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.

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$(5pi)/6 = pi - (5pi)/6 = pi/6$

Angle is in II quadrant.

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

$sin ((5pi)/6) = sin (pi/6) = 1/2$

$csc ((5pi)/6) = csc (pi/6) = 1 / sin (pi/6) = 2$

$tan ((5pi)/6) = - tan (pi/6) = -1/sqrt3$

$cot ((5pi)/6) = - cot (pi/6) = - sqrt3$

$cos ((5pi)/6) = -cos (pi/6) = -sqrt3/2$

$sec ((5pi)/6) = -sec (pi/6) = -1/ cos (pi/6) = -2/sqrt3$