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Find the value of arcsin(1/2) using the unit circle
To find the value of $ \arcsin(\frac{1}{2}) $ using the unit circle, we need to determine the angle whose sine is $ \frac{1}{2} $.
On the unit circle, the sine of an angle is the y-coordinate of the corresponding point.
The angle that has a sine of $ \frac{1}{2} $ in the range $ -\frac{\pi}{2} $ to $ \frac{\pi}{2} $ is $ \frac{\pi}{6} $.
Therefore, the value of $ \arcsin(\frac{1}{2}) $ is:
$$ \frac{\pi}{6} $$
Determine the coordinates of a point on the unit circle where the sine value is 1/2 and the tangent value is positive
To find the coordinates where $\sin(\theta) = \frac{1}{2}$ and $\tan(\theta)$ is positive, we analyze the unit circle.
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The sine function equals $\frac{1}{2}$ at two angles: $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}$.
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Since the tangent function is positive when both sine and cosine have the same sign, we consider the angles in the first and third quadrants.
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For $\theta = \frac{\pi}{6}$, the coordinates on the unit circle are:
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$$ (\cos(\frac{\pi}{6}), \sin(\frac{\pi}{6})) = (\frac{\sqrt{3}}{2}, \frac{1}{2}) $$
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Thus, the coordinates are:
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$$ (\frac{\sqrt{3}}{2}, \frac{1}{2}) $$
Find the value of tan(θ) using the unit circle when θ is in the third quadrant
To find the value of $ \tan(θ) $ using the unit circle, we need to determine the coordinates where $ θ $ intersects the unit circle in the third quadrant.
In the third quadrant, both the x and y coordinates are negative. Suppose $ θ = 225° $ (or $ \frac{5π}{4} $ in radians). In this case, the coordinates on the unit circle are $ ( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} ) $.
The tangent of $ θ $ is given by the ratio of the y-coordinate to the x-coordinate:
$$ \tan(225°) = \frac{y}{x} = \frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = 1 $$
Find the cosine of the angle pi/4 on the unit circle
The unit circle defines the standard positions and values of trigonometric functions. For the angle $ \frac{\pi}{4} $ (or 45 degrees), we use the unit circle definition:
The coordinates of the point on the unit circle corresponding to the angle $ \frac{\pi}{4} $ are:
$$ ( \cos( \frac{\pi}{4} ), \sin( \frac{\pi}{4} )) $$
Since the unit circle has radius 1, we get:
$$ \cos( \frac{\pi}{4} ) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} $$
Determine the coordinates of a point in the first quadrant of the unit circle given its angle
To determine the coordinates of a point in the first quadrant on the unit circle given its angle $ \theta $, we use the trigonometric identities for sine and cosine:
$$ x = \cos(\theta) $$
$$ y = \sin(\theta) $$
For example, if $ \theta = \frac{\pi}{4} $:
$$ x = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} $$
$$ y = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} $$
So the coordinates are $ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $.
Find the exact value of the inverse trig function expressions
Consider the expression $$ \sin^{-1}\left( \frac{\sqrt{3}}{2} \right) $$.
We know that $$ \sin\left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} $$.
Therefore, $$ \sin^{-1}\left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{3} $$.
Next, consider the expression $$ \tan^{-1}(1) $$.
We know that $$ \tan\left( \frac{\pi}{4} \right) = 1 $$.
Therefore, $$ \tan^{-1}(1) = \frac{\pi}{4} $$.
Finally, consider the expression $$ \cos^{-1}\left( -\frac{1}{2} \right) $$.
We know that $$ \cos\left( \pi – \frac{\pi}{3} \right) = -\frac{1}{2} $$.
Therefore, $$ \cos^{-1}\left( -\frac{1}{2} \right) = \frac{2\pi}{3} $$.
In summary:
$$ \sin^{-1}\left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{3} $$
$$ \tan^{-1}(1) = \frac{\pi}{4} $$
$$ \cos^{-1}\left( -\frac{1}{2} \right) = \frac{2\pi}{3} $$
Find the points where the ellipse intersects the empty unit circle
To find the points where the ellipse intersects the empty unit circle, we start with the equations of the ellipse and the empty unit circle:
Ellipse: $$\x0crac{x^2}{a^2} + \x0crac{y^2}{b^2} = 1$$
Empty unit circle: $$x^2 + y^2 = 1$$
We solve these equations simultaneously. First, we solve the ellipse equation for $$x^2$$:
$$x^2 = a^2(1 – \x0crac{y^2}{b^2})$$
Substitute this into the unit circle equation:
$$a^2(1 – \x0crac{y^2}{b^2}) + y^2 = 1$$
Simplify and solve for $$y^2$$:
$$a^2 – \x0crac{a^2y^2}{b^2} + y^2 = 1$$
$$\x0crac{y^2(a^2 – b^2)}{b^2} = 1 – a^2$$
$$y^2 = \x0crac{b^2(1 – a^2)}{a^2 – b^2}$$
We then find the corresponding $$x$$ values using the unit circle equation:
$$x^2 = 1 – y^2$$
With $$y^2 = \x0crac{b^2(1 – a^2)}{a^2 – b^2}$$, we find:
$$x^2 = 1 – \x0crac{b^2(1 – a^2)}{a^2 – b^2}$$
Thus, the points of intersection are solutions to these values of $$x$$ and $$y$$.
Determine the points of intersection between the unit circle and the curve y = x^3 – x
To find the points of intersection between the unit circle $x^2 + y^2 = 1$ and the curve $y = x^3 – x$, we substitute $y$ from the second equation into the first equation:
$$x^2 + (x^3 – x)^2 = 1$$
Expanding and simplifying, we get:
$$x^2 + (x^6 – 2x^4 + x^2) = 1$$
Combining like terms, we have:
$$x^6 – 2x^4 + 2x^2 – 1 = 0$$
This is a polynomial equation of degree 6, which we need to solve for $x$. Let us solve this numerically as it is not straightforward to solve algebraically. Assuming we find the solutions $x_1, x_2, x_3, …, x_6$, the corresponding $y$ values can be found by plugging each $x_i$ into $y = x_i^3 – x_i$. Therefore, the points of intersection are: $(x_1, x_1^3 – x_1), (x_2, x_2^3 – x_2), …, (x_6, x_6^3 – x_6)$.
Find the coordinates of the point on the unit circle at angle θ = π/4
The coordinates of the point on the unit circle at angle $ \theta = \frac{\pi}{4} $ can be found using the sine and cosine functions:
The x-coordinate is:
$$ x = \cos\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} $$
The y-coordinate is:
$$ y = \sin\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} $$
Therefore, the coordinates are:
$$ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $$
How to remember the unit circle using trigonometric identities
To remember the unit circle, you can leverage trigonometric identities and properties:
1. Know the key angles and their corresponding coordinates: $0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}$, etc.
2. Understand that for any angle $\theta$, the coordinates on the unit circle are $(\cos\theta, \sin\theta)$.
3. Remember the symmetry properties: $\cos(-\theta) = \cos(\theta)$ and $\sin(-\theta) = -\sin(\theta)$.
4. Utilize special triangles (like $30^\circ-60^\circ-90^\circ$ and $45^\circ-45^\circ-90^\circ$) to derive coordinates.
With these strategies, you can reconstruct the unit circle efficiently.
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