what would the SIX trig ratios be if COTX=-1, and SinX=√2/2?

Question

Thank you in advance!

Answer 1

sinx = sqr2/2 = opposite side/hypotenuse = o/h

cotx = -1 = adjacent side/opposite side = a/o

cscx = 1/sinx = h/o = 2/sqr2 = sqr2

tanx = 1/cotx = o/a = 1/-1 = -1

cosx = a/h =-sqr2/2

secx = h/a = 1/cosx = - 2/sqr2= -sqr2

you could just notice that x = 135 degrees

and calculate the other 4 trig functions of 135

Answer 2

Based on the sin being positive and cot being negative, this must be in the second quadrant. The adjacent side is -1, the opposite side is 1 based on the cot value given and the hypotenuse is √2 based on the sin values given. Therefore using SOH CAH TOA,

sin x = √2/2 (given)

cos x = -√2/2

tan x = -1

cscx = √2

sec x = -√2

cot x = -1 (given)