A point on the terminal side of an angle θ in standard position is given. Find the exact values of the six trigonometric functions of (-1/6, 1/8). sint = cost = tant = csct= sect = cott =

Answer 1

(-1`/6, 1/8) is a point in the 2nd quadrant, with all trig functions negative except sines and cosecants.

sint = opposite over hypotenuse = o/h where o=1/8= .125 and h = sqr(1/6^2 + 1/8^) = .54

sint = (1/8)/sqr(7/24) = .125/(.54) =.23 rounded off to two decimal places

csct = 1/.23 =4.3

cost = adjacent side over hypotenuse =a/h = (-1/6)/(.54) = -.3

sect = 1/-.3 = -3.3

tant = o/a = (1/8)/(-1/6) = -3/4 =-.75

cott =1/tant a/o = -4/3 = -1.3

but if you want only "exact' solutions, then

sint= 0.125/sqr(7/24) = 1/8sqr(7/24)

csct = 8sqr(7/24)

cost=-1/6/sqr(7/24)= -1/6sqr(7/24)

sect = -6sqr(7/24)

tant= -3/4

cott = -4/3