Unit Circle

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.

Find the sine and cosine of 45 degrees in radians

To find the sine and cosine of $ \frac{\pi}{4} $, we use the unit circle. Since $ \frac{\pi}{4} $ corresponds to 45 degrees:

$$ \sin\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} $$

$$ \cos\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} $$

Prove the identity of sin(θ) on the unit circle

To prove the identity of $ \sin(\theta) $ on the unit circle, we start by considering a point on the unit circle at angle $ \theta $. The coordinates of this point can be represented as $ (\cos(\theta), \sin(\theta)) $.

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Using the Pythagorean identity for the unit circle, we have:

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$$ \cos^2(\theta) + \sin^2(\theta) = 1 $$

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Now consider a right triangle with the hypotenuse being the radius of the unit circle (which is 1). The opposite side of angle $ \theta $ is $ \sin(\theta) $ and the adjacent side is $ \cos(\theta) $.

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By the definition of sine in a right triangle, we get:

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$$ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} $$

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Since the hypotenuse is 1, the opposite side is $ \sin(\theta) $, thus:

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$$ \sin(\theta) = \sin(\theta) $$

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This completes the proof.

How to calculate points on the unit circle for specific angles

To calculate points on the unit circle for a specific angle $ \theta $, follow these steps:

1. Recall that the unit circle is a circle with radius 1 centered at the origin (0,0).

2. Points on the unit circle are given by the coordinates $( \cos(\theta), \sin(\theta) )$, where $ \theta $ is the angle in radians measured from the positive x-axis.

3. For example, for $ \theta = \frac{ \pi }{4} $, the coordinates are:

$$ ( \cos( \frac{ \pi }{4} ), \sin( \frac{ \pi }{4} ) ) = ( \frac{ \sqrt{2} }{2}, \frac{ \sqrt{2} }{2} ) $$

Given that tan(θ) = 2 and θ is in the second quadrant, find the exact values of sin(θ) and cos(θ)

1. Given that $ \tan(\theta) = 2 $, we can write:

$$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = 2 $$

Let $ \sin(\theta) = 2k $ and $ \cos(\theta) = -k $ (since $ \theta $ is in the second quadrant where cosine is negative). Then:

$$ \frac{2k}{-k} = 2 $$

2. From the Pythagorean identity:

$$ \sin^2(\theta) + \cos^2(\theta) = 1 $$

Substitute the values:

$$ (2k)^2 + (-k)^2 = 1 $$

$$ 4k^2 + k^2 = 1 $$

3. Solving for $ k $:

$$ 5k^2 = 1 $$

$$ k^2 = \frac{1}{5} $$

$$ k = \pm \frac{1}{\sqrt{5}} $$

4. Since $ \sin(\theta) = 2k $ and $ \cos(\theta) = -k $, we have:

$$ \sin(\theta) = 2 \left( \frac{1}{\sqrt{5}} \right) = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} $$

$$ \cos(\theta) = – \left( \frac{1}{\sqrt{5}} \right) = -\frac{\sqrt{5}}{5} $$