How do you evaluate the six trigonometric functions given t=π?

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Answer 1 · 2018-02-20T11:52:29

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#pi^c = 180^@# is in second quadrant.

Only sine and cosecant are positive.

#sin (pi) = sin (180) = sin (pi - pi) = sin 0 = 0#
as (#sin (pi - theta) = sin theta#)

#cos pi = cos 180 = cos (pi - pi) = -cos 0 = -1#
as (#cos (pi - theta) = - cos theta#)

#tan pi = tan (pi - pi) = - tan0 = -0# or #0#
as (#tan (pi-theta) = - tan theta#)

#csc pi = 1 / sin pi = 1 / 0 = oo#

#sec pi = 1 / cos pi = 1 / -1 = -1#

#cot pi = 1 / tan pi = 1 / -0 = -oo#