Question 2
Note: Let t = theta since I there is no theta symbol on my keyboard.
(0, -1) = (cos t, sin t)
cos t = 0
sin t = -1
I read the answer from the given point.
On the unit circle, the point (0, -1) lies at 3pi/2. So, angle theta is 3pi/2.
tan t = sin t/cos t
tan t = cos t ÷ sin t
Therefore, tan t = what?
You finish....
We know that any point located at (x,y)
Also, if the x-value and y-value both are positive then the point lie in first quadrant.
if both are negative then it lie in third quadrant.
If x-value is positive and y-value is negative then it lie in the fourth quadrant.
If x-value is negative and y-value is positive then it lie in the second quadrant.
The [tex]\sin \theta=\dfrac{y}{\sqrt{x^2+y^2}}[/tex]
[tex]\cos \theta=\dfrac{x}{\sqrt{x^2+y^2}}[/tex]
and [tex]\tan \theta=\dfrac{y}{x}[/tex]
1)
[tex]P(-\dfrac{1}{2},\dfrac{\sqrt{3}}{2})[/tex]
as x-value is negative and y-value is positive.
Hence, the point lie in the second quadrant.
Also,
[tex]x=-\dfrac{1}{2}\ ,\ y=\dfrac{\sqrt{3}}{2}[/tex]
Hence,
[tex]\sqrt{x^2+y^2}=\sqrt{(\dfrac{-1}{2})^2+(\dfrac{\sqrt{3}}{2})^2}\\\\\\\sqrt{x^2+y^2}=\sqrt{\dfrac{1}{4}+\dfrac{3}{4}}\\\\\\\sqrt{x^2+y^2}=1[/tex]
Hence, we have:
2)
P(0,-1)
As x-value is zero and y-value is negative.
Hence, the point lie on negative y-axis.
Also,
[tex]x=0\ ,\ y=-1[/tex]
Hence,
[tex]\sqrt{x^2+y^2}=\sqrt{(0)^2+(-1)^2}\\\\\\\sqrt{x^2+y^2}=1[/tex]
Hence, we have:
Hence, we have: tan θ= undefined
3)
[tex]P(-\dfrac{\sqrt{2}}{2},-\dfrac{\sqrt{2}}{2})[/tex]
which is same as:
[tex]P(-\dfrac{1}{\sqrt{2}},-\dfrac{1}{\sqrt{2}})[/tex]
As both the points are negative, hence the point lie in the third quadrant.
Also,
[tex]x=\dfrac{-1}{\sqrt{2}}\ ,\ y=\dfrac{-1}{\sqrt{2}}[/tex]
Hence, we have:
[tex]\sqrt{x^2+y^2}=\sqrt{(\dfrac{-1}{\sqrt{2}})^2+(\dfrac{-1}{\sqrt{2}})^2}\\\\\\\sqrt{x^2+y^2}=\sqrt{\dfrac{1}{2}+\dfrac{1}{2}}\\\\\\\sqrt{x^2+y^2}=1[/tex]
Hence, we have:
