Question #d9cff

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Question #d9cff

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Answer 1 · 2017-03-06T02:15:28

$\pm \frac{3 + 2 \sqrt{3}}{2 \sqrt{13}}$ $+- (3 + 2sqrt3)/(2sqrt13)$

Explanation:

Call x the $arctan (\frac{2}{3})$ $arctan (2/3)$ --> $tan x = \frac{2}{3}$ $tan x = 2/3$
${cos}^{2} x = \frac{1}{1 + {tan}^{2} x} = \frac{1}{1 + \frac{4}{9}} = \frac{9}{13}$ $cos^2 x = 1/(1 + tan^2 x) = 1/( 1 + 4/9) = 9/13$
$cos x = \pm \frac{3}{\sqrt{13}}$ $cos x = +- 3/sqrt13$
$sin x = 1 - {cos}^{2} x = 1 - \frac{9}{13} = \frac{4}{13}$ $sin x = 1 - cos^2 x = 1 - 9/13 = 4/13$
$sin x = \pm \frac{2}{\sqrt{13}}$ $sin x = +- 2/sqrt13$
Since $tan x = \frac{2}{3}$ $tan x = 2/3$ --> x could be in Quadrant 1 or Quadrant 3.
If x is in Q. 1, sin x and cos x are both positive
If x is in Q.3, sin x and cos x are both negative.
Use trig identity sin ( a - b) = sin a.cos b - sin b.cos a
a. If sin x and cos x are both positive
$sin (\frac{5 π}{6} - x) = sin (\frac{5 π}{6}) . cos x - sin x . cos (\frac{5 π}{6}) =$ $sin ((5pi)/6 - x) = sin ((5pi)/6).cos x - sin x.cos ((5pi)/6) =$
$= (\frac{1}{2}) (\frac{3}{\sqrt{13}}) - (\frac{2}{\sqrt{13}}) (- \frac{\sqrt{3}}{2}) = \frac{3 + 2 \sqrt{3}}{2 \sqrt{13}}$ $= (1/2)(3/sqrt13) - (2/sqrt13)(-sqrt3/2) = (3 + 2sqrt3)/(2sqrt13)$
b. If sin x and cos x are both negative:
$sin (\frac{5 π}{6} - x) = (\frac{1}{2}) (- \frac{3}{\sqrt{13}}) - (- \frac{2}{\sqrt{13}}) (- \frac{\sqrt{3}}{2}) =$ $sin ((5pi)/6 - x) = (1/2)(-3/sqrt13) - (-2/sqrt13)(- sqrt3/2) =$
$= - \frac{3 + 2 \sqrt{3}}{2 \sqrt{13}}$ $= - (3 + 2sqrt3)/(2sqrt13)$

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Question #d9cff

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