Question
find tan 0 on the unit circle
I will describe the image there is a 0 on the bottom of the line through the image and above the line there is (1/2,sqrt 3/2)
I will describe the image there is a 0 on the bottom of the line through the image and above the line there is (1/2,sqrt 3/2)
Answers
Answered by
GPT-4o mini
To find \(\tan(0)\) on the unit circle, let's first clarify the position of the point based on your description of the unit circle.
On the unit circle, the angle \(0\) radians corresponds to the point \((1, 0)\) on the coordinate plane. The tangent function is defined as the ratio of the sine to the cosine of an angle, which corresponds to the \(y\)-coordinate to the \(x\)-coordinate of a point on the unit circle.
For \( \theta = 0 \):
- The coordinates are \((\cos(0), \sin(0)) = (1, 0)\)
- Therefore, \(\tan(0) = \frac{\sin(0)}{\cos(0)} = \frac{0}{1} = 0\)
Thus, \(\tan(0) = 0\).
Regarding the coordinates you mentioned, \((\frac{1}{2}, \frac{\sqrt{3}}{2})\) corresponds to an angle of \(\frac{\pi}{3}\) radians (or 60 degrees) in the unit circle, which is not relevant for calculating \(\tan(0)\).
In summary, \(\tan(0) = 0\).
On the unit circle, the angle \(0\) radians corresponds to the point \((1, 0)\) on the coordinate plane. The tangent function is defined as the ratio of the sine to the cosine of an angle, which corresponds to the \(y\)-coordinate to the \(x\)-coordinate of a point on the unit circle.
For \( \theta = 0 \):
- The coordinates are \((\cos(0), \sin(0)) = (1, 0)\)
- Therefore, \(\tan(0) = \frac{\sin(0)}{\cos(0)} = \frac{0}{1} = 0\)
Thus, \(\tan(0) = 0\).
Regarding the coordinates you mentioned, \((\frac{1}{2}, \frac{\sqrt{3}}{2})\) corresponds to an angle of \(\frac{\pi}{3}\) radians (or 60 degrees) in the unit circle, which is not relevant for calculating \(\tan(0)\).
In summary, \(\tan(0) = 0\).