The point ( 1/5 , − 1/5 ) lies on the terminal arm of angle 𝛽 in standard position. Determine the exact values of csc 𝛽.

Question

The point ( 1/5 , − 1/5 ) lies on the terminal arm of angle 𝛽 in standard position. Determine the exact values of csc 𝛽.

Answer 1

Angle β is a 45° angle to point (1/5,-1/5)

Cosine of β = 0.707107

Cosecant of β = 1/Cosine β = 1.4142 or √2

Answer 2

B = 315 degrees = 7pi/4

reference angle = 45 degrees = pi/4

cscB = 1/sinB = 1/-sqr2/2 = -2/sqr2

(1/5, -1/5) has tanB = opposite side over adjacent side = -1/5 over 1/5 = -1

tan B = -1

opposite side =-1, adjacent side = 1 hypotenuse = sqr(1+1) = sqr2

sinB = -1/sqr2 = -sqr2/2

cscB = -sqr2= about - 1.414 rounded off to 3 decimal places

csc Beta = exactly negative square root of two