How do you find the 6 trigonometric functions for -45 degrees?

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Answer 1 · 2018-03-17T06:23:08

As below

$hat (-45)$ is in IV quadrant where only $cos, sec$ are positive.

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$sin (-45) = - sin 45 = - (1/sqrt2)$

$csc (-45) = - csc 45 = - 1/ sin 45 = - sqrt2)$

$cos (-45) = cos 45 = - (1/sqrt2)$

$sec (-45) = sec 45 = 1/cos 45 = sqrt2)$

$tan (-45) = - tan 45 = - 1$

$cot (-45) = - cot 45 = -1/ tan 45 = - 1$