y$
${sin}^{2} y = \frac{2 + \sqrt{3}}{4}$ $sin^2 y = (2 + sqrt3)/4$
$sin y = sin (\frac{5 π}{12}) = \pm \frac{\sqrt{2 + \sqrt{3}}}{2}$ $sin y = sin ((5pi)/12) = +- sqrt(2 + sqrt3)/2$

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How do you find the exact values of tan(5pi/12) & sin(5pi/12) using the half angle formula?

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