Explore the unit circle and its relationship to angles, radians, trigonometric ratios, and coordinates in the coordinate plane.
Find the image of the unit circle under the transformation f(z)=z^2
The unit circle in the complex plane is given by $ |z| = 1 $, meaning any point $ z $ on the unit circle can be written as $ z = e^{i\theta} $ for some real number $ \theta $.
Under the transformation $ f(z) = z^2 $, the image of $ z $ is:
$$ f(z) = (e^{i\theta})^2 = e^{i(2\theta)} $$
Since $ e^{i(2\theta)} $ is still a point on the unit circle, the image of the unit circle under $ f(z) = z^2 $ is the unit circle itself.
Identify the quadrant in which the angle lies
To identify the quadrant in which the angle $ \theta $ lies, follow these steps:
1. If $ 0 \leq \theta < \frac{\pi}{2} $, then the angle is in the first quadrant.
2. If $ \frac{\pi}{2} \leq \theta < \pi $, then the angle is in the second quadrant.
3. If $ \pi \leq \theta < \frac{3\pi}{2} $, then the angle is in the third quadrant.
4. If $ \frac{3\pi}{2} \leq \theta < 2\pi $, then the angle is in the fourth quadrant.
Find the value of cos(π/3)
The value of $ \cos\left( \frac{\pi}{3} \right) $ can be found using the unit circle. The angle $ \frac{\pi}{3} $ corresponds to 60 degrees. On the unit circle, the coordinates for the angle 60 degrees are:
$$ \left( \cos\left( \frac{\pi}{3} \right), \sin\left( \frac{\pi}{3} \right) \right) = \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) $$
Therefore, the value of $ \cos\left( \frac{\pi}{3} \right) $ is:
$$ \frac{1}{2} $$
Find the coordinates of a point on the unit circle at angle π/4
To find the coordinates of a point on the unit circle at angle $\frac{\pi}{4}$, we use the trigonometric functions:
$$ x = \cos\left(\frac{\pi}{4}\right) $$
$$ y = \sin\left(\frac{\pi}{4}\right) $$
Since:
$$ \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} $$
$$ \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} $$
The coordinates are:
$$ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $$
Find the values of sine, cosine, and tangent for the angle 5π/6 using the unit circle
To find the values of $ \sin \frac{5\pi}{6} $, $ \cos \frac{5\pi}{6} $, and $ \tan \frac{5\pi}{6} $ using the unit circle, we first locate the angle $ \frac{5\pi}{6} $ on the unit circle.
The angle $ \frac{5\pi}{6} $ is in the second quadrant, where sine is positive and cosine is negative. The reference angle is $ \pi – \frac{5\pi}{6} = \frac{\pi}{6} $.
From the unit circle, we know:
$$ \sin \frac{\pi}{6} = \frac{1}{2} $$ and $$ \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} $$
Since we are in the second quadrant:
$$ \sin \frac{5\pi}{6} = \frac{1}{2} $$
$$ \cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} $$
$$ \tan \frac{5\pi}{6} = \frac{ \sin \frac{5\pi}{6} }{ \cos \frac{5\pi}{6} } = \frac{ \frac{1}{2} }{ -\frac{\sqrt{3}}{2} } = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} $$
Find the points where the line y = mx + b is tangent to the unit circle
To find the points where the line $ y = mx + b $ is tangent to the unit circle, we start with the equation of the unit circle:
$$ x^2 + y^2 = 1 $$
Substitute $ y = mx + b $ into the unit circle equation:
$$ x^2 + (mx + b)^2 = 1 $$
Expand the equation:
$$ x^2 + m^2x^2 + 2mxb + b^2 = 1 $$
Combine like terms:
$$ (1 + m^2)x^2 + 2mxb + b^2 = 1 $$
This is a quadratic equation in $ x $:
$$ (1 + m^2)x^2 + 2mxb + (b^2 – 1) = 0 $$
For the line to be tangent to the circle, the discriminant must be zero:
$$ (2mb)^2 – 4(1 + m^2)(b^2 – 1) = 0 $$
Simplify the discriminant:
$$ 4m^2b^2 – 4(1 + m^2)(b^2 – 1) = 0 $$
Solving this will give the condition on $ b $.
Find the value of tan(θ) when θ is at the angle π/4 on the unit circle
To find the value of $ \tan(\theta) $ when $ \theta $ is at the angle $ \frac{\pi}{4} $ on the unit circle, we use the definition of tangent in terms of sine and cosine:
$$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $$
At $ \theta = \frac{\pi}{4} $, we know:
$$ \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} $$
$$ \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} $$
So:
$$ \tan(\frac{\pi}{4}) = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 $$
Find the exact values of sin(θ) and cos(θ) for θ = π/4
To find the exact values of $ \sin(\theta) $ and $ \cos(\theta) $ for $ \theta = \frac{\pi}{4} $, we use the unit circle definition:
On the unit circle, the coordinates of the point corresponding to $ \theta = \frac{\pi}{4} $ are:
$$ ( \cos(\frac{\pi}{4}), \sin(\frac{\pi}{4}) ) $$
For $ \theta = \frac{\pi}{4} $, both $ \sin(\frac{\pi}{4}) $ and $ \cos(\frac{\pi}{4}) $ are:
$$ \frac{\sqrt{2}}{2} $$
Thus, we have:
$$ \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} $$
$$ \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} $$
Find the tangent of an angle $\theta$ on the unit circle in different quadrants
To find the tangent of an angle $\theta$ on the unit circle, note that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$. Consider the following angles:
1. For $\theta = \frac{\pi}{4}$ in the first quadrant:
$$\tan\left(\frac{\pi}{4}\right) = \frac{\sin\left(\frac{\pi}{4}\right)}{\cos\left(\frac{\pi}{4}\right)} = 1$$
2. For $\theta = \frac{3\pi}{4}$ in the second quadrant:
$$\tan\left(\frac{3\pi}{4}\right) = \frac{\sin\left(\frac{3\pi}{4}\right)}{\cos\left(\frac{3\pi}{4}\right)} = -1$$
3. For $\theta = \frac{5\pi}{4}$ in the third quadrant:
$$\tan\left(\frac{5\pi}{4}\right) = \frac{\sin\left(\frac{5\pi}{4}\right)}{\cos\left(\frac{5\pi}{4}\right)} = 1$$
4. For $\theta = \frac{7\pi}{4}$ in the fourth quadrant:
$$\tan\left(\frac{7\pi}{4}\right) = \frac{\sin\left(\frac{7\pi}{4}\right)}{\cos\left(\frac{7\pi}{4}\right)} = -1$$
Determine the tangent values for points on the unit circle at angles θ = π/4, θ = 2π/3, and θ = 5π/6
To find the tangent values for the given angles on the unit circle, we use the definition of the tangent function, which is $ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $.
For $ \theta = \frac{\pi}{4} $:
$$ \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \text{ and } \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} $$
Therefore,
$$ \tan\left(\frac{\pi}{4}\right) = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 $$
For $ \theta = \frac{2\pi}{3} $:
$$ \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \text{ and } \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} $$
Therefore,
$$ \tan\left(\frac{2\pi}{3}\right) = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = -\sqrt{3} $$
For $ \theta = \frac{5\pi}{6} $:
$$ \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} \text{ and } \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} $$
Therefore,
$$ \tan\left(\frac{5\pi}{6}\right) = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} $$
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