Unit circle

[Baron]

Point P is the intersection of the terminal arm of angle ? in standard position and the unit circle with centre (0,0). If P is quadrant 3 and cos ? = m. Find the coordinates of P in terms of m.

My work

The x coordinate is given (cos ? = m)
Draw a right triangle along the Cartesian plane in the third quadrant. M is on the x axis. Hypotenuse is 1

sin^2 (x) + cos^2 (x) = 1

m^2 + cos^2 (x) = 1
(±sqrt(1-m^2) = cos (x)

Coordinates of point P are (cos ?, sin ?)

My answer is either [m, sqrt(1-m^2)] or [m, -sqrt(1-m^2)]. Which one is correct? Why?

I'm thinking [m, -sqrt(1-m^2)] because it's in quadrant 3, so both the x and y coordinates are both negative. I'm confused because there is no negative sign in front of the m.

 

[tkhunny]

m^2 + cos^2 (x) = 1

No. Why did you substitute for sine when you were given cosine?

(±sqrt(1-m^2) = cos (x)

No. We know the cosine is negative. Why would the positive square root get in there?

My answer is either [m, sqrt(1-m^2)] or [m, -sqrt(1-m^2)]. Which one is correct? Why?

That's right, but I don't know how you got here.

I'm thinking [m, -sqrt(1-m^2)] because it's in quadrant 3, so both the x and y coordinates are both negative.

That's right. Very good.

I'm confused because there is no negative sign in front of the m.

What? Reread your definition. Why does there have to be a negative sign? It's already negative. If you add a negative sign, it will be positive.

 

[Baron]

My bad ... I made a typo.

My work should have been:
Given cos(?) = m ... so the x coordinate is m

sin^2 (?) + cos^2 (?) = 1
sin^2 (?) + m^2 = 1
sin^2 (?) = 1- m^2
sin (?) = ±sqrt(1-m^2)

Since on a unit circle, the coordinates (x,y) are equal to (cos ?, sin ?), the P is either [m, sqrt(1-m^2)] or [m, -sqrt(1-m^2)]

Since it is in quadrant 3, I think the answer is [m, -sqrt(1-m^2)]
Is my thinking correct?

If the question said the angle was in the second or first quadrant, would the answer be [m, sqrt(1-m^2)]?
If the question said the angle was in the third or fourth, would the answer be [m, -sqrt(1-m^2)]?

 

[tkhunny]

sin (?) = ±sqrt(1-m^2)

Still no. We know already that it is Quadrant III. The sine is negative. Don't even write the positive. Everyone will be less confused.

Since it is in quadrant 3, I think the answer is [m, -sqrt(1-m^2)]
Is my thinking correct?

You have it.

If the question said the angle was in the second or first quadrant, would the answer be [m, sqrt(1-m^2)]?

Quadrant II - Yes.
Quadrant I - No - That would be -m and the positive square root. "m" is negative. "-m" is positive.

If the question said the angle was in the third or fourth, would the answer be [m, -sqrt(1-m^2)]?

Quadrant III - Yes
Quadrant IV - No - That would be -m and the negative square root. "m" is negative. "-m" is positive.

Why would the result EVER be the same in Quadrants III and IV (or any other pair of adjacent Quadrants)? The sign of one coordinate or the other MUST change to get to another adjacent Quadrant.

 

[Baron]

If the question said the angle was in the second or first quadrant, would the answer be [m, sqrt(1-m^2)]?

Quadrant II - Yes.
Quadrant I - No - That would be -m and the positive square root. "m" is negative. "-m" is positive.

I'm confused. I meant if the question read:

Point P is the intersection of the terminal arm of angle ? in standard position and the unit circle with centre (0,0). If P is quadrant 1 and cos ? = m. Find the coordinates of P in terms of m.

Would the answer be [m, sqrt(1-m^2)]. Because in this case "m" is positive.

If the question read .

Point P is the intersection of the terminal arm of angle ? in standard position and the unit circle with centre (0,0). If P is quadrant 2 and cos ? = m. Find the coordinates of P in terms of m.

Would the answer be [m, sqrt(1-m^2)]. Because in the case "m" is negative

 

[tkhunny]

If you are given cos(x) = m, then the x-coordinate will always be 'm'. It's only the sign on the y-coordinate that needs some figuring. This is as you have stated it. Good work.