How do you use a half-angle formula to simplify $tan (\frac{7 π}{8})$ $tan ((7pi)/8)$ ?

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How do you use a half-angle formula to simplify $tan (\frac{7 π}{8})$ $tan ((7pi)/8)$ ?

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Answer 1 · 2015-06-30T00:59:45

Simplify $tan (\frac{7 π}{8})$ $tan ((7pi)/8)$

Explanation:

Call $tan (\frac{7 π}{8}) = t$ $tan ((7pi)/8) = t$
Use trig identity: $tan 2 a = \frac{2 tan a}{1 - {tan}^{2} a}$ $tan 2a = (2tan a)/(1 - tan^2 a)$

$tan (\frac{14 π}{8}) = \frac{2 t}{1 - t^{2}}$ $tan ((14pi)/8) = (2t)/(1 - t^2)$

Since $tan (\frac{14 π}{8}) = tan (\frac{6 π}{8} + π) = tan (\frac{3 π}{4} + π)$ $tan ((14pi)/8) = tan ((6pi)/8 + pi) = tan ((3pi)/4 + pi)$ =

$= - tan (\frac{3 π}{4}) = - 1$ $= - tan ((3pi)/4) = -1$

Therefor, $\frac{2 t}{1 - t^{2}} = - 1$ $(2t)/(1 - t^2) = -1$ -> $t^{2} - 2 t - 1 = 0$ $t^2 - 2t - 1 = 0$
Solve this quadratic equation.
$D = d^{2} = b^{2} - 4 a c = 4 + 4 = 8 \to d = \pm 2 \sqrt{2}$ $D = d^2 = b^2 - 4ac = 4 + 4 = 8 -> d = +- 2sqrt2$
$t = 1 \pm \sqrt{2}$ $t = 1 +- sqrt2$

$t = tan (\frac{7 π}{8}) = 1 - \sqrt{2}$ $t = tan ((7pi)/8) = 1 - sqrt2$ (Quadrant II)

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How do you use a half-angle formula to simplify $tan (\frac{7 π}{8})$ $tan ((7pi)/8)$ ?

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How do you use a half-angle formula to simplify tan(7π8)tan ((7pi)/8)?

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How do you use a half-angle formula to simplify $tan (\frac{7 π}{8})$ $tan ((7pi)/8)$ ?