If cot x = ( - square root 2) and sin x > 0, find cos x
3 answers
cotx = (cosx)/ (sinx). Since sin>0, it must be in quadrant I or II. Since cotx is negative, cosx<0 or negative. The only quadrant where sin>0 and cosx<0 is quadrant II. Using exact values of trigonometric functions, you can know that cosx = -(sqrt(2))/ 2 and sinx = 2/1.
cotØ = -√2 and sinØ > 0
Ø must be in quad II by the CAST rule
cotØ = -√2 = x/y , so x = -√2 and y = 1 (in quad II)
x^2 + y^2 = r^2
2 + 1 = r^2
r = √3
cosØ = x/r = -√2/√3 or if you rationalize it, -√6/3
Ø must be in quad II by the CAST rule
cotØ = -√2 = x/y , so x = -√2 and y = 1 (in quad II)
x^2 + y^2 = r^2
2 + 1 = r^2
r = √3
cosØ = x/r = -√2/√3 or if you rationalize it, -√6/3
as above, we know that θ is in QII. Unfortunately, sinθ is never 2/1. and cosθ is also wrong.
cotθ = x/y = -√2/1
r = √3
cosθ = x/r = -√2/√3
cotθ = x/y = -√2/1
r = √3
cosθ = x/r = -√2/√3
1 answer