If cot θ = −2, and θ terminates in QII, find the remaining trigonometric functions of θ.

Answer 1

cotx = -2

tanx = 1/cotx = 1/-2 = -1/2

hypotenuse = sqr(2^2 +1^2) = sqr5

adjacent side = -2

opposite side = 1

cotx = adjacent side over opposite side

tanx = opposite side over adjacent side

sinx = opposite side over hypotenuse

cosx = adjacent side over hypotenuse

sinx = 1/sqr5 = (1/5)sqr5

cosx = -2/sqr5 = (-2/5)sqr5

cscx = 1/sinx = sqr5

secx = 1/cosx =-(1/2)sqr5

one way to partially at least remember this is an old Indian word "SOHCAHTOA" It rhymes, at least 1st and last syllables do. It may not be PC lately, but it's a pneumatic device to remember sines, cosines and tangents

SOH

sine = o/h, opposite side divided by hypotenuse

CAH

cosine = a/h, adjacent side divided by hypotenuse

TOA

tangent = o/a, opposite side divided by adjacent side