Question

Find the exact values of the six trigonometric functions of theta given the point (-4, 5) on the terminal side of theta in standard position.

Answer 1

You have x = -5 and y = -4 ; solve for "r".

r^2 = 5^2 + 4^2 = 41
r = sqrt(41) ?

Answer 2

sin(theta) = y/r = -5/sqrt(41)
cos(theta) = x/r = -4/sqrt(41)
tan(theta) = y/x = 5/4

Answer 3

Make a right triangle with legs 4 and 5. The hypotenuse will be sqrt(41) using pythagoras.
Sin = y / hyp = 5 / sqrt(41)
cos = x / hyp = 4 / sqrt(41-
tan = y / x = 5 / 4
cot = x / y = 4 / 5
sec =hyp / x = sqrt(41) / 4
csc = hyp / y = sqrt(41) / 5

Answer 4

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