Asked by Wolferal
                If cot x = ( - square root 2) and sin x > 0, find cos x
            
            
        Answers
                    Answered by
            Amy
            
    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.  
    
                    Answered by
            Reiny
            
    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
                    Answered by
            oobleck
            
    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
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