of course not. You know that cosθ is never greater than 1 or less than -1.
Similarly, cscθ is never less than 1.
Given a point (x,y),
r^2 = x^2+y^2
cosθ = x/r
cscθ = r/y
Also, given that cosθ<0 and cscθ>0. tanθ cannot be positive.
the following is a point not on a unit circle. θ → (-5,4) use it to find:
1. cos θ = -5
2. csc θ = 1/4
am i correct?
also, sec θ is negative, tan θ is positive. in what quadrant is < θ?
4 answers
1. cos θ = .284
2. csc θ = 1.043
is this correct?
and would it be quadrant 3?
2. csc θ = 1.043
is this correct?
and would it be quadrant 3?
For the point (-5,4)
x = -5
y = 4
r = √41
cosθ = x/r = -5/√41 = -.781
cscθ = r/y = √41/4 = 1.601
Clearly the point is in QII !
How did you come up with your values?
x = -5
y = 4
r = √41
cosθ = x/r = -5/√41 = -.781
cscθ = r/y = √41/4 = 1.601
Clearly the point is in QII !
How did you come up with your values?
i wrote quadrant 2 but my teacher said it was wrong