Question

the given point p is located on the unit circle State the quadrant and find the angle also sin,cos,tan

Answer 1

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....

Answer 2

Answer with explanation:

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 $\sin \theta=\dfrac{y}{\sqrt{x^2+y^2}}$

$\cos \theta=\dfrac{x}{\sqrt{x^2+y^2}}$

and $\tan \theta=\dfrac{y}{x}$

1)

 $P(-\dfrac{1}{2},\dfrac{\sqrt{3}}{2})$

as x-value is negative and y-value is positive.

Hence, the point lie in the second quadrant.

Also,

$x=-\dfrac{1}{2}\ ,\ y=\dfrac{\sqrt{3}}{2}$

Hence,

$\sqrt{x^2+y^2}=\sqrt{(\dfrac{-1}{2})^2+(\dfrac{\sqrt{3}}{2})^2}\\\\\\\sqrt{x^2+y^2}=\sqrt{\dfrac{1}{4}+\dfrac{3}{4}}\\\\\\\sqrt{x^2+y^2}=1$

Hence, we have:

• $\sin \theta=\dfrac{\dfrac{\sqrt{3}}{2}}{1}\\\\\\\sin\theta=\dfrac{\sqrt{3}}{2}$

• $\cos \theta=\dfrac{-1}{2}$

• and $\tan \theta=\dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}}\\\\\\\tan \theta=\sqrt{3}$

2)

P(0,-1)

As x-value is zero and y-value is negative.

Hence, the point lie on negative y-axis.

Also,

$x=0\ ,\ y=-1$

Hence,

$\sqrt{x^2+y^2}=\sqrt{(0)^2+(-1)^2}\\\\\\\sqrt{x^2+y^2}=1$

Hence, we have:

• $\sin \theta=\dfrac{-1}{1}\\\\\\\sin \theta=-1$

• $\cos \theta=\dfrac{0}{1}=0$

• and $\tan \theta=\dfrac{-1}{0}$,

       Hence, we have:  tan θ= undefined

3)

$P(-\dfrac{\sqrt{2}}{2},-\dfrac{\sqrt{2}}{2})$

which is same as:

       $P(-\dfrac{1}{\sqrt{2}},-\dfrac{1}{\sqrt{2}})$

As both the points are negative, hence the point lie in the third quadrant.

Also,

$x=\dfrac{-1}{\sqrt{2}}\ ,\ y=\dfrac{-1}{\sqrt{2}}$

Hence, we have:

$\sqrt{x^2+y^2}=\sqrt{(\dfrac{-1}{\sqrt{2}})^2+(\dfrac{-1}{\sqrt{2}})^2}\\\\\\\sqrt{x^2+y^2}=\sqrt{\dfrac{1}{2}+\dfrac{1}{2}}\\\\\\\sqrt{x^2+y^2}=1$

Hence, we have:

• $\sin \theta=\dfrac{-1}{\sqrt{2}}$

• $\cos \theta=\dfrac{-1}{\sqrt{2}}$

• $\tan \theta=1$