129 lines
6.3 KiB
Markdown
129 lines
6.3 KiB
Markdown
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# Sine/Cosine
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Given the above graph:
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- At the origin, $sin(x) = 0$ and $cos(x) = 1$
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- A full wavelength takes $2\pi$
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# Manipulation
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| Formula | Movement |
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| ---------------- | ---------------------------------- |
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| $y = cos(x) - 1$ | Vertical shift down by 1 |
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| $y = 2cos(x)$ | Vertical stretch by a factor of 2 |
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| $y = -cos(x)$ | Flip over x axis |
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| $y = cos(2x)$ | Horizontal shrink by a factor of 2 |
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# Periodic Functions
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A function is considered periodic if it repeats itself at even intervals, where each interval is a complete cycle, referred to as a *period*.
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# Sinusoidal Functions
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A function that has the same shape as a sine or cosine wave is known as a sinusoidal function.
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There are 4 general functions:
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| $$A * sin(B*x - C) + D$$ | $$ y = A * cos(B*x -c) + D$$ |
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| ----------------------------------------- | -------------------------------------- |
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| $$ y = A * sin(B(x - \frac{C}{B})) + D $$ | $$ y = A*cos(B(x - \frac{C}{B})) + D$$ |
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How to find the:
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- Amplitude: $|A|$
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- Period: $\frac{2\pi}{B}$
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- Phase shift: $\frac{C}{|B|}$
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- Vertical shift: $D$
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$$ y = A * \sin(B(x-\frac{C}{B})) $$
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# Tangent
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$$ y = tan(x) $$
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To find relative points to create the above graph, you can use the unit circle:
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If $tan(x) = \frac{sin(x)}{cos(x})$, then:
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| $sin(0) = 0$ | $cos(0) = 1$ | $tan(0) = \frac{cos(0)}{sin(0)} = \frac{0}{1} =0$ |
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| ----------------------------------------- | ----------------------------------------- | ---------------------------------------------------------------- |
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| $sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ | $cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ | $tan(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}/\frac{\sqrt{2}}{2} = 1$ |
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| $sin(\frac{\pi}{2}) = 1$ | $cos(\frac{\pi}{2}) = 0$ | $tan(\frac{\pi}{2}) = \frac{1}{0} = DNF$ |
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Interpreting the above table:
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- When $x = 0$, $y = 0$
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- When $x = \frac{\pi}{4}$, $y = 1$
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- When $x = \frac{\pi}{2}$, there's an asymptote
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Without any transformations applied, the period of $tan(x) = \pi$. Because $tan$ is an odd function, $tan(-x) = -tan(x)$.
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# Cotangent
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$$ y = cot(x) $$
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To find relative points to create the above graph, you can use the unit circle:
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If $cot(x) = \frac{cos(x)}{sin(x)}$, then:
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| $sin(0) = 0$ | $cos(0) = 1$ | $cot(0) = \frac{sin(0)}{cos(0)} = \frac{1}{0} = DNF$ |
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| ----------------------------------------- | ----------------------------------------- | ---------------------------------------------------------------- |
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| $sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ | $cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ | $cot(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}/\frac{\sqrt{2}}{2} = 1$ |
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| $sin(\frac{\pi}{2}) = 1$ | $cos(\frac{\pi}{2}) = 0$ | $tan(\frac{\pi}{2}) = \frac{1}{0} = DNF$ |
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Without any transformations applied, the period of $cot(x) = \pi$. Because $cot$ is an odd function, $cot(-x) = -cot(x)$.
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# Features of Tangent and Cotangent
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Given the form $y = A\tan(Bx - C) + D$ (the same applies for $\cot$)
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- The stretching factor is $|A|$
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- The period is $\frac{\pi}{|B|}$
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- The domain of $tan$ is all of $x$, where $x \ne \frac{C}{B} + \frac{\pi}{2} + {\pi}{|B|}k$, where $k$ is an integer. (everywhere but the asymptotes)
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- The domain of $cot$ is all of $x$, where $x \ne \frac{C}{B} + \frac{\pi}{|B|}k$, where $k$ is an integer (everywhere but the asymptotes)
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- The range of both is $(-\infty, \infty)$
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- The phase shift is $\frac{C}{B}$
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- The vertical shift is $D$
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# Secant
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$$ y = \sec(x) $$
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$$ sec(x) = \frac{1}{\cos{x}} $$
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Because secant is the reciprocal of cosine, when $\cos{x} = 0$, then secant is undefined. $|\cos$| is never *greater than* 1, so secant is never *less than* 1 in absolute value. When the graph of cosine crosses the x axis, an asymptote for a matching graph of secant will appear there.
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The general form of secant is:
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$$ y = A\sec(B{x} - C) + D $$
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$A$, $B$, $C$, and $D$ will have similar meanings to the secant function as they did to the sine and cosine functions.
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# Cosecant
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$$ y = \csc(x) $$
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$$ \csc(x) = \frac{1}{\sin(x)} $$
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Because cosecant is the reciprocal of sine, when $\sin{x} = 0$, then cosecant is undefined. $|\sin$| is never *greater than* 1, so secant is never *less than* 1 in absolute value. When the graph of sine crosses the x axis, an asymptote for a matching graph of cosecant will appear there.
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The general form of cosecant is:
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$$ y = A\csc(B{x} - C) + D $$
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$A$, $B$, $C$, and $D$ will have similar meanings to the cosecant function as they did to the sine and cosine functions.
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# Features of Secant and Cosecant
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- The stretching factor is $|A|$
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- The period is $\frac{2\pi}{|B|}$
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- The domain of secant is all $x$, where $x \ne \frac{C}{B} + \frac{\pi}{2} + \frac{\pi}{|B|}k$, where $k$ is an integer. (Every half period + phase shift is where asymptotes appear)
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- The domain of cosecant is all $x$, where $x \ne \frac{C}{B} + \frac{\pi}{|B|}k$, where $k$ is an integer.
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- The range is $(\infty, -|A| +D]\cup [|A| + D], \infty)$
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- The vertical asymptotes of secant occur at $x = \frac{C}{B} + \frac{\pi}{2} + \frac{\pi}{2} + \frac{\pi}{|B|}k$, where $k$ is an integer.
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- The vertical asymptotes of cosecant occur at $x = \frac{C}{B} + \frac{\pi}{|B|}k$, where $k$ is an integer.
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- The vertical shift is $D$.
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# Examples
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> Given $-2\tan(\pi*x + \pi) - 1$
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$A = -2$, $B = \pi$, $C = -\pi$, $D = -1$
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> Identify the vertical stretch/compress factor, period, phase shift, and vertical shift of the function $y = 4\sec(\frac{\pi}{3}x - \frac{\pi}{2}) + 1$
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$A = 4$, $B = \frac{\pi}{3}$, $C = \frac{\pi}{2}$, $D = 4$
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Vertical stretch: $|4| = 4$
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Period: $\frac{2\pi}{\frac{\pi}{3}} = \frac{2\pi}{1} * \frac{3}{\pi} = 6$
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Phase shift: $\frac{}
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| Transformation | Equation |
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| -------------- | ------------------------- |
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| Stretch | $\|-2\| = 2$ |
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| Period | $\frac{\pi}{\|\pi\|} = 1$ |
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| Phase shift | $\frac{-\pi}{\pi} = -1$ |
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| Vertical shift | $-1$ |
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