Math

Find the tangent values at 0, π/4, and π/2 on the unit circle

To find the tangent values at specific points on the unit circle, we use the definition of the tangent function, which is $$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $$.

1. At $ \theta = 0 $:

$$ \tan(0) = \frac{\sin(0)}{\cos(0)} = \frac{0}{1} = 0 $$

2. At $ \theta = \frac{\pi}{4} $:

$$ \tan\left(\frac{\pi}{4}\right) = \frac{\sin\left(\frac{\pi}{4}\right)}{\cos\left(\frac{\pi}{4}\right)} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 $$

3. At $ \theta = \frac{\pi}{2} $:

$$ \tan\left(\frac{\pi}{2}\right) = \frac{\sin\left(\frac{\pi}{2}\right)}{\cos\left(\frac{\pi}{2}\right)} = \frac{1}{0} $$

Since division by zero is undefined, $ \tan\left(\frac{\pi}{2}\right) $ does not exist.

Find the coordinates of the point on the unit circle where the angle is pi over 4 radians

To find the coordinates of the point on the unit circle where the angle is $ \frac{\pi}{4} $ radians, we use the unit circle definition:

The coordinates at this angle are:

$$ \left( \cos \frac{\pi}{4} , \sin \frac{\pi}{4} \right) $$

Using known values:

$$ \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} $$

$$ \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} $$

So, the coordinates are:

$$ \left( \frac{\sqrt{2}}{2} , \frac{\sqrt{2}}{2} \right) $$

Identify the coordinates on the unit circle for angle θ

The coordinates of a point on the unit circle for a given angle $ \theta $ are given by:

$$ ( \cos(\theta), \sin(\theta) ) $$

For example, if $ \theta = \frac{\pi}{4} $:

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

Find the values of tan(θ) for θ in the interval [0, 2π] that satisfy the equation tan(θ) = 2

To find the values of $ \tan(\theta) $ that satisfy the equation $ \tan(\theta) = 2 $ in the interval $ [0, 2\pi] $, we need to determine the angles where the tangent function equals 2.

First, recall that the tangent function is periodic with period $ \pi $, and the angles where $ \tan(\theta) = 2 $ are:

$$ \theta_1 = \arctan(2) $$

and

$$ \theta_2 = \arctan(2) + \pi $$

Because the tangent function repeats every $ \pi $ radians, we only need to check within one period:

$$ \theta_1 = \arctan(2) $$

$$ \theta_2 = \arctan(2) + \pi $$

Thus, the solutions within $ [0, 2\pi] $ are:

$$ \theta = \arctan(2) $$

and

$$ \theta = \arctan(2) + \pi $$

Solve the equation to find all distinct points where the ellipse intersects the unit circle

To find where the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ intersects the unit circle $x^2 + y^2 = 1$, we need to solve the system of equations:

$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$

$$ x^2 + y^2 = 1 $$

First, substitute $y^2 = 1 – x^2$ into the ellipse equation:

$$ \frac{x^2}{a^2} + \frac{1 – x^2}{b^2} = 1 $$

Multiply through by $a^2b^2$ to clear the denominators:

$$ b^2x^2 + a^2(1 – x^2) = a^2b^2 $$

Simplify and solve for $x^2$:

$$ b^2x^2 + a^2 – a^2x^2 = a^2b^2 $$

$$ (b^2 – a^2)x^2 = a^2(b^2 – 1) $$

$$ x^2 = \frac{a^2(b^2 – 1)}{b^2 – a^2} $$

Then solve for $y^2$ using $y^2 = 1 – x^2$.

Determine the sine and cosine values of 5π/6

To determine the sine and cosine values of $ \frac{5\pi}{6} $, we refer to the unit circle.

The angle $ \frac{5\pi}{6} $ is located in the second quadrant.

In the second quadrant, sine is positive, and cosine is negative.

The reference angle for $ \frac{5\pi}{6} $ is $ \frac{\pi}{6} $.