Question 1080829: The terminal arm of angle θ passes through the point (a.b) in quadrant I. Leave your answers in terms of a and b. Determine the following:
a) sec θ
I got (sqrt a^2+b^2)/(a)
Correct?
b) cot θ
I got a/b
Correct?
c) sin (θ-pi)
Why do you switch the a and b values here, so like it becomes the x value becomes -b and y value becomes a?
Whats the solution then?
d) cos (θ +pi/2)
e) csc (θ-pi/2)
Please answer fast! Thank you very much!
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! The terminal arm of angle θ passes through the point (a.b) in quadrant I. Leave your answers in terms of a and b. Determine the following:
a) sec θ
I got (sqrt a^2+b^2)/(a)
Correct?
----
Correct
---
b) cot θ
I got a/b
Correct?
Correct
--------------------
c) sin (θ-pi)
Why do you switch the a and b values here, so like it becomes the x value becomes -b and y value becomes a?
Whats the solution then?
---
The terminal point is rotated to (-a,-b) in Q3
sin = -a/sqrt(a^2+b^2)
-------
d) cos (θ +pi/2)
TP is rotated to (-b,a)
cos = -b/(\sqrt(a^2+b^2)
------------------
e) csc (θ-pi/2)
TP is (b,-a)
csc = -sqrt(a^2+b^2)/a
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