The point (-1,1) is on the terminal side of an angle in standard position, how do you determine the exact values of the six trigonometric functions of the angle?

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Answer 1 · 2018-05-25T17:24:12

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Given: terminal side point #(-1,1)#

The point #(-1, 1)# is in the 2nd quadrant.

It has the #x# value #= -1# and the #y#-value #= 1#. Drop the point to the #x#-axis and form a reference triangle.

Using the Pythagorean Theorem #r = sqrt((-1)^2 + 1^2) = sqrt(2)#

#sin theta = y/r = 1/sqrt(2) = 1/sqrt(2) * sqrt(2)/sqrt(2) = sqrt(2)/2#

#cos theta = x/r = -1/sqrt(2) = -1/sqrt(2) * sqrt(2)/sqrt(2) = -sqrt(2)/2#

#tan theta = y/x = 1/-1 = -1#

#csc theta = 1/(sin theta) = r/y = sqrt(2)/1 = sqrt(2)#

#sec theta = 1/(cos theta) = r/x = sqrt(2)/-1 = -sqrt(2)#

#cot theta = 1/(tan theta) = x/y = -1/1 = -1#