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 ( $tan (pi-theta) = - tan theta$ )

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

$sec pi = 1 / cos pi = 1 / -1 = -1$

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