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
Tan theta =2/3 and cos theta <0, find cos Theta and csc theta
3 answers
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.
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
the cscθ should be -√13/2
2 answers