How do you find the exact value of the six trigonometric functions of θ when your given a point 5π/6?

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How do you find the exact value of the six trigonometric functions of θ when your given a point 5π/6?

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Answer 1 · 2018-03-12T01:52:46

As below.

Explanation:

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$\frac{5 π}{6} = π - \frac{5 π}{6} = \frac{π}{6}$ $(5pi)/6 = pi - (5pi)/6 = pi/6$

Angle is in II quadrant.

Only sin, csc are positive in II quadrant and cos, sec, tan and cot are negative.

$sin (\frac{5 π}{6}) = sin (\frac{π}{6}) = \frac{1}{2}$ $sin ((5pi)/6) = sin (pi/6) = 1/2$

$csc (\frac{5 π}{6}) = csc (\frac{π}{6}) = \frac{1}{sin (\frac{π}{6})} = 2$ $csc ((5pi)/6) = csc (pi/6) = 1 / sin (pi/6) = 2$

$tan (\frac{5 π}{6}) = - tan (\frac{π}{6}) = - \frac{1}{\sqrt{3}}$ $tan ((5pi)/6) = - tan (pi/6) = -1/sqrt3$

$cot (\frac{5 π}{6}) = - cot (\frac{π}{6}) = - \sqrt{3}$ $cot ((5pi)/6) = - cot (pi/6) = - sqrt3$

$cos (\frac{5 π}{6}) = - cos (\frac{π}{6}) = - \frac{\sqrt{3}}{2}$ $cos ((5pi)/6) = -cos (pi/6) = -sqrt3/2$

$sec (\frac{5 π}{6}) = - sec (\frac{π}{6}) = - \frac{1}{cos (\frac{π}{6})} = - \frac{2}{\sqrt{3}}$ $sec ((5pi)/6) = -sec (pi/6) = -1/ cos (pi/6) = -2/sqrt3$

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How do you find the exact value of the six trigonometric functions of θ when your given a point 5π/6?

1 Answer

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