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?
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.
1. cos θ = .284
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?
i wrote quadrant 2 but my teacher said it was wrong
To determine the values of trigonometric functions using the given point (-5, 4), we need to first find the radius of the circle that the point lies on. Then, we can use this information to find the values of the trigonometric functions.
The formula for the distance between the origin (0, 0) and a point (x, y) is given by the Pythagorean theorem: r = √(x^2 + y^2). Let's calculate this for the given point:
r = √((-5)^2 + 4^2)
= √(25 + 16)
= √41
Now, we can find the trigonometric functions using the following definitions:
1. cos θ = x / r
= -5 / √41
≈ -0.780
So, the value of cos θ is approximately -0.780.
2. csc θ = 1 / y
= 1 / 4
= 0.25
So, the value of csc θ is 0.25.
Based on the given information, sec θ is negative. The secant function is negative in the third and fourth quadrants. Since tan θ is positive, it is only positive in the first and third quadrants.
Therefore, based on the given restrictions, θ lies in the third quadrant (Q3).