Find the values of cos(θ) and sin(θ) at specific angles on the unit circle
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Find the values of cos(θ) and sin(θ) at specific angles on the unit circle
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Find the value of tan(7π/6) and explain using the unit circle
To find the value of $ \tan(\frac{7\pi}{6}) $ using the unit circle:
1. Locate the angle $\frac{7\pi}{6}$ on the unit circle. This angle is in the third quadrant.
2. The reference angle for $\frac{7\pi}{6}$ is $\frac{\pi}{6}$.
3. In the third quadrant, both sine and cosine are negative. Knowing the coordinates for $\frac{\pi}{6}$ are $(\frac{\sqrt{3}}{2}, \frac{1}{2})$:
The coordinates for $\frac{7\pi}{6}$ are $(-\frac{\sqrt{3}}{2}, -\frac{1}{2})$.
4. Finally, calculate the tangent value:
$$ \tan(\frac{7\pi}{6}) = \frac{\sin(\frac{7\pi}{6})}{\cos(\frac{7\pi}{6})} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} $$
Finding the sine, cosine, and tangent of 45 degrees using the unit circle
To find the trigonometric functions of $45^\circ$ using the unit circle, note that $45^\circ$ corresponds to an angle in the first quadrant where both the x and y coordinates are equal.
Since the radius of the unit circle is 1, the coordinates of the point on the circle at $45^\circ$ are $\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$.
Thus:
$$ \sin(45^\circ) = \frac{\sqrt{2}}{2} $$
$$ \cos(45^\circ) = \frac{\sqrt{2}}{2} $$
$$ \tan(45^\circ) = 1 $$
Find the values of sin, cos, and tan for 45 degrees using the unit circle
To find the values of $ \sin $, $ \cos $, and $ \tan $ for $ 45^{\circ} $ using the unit circle, we start with:
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$$ \sin(45^{\circ}) = \frac{\sqrt{2}}{2} $$
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$$ \cos(45^{\circ}) = \frac{\sqrt{2}}{2} $$
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$$ \tan(45^{\circ}) = 1 $$
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Determine the cosine of an angle corresponding to the point (1/2, sqrt(3)/2) on the unit circle
To determine the cosine of the angle corresponding to the point $ \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) $ on the unit circle, we must recognize the coordinates $(x, y)$ represent $(\cos(\theta), \sin(\theta))$.
In this case, the point is:
$$( \cos(\theta), \sin(\theta) ) = \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)$$
Thus, the cosine of the angle is:
$$ \cos(\theta) = \frac{1}{2} $$
Find the coordinates of the point at an angle of pi/3 on the unit circle
To find the coordinates of the point at an angle of $ \frac{\pi}{3} $ on the unit circle, we use the unit circle definition where the coordinates are given by $ (\cos\theta, \sin\theta) $.
For $ \theta = \frac{\pi}{3} $, we have:
$$ \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} $$
$$ \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} $$
So, the coordinates are:
$$ \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) $$
Determine the sine and cosine values of an angle 𝜃 in radians on the unit circle given that 𝜃 = 5𝜋/4
Given $\theta = \frac{5\pi}{4}$, we determine the sine and cosine values by examining the unit circle.
The angle $\frac{5\pi}{4}$ is located in the third quadrant, where sine and cosine values are negative. Specifically:
$$ \sin \left( \frac{5\pi}{4} \right) = -\frac{\sqrt{2}}{2} $$
$$ \cos \left( \frac{5\pi}{4} \right) = -\frac{\sqrt{2}}{2} $$
Find the exact values of the inverse trigonometric functions on the unit circle
Consider the point $$P(-\frac{1}{2}, -\frac{\sqrt{3}}{2})$$ on the unit circle. Determine the exact values for the following inverse trigonometric functions:
1. $$\arcsin(-\frac{\sqrt{3}}{2})$$
2. $$\arccos(-\frac{1}{2})$$
3. $$\arctan(\frac{\sqrt{3}}{3})$$
Answer:
1. $$\arcsin(-\frac{\sqrt{3}}{2}) = -\frac{\pi}{3}$$
2. $$\arccos(-\frac{1}{2}) = \frac{2\pi}{3}$$
3. $$\arctan(\frac{\sqrt{3}}{3}) = \frac{\pi}{6}$$
Find the value of tan(θ) at θ = 3π/4 on the unit circle
To find the value of $ \tan(θ) $ at $ θ = \frac{3π}{4} $, we first identify the coordinates on the unit circle:
At $ θ = \frac{3π}{4} $, the coordinates are $ (-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) $.
So, $ \tan(θ) $ is given by:
$$ \tan(θ) = \frac{y}{x} = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1 $$
Therefore, $ \tan(θ) = -1 $.
Determine the values of sin, cos, and tan for an angle of 7π/6 on the unit circle
First, locate the angle $\frac{7\pi}{6}$ on the unit circle. This angle corresponds to 210 degrees.
The coordinates of the point on the unit circle at this angle are:
$$ \left( -\frac{\sqrt{3}}{2}, -\frac{1}{2} \right) $$
Thus:
$$ \sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2} $$
$$ \cos\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2} $$
$$ \tan\left(\frac{7\pi}{6}\right) = \frac{\sin\left(\frac{7\pi}{6}\right)}{\cos\left(\frac{7\pi}{6}\right)} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} $$
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