cscØ = -3 , so sinØ = -1/3
the sine is negative in III and IV
cosine is positive in I and IV
so you are correct to say that Ø is in IV
make yourself familiar with the CAST rule, here is an explanation for it
http://www.mathsrevision.net/alevel/pages.php?page=36
So your triangle in IV should be a 1-√8 -3 triangle
sinØ = -1/3 , cscØ = -3 (that was the given)
cos = √8/3 , secØ = 3/√8
tanØ = -1/√8 , cotØ = -√8
(Your textbook answer might have written the √8 as 2√2 )
given that csc(theta) =-3 and cos(theta)>0, find the remaining 5 trigonometric functions of theta.
i have my triangle graphed in the fourth quadrant but I'm getting confused because of the negatives...
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