On the unit circle, $\tan(\theta)$ represents the slope of the terminal ray. It's the ratio of the y-coordinate (sine) to the x-coordinate (cosine).

Tangent has vertical asymptotes where $\cos(\theta) = 0$, because $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$. Division by zero is undefined.

The slope of the terminal ray on the unit circle repeats every $\pi$ radians (half-rotation), unlike sine and cosine which repeat every $2\pi$ radians.

Quadrant I: Positive (both sine and cosine are positive). Quadrant II: Negative (sine is positive, cosine is negative). Quadrant III: Positive (both sine and cosine are negative). Quadrant IV: Negative (sine is negative, cosine is positive).

As $x$ approaches an asymptote, $\tan(x)$ approaches either positive infinity or negative infinity. The function becomes infinitely steep.

A negative 'a' value reflects the graph of $y = \tan(x)$ over the x-axis.

Changing 'b' in $y = \tan(bx)$ alters the period of the tangent function. A larger 'b' compresses the graph horizontally, decreasing the period, while a smaller 'b' stretches the graph horizontally, increasing the period.

The range of the standard tangent function, $y = \tan(x)$, is all real numbers, or $(-\infty, \infty)$.

Phase shift is the horizontal translation of a tangent function. In the equation $y = a \tan(b(x - c)) + d$, 'c' represents the phase shift. A positive 'c' shifts the graph to the right, while a negative 'c' shifts it to the left.

The 'd' value represents the vertical shift of the tangent function. A positive 'd' shifts the graph upward, while a negative 'd' shifts it downward.

1. Rewrite as $y = 3\tan(2(x + \frac{\pi}{2})) - 1$. 2. Identify $b = 2$. 3. Use the formula $T = \frac{\pi}{|b|} = \frac{\pi}{2}$.

1. Identify 'c' in the general equation $y = a \tan(b(x - c)) + d$. 2. In this case, $c = \frac{\pi}{4}$. 3. The phase shift is $\frac{\pi}{4}$ to the right.

1. Set $2x = \frac{\pi}{2} + k\pi$. 2. Solve for $x$: $x = \frac{\pi}{4} + \frac{k\pi}{2}$. 3. Find values of k that place x in $[0, \pi]$. For k=0, $x = \frac{\pi}{4}$. For k=1, $x = \frac{3\pi}{4}$.

1. Start with the graph of $y = \tan(x)$. 2. Reflect the graph over the x-axis. This means every y-value becomes its opposite.

1. Identify 'd' in the general equation $y = a \tan(b(x - c)) + d$. 2. In this case, $d = 2$. 3. The vertical shift is 2 units upward.

1. Identify that $b = \frac{1}{2}$, which means the period will be $\frac{\pi}{1/2} = 2\pi$. 2. Stretch the graph horizontally by a factor of 2. The asymptotes will now be at $x = \pi + k2\pi$.

1. Use the formula $T = \frac{\pi}{|b|}$. 2. Set $\frac{\pi}{3} = \frac{\pi}{|b|}$. 3. Solve for $b$: $b = 3$. 4. The equation is $y = \tan(3x)$ (assuming no phase or vertical shift).

1. Identify where $\cos(x) = 0$. 2. These are the points where the tangent function is undefined. 3. The domain is all real numbers except $x = \frac{\pi}{2} + k\pi$, where k is an integer.

1. The range of the tangent function is always all real numbers, unless there are restrictions given in the problem. 2. Therefore the range is $(-\infty, \infty)$.

1. Recognize that $\tan(\frac{\pi}{4}) = 1$. 2. Since the period of tangent is $\pi$, the general solution is $x = \frac{\pi}{4} + k\pi$, where k is an integer.

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

$y = a \tan(b(x - c)) + d$

$T = \frac{\pi}{|b|}$

$x = \frac{\pi}{2} + k\pi$, where $k$ is an integer.

'a' controls the vertical dilation. If 'a' is negative, the function is reflected over the x-axis.

'b' affects the period of the function. The period is $T = \frac{\pi}{b}$.

'c' shifts the graph horizontally. Positive 'c' shifts right, negative 'c' shifts left.

'd' shifts the graph vertically. Positive 'd' shifts up, negative 'd' shifts down.

$x = cos(\theta)$

$y = sin(\theta)$