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Question #68f30

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t0hierry · Stefan V.

Apr 26, 2016

$csc (\frac{2 π}{3} + \frac{5 π}{4}) = - (\sqrt{2} + \sqrt{6})$ $csc((2pi)/3+(5pi)/4) = -(sqrt2 + sqrt6)$

Explanation:

That is $csc (\frac{2 π}{3} + \frac{5 π}{4})$ $csc((2pi)/3+(5pi)/4)$

$sin (\frac{2 π}{3} + \frac{5 π}{4}) = sin (\frac{2 π}{3}) cos (\frac{5 π}{4}) + cos (\frac{2 π}{3}) sin (\frac{5 π}{4})$ $sin((2pi)/3+(5pi)/4) = sin((2pi)/3)cos((5pi)/4) + cos((2pi)/3)sin((5pi)/4)$

$= \frac{\sqrt{3}}{2} (- \frac{\sqrt{2}}{2}) + (- \frac{1}{2}) (- \frac{\sqrt{2}}{2})$ $= sqrt3/2 (-sqrt2/2) + (-1/2)(-sqrt2/2)$

$= \frac{\sqrt{2} - \sqrt{6}}{4}$ $= (sqrt2 - sqrt6)/4$

thus

$csc (\frac{2 π}{3} + \frac{5 π}{4}) = \frac{4}{\sqrt{2} - \sqrt{6}} = - 4 \cdot \frac{\sqrt{2} + \sqrt{6}}{4} = - \sqrt{2} - \sqrt{6}$ $csc((2pi)/3+(5pi)/4) = 4/(sqrt2 - sqrt6) = -4 * (sqrt2 + sqrt6)/4 = -sqrt2 - sqrt6$

Note

$\frac{23}{12} = \frac{2}{3} + \frac{5}{4}$ $23/12 = 2/3 + 5/4$

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