Tan theta =2/3 and cos theta <0, find cos Theta and csc theta
tan>0 and cos<0 means your are in QIII. So, since tanθ = y/x,
y = -2
x = -3
r = √13
cosθ = x/r
cscθ = r/y
For these type of questions, sketch yourself a right-angled triangle and
use the basic definitions of the trig functions
e.g. tanθ = opposite/adjacent = y/x = 2/3
It also said that cosθ is negative, we know that if sinθ=+ and cosθ=- , θ must be in quad II
so x = -3, and y = 2
r^2 = 9 + 4 = 13
r = √13
Now you can form any of the 6 trig functions
cosθ = x/r = -3/√13
cscθ = r/y = √13/2
You should always test the results with a calculator.
oops, as oobleck has shown, I am in the wrong quadrant, should be in III
the cscθ should be -√13/2
To find the value of cos(theta), we need to determine whether theta is in the first or second quadrant. Since cos(theta) is negative, we can conclude that theta is in the second quadrant.
Since tan(theta) = 2/3, we can express this as the ratio of the opposite side and the adjacent side in a right triangle. Let's assume that the opposite side is length 2k and the adjacent side is length 3k, where k is a positive constant.
Using the Pythagorean theorem, we can determine the length of the hypotenuse:
(2k)^2 + (3k)^2 = (2k)^2 + (9k^2) = 4k^2 + 9k^2 = 13k^2
√(13k^2) = √13k
Now, we can determine the values of cos(theta) and csc(theta):
cos(theta) = adjacent / hypotenuse
= (3k) / √13k
= 3√13 / 13
csc(theta) = hypotenuse / opposite
= √13k / (2k)
= √13 / 2
Therefore, cos(theta) = 3√13 / 13 and csc(theta) = √13 / 2.