The Terminal arm of angle θ passes through the point (a,b) in quadrant 1.

[ZMS]

The Terminal arm of angle θ passes through the point (a,b) in quadrant 1. Leaving your answers in terms of a and b, determine the following: (θ=theta)
a) sec θ
At first I thought it would be r/x, but then I read that it says leave your answers in terms of a and b. How would you do that then?

b) cot θ
That's just a/b, right?

c) sin (θ-pi)

d) cos (θ+pi/2)

e) csc (θ-pi/2)

 

[ZMS]

I think I may have figured it out, can someone please verify my solutions are correct?
4a) Radius = sqrt (a^2+b^2)
Since sec is r/x, it'd be sqrt a^2+b^2/a. Is that right?

c) This is where I'm confused, so when it's being shifted by -180 (pi) does it start from a positive x and y value in quad 1 and go -180 all the way to quad 3 where sin is negative or does it start at the x-axis of quad 1 and shift -180 making it be in quad 2 where sin is positive.

If it was option 1, then it'd -b (sqrt)a^2+b^2/a^2+b^2. If option 2 it would then be positive instead of negative.

 

[stapel]

I think I may have figured it out, can someone please verify my solutions are correct?
4a) Radius = sqrt (a^2+b^2)
Since sec is r/x, it'd be sqrt[a^2+b^2]/a. Is that right?

Yes.

c) This is where I'm confused, so when it's being shifted by -180 (pi) does it start from a positive x and y value in quad 1 and go -180 all the way to quad 3 where sin is negative...

Yes. But the lengths of the sides and of the hypotenuse will remain the same. Only the signs on "a" and "b" will have changed.

...or does it start at the x-axis of quad 1 and shift -180 making it be in quad 2 where sin is positive.

I'm not sure how you're getting this...?

If it was option 1, then it'd -b (sqrt)a^2+b^2/a^2+b^2.

I'm not sure what the above means...? Because (a^2 + b^2)/(a^2 + b^2), which I'm assuming is what you meant, equals 1, which isn't very helpful.

Instead, try taking the Quadrant I value of sine, being b/sqrt[a^2 + b^2], and inverting it, changing the sign as appropriate.

 

[ZMS]

Yes.


Yes. But the lengths of the sides and of the hypotenuse will remain the same. Only the signs on "a" and "b" will have changed.


I'm not sure how you're getting this...?


I'm not sure what the above means...? Because (a^2 + b^2)/(a^2 + b^2), which I'm assuming is what you meant, equals 1, which isn't very helpful.

Instead, try taking the Quadrant I value of sine, being b/sqrt[a^2 + b^2], and inverting it, changing the sign as appropriate.

Nvm about that confusion I had earlier, I was being dumb. Also, why do a and b switch?

 

[stapel]

...why do a and b switch?

How are they "switching"? Where?

Please be specific. Thank you!