How do you find the exact value of the six trigonometric functions of θ when your given a point (-4,-6)?

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Answer 1 · 2016-07-15T17:01:45

#sintheta = -3sqrt13/13#

#costheta = -2sqrt13/13#

#tantheta = 3/2#

#sectheta = -sqrt13/2#

#csctheta = -sqrt13/3#

#ctntheta = 2/3#

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For the point #(-4,-6)# we can create a right triangle with legs of #x = -4# and #y =-6#

Using the pythagorean theorem we get the hypotenuse.

#a^2 + b^2 = c^2#

#(-4)^2 + (-6)^2 = hyp^2#

#16+36=hyp^2#

#52 = hyp^2#

#sqrt52 = hyp#

#2sqrt13 = hyp#

enter image source here

The six trig functions are

#sintheta = (opp)/(hyp)#

#costheta = (adj)/(hyp)#

#tantheta = (opp)/(adj)#

#sectheta = (hyp)/(adj)#

#csctheta = (hyp)/(opp)#

#ctntheta = (adj)/(opp)#

#sintheta = (-6)/(2sqrt13) = -3/sqrt13 * sqrt13/sqrt13 = -3sqrt13/13#

#costheta = (-4)/(2sqrt13) = -2/sqrt13 * sqrt13/sqrt13 = -2sqrt13/13#

#tantheta = (-6)/-4 = 3/2#

#sectheta = (2sqrt13)/-4 = -sqrt13/2#

#csctheta = (2sqrt13)/-6 = -sqrt13/3#

#ctntheta = (-4)/-6 = 2/3#