# Sine/Cosine ![A graph of sine and cosine](./assets/graphsincos.png) Given the above graph: - At the origin, $sin(x) = 0$ and $cos(x) = 1$ - A full wavelength takes $2\pi$ # Manipulation | Formula | Movement | | ---------------- | ---------------------------------- | | $y = cos(x) - 1$ | Vertical shift down by 1 | | $y = 2cos(x)$ | Vertical stretch by a factor of 2 | | $y = -cos(x)$ | Flip over x axis | | $y = cos(2x)$ | Horizontal shrink by a factor of 2 | # Periodic Functions A function is considered periodic if it repeats itself at even intervals, where each interval is a complete cycle, referred to as a *period*. # Sinusoidal Functions A function that has the same shape as a sine or cosine wave is known as a sinusoidal function. There are 4 general functions: | $$A * sin(B*x - C) + D$$ | $$ y = A * cos(B*x -c) + D$$ | | ----------------------------------------- | -------------------------------------- | | $$ y = A * sin(B(x - \frac{C}{B})) + D $$ | $$ y = A*cos(B(x - \frac{C}{B})) + D$$ | How to find the: - Amplitude: $|A|$ - Period: $\frac{2\pi}{B}$ - Phase shift: $\frac{C}{|B|}$ - Vertical shift: $D$ $$ y = A * \sin(B(x-\frac{C}{B})) $$ # Tangent $$ y = tan(x) $$ ![Graph of tangent](assets/graphtan.png) To find relative points to create the above graph, you can use the unit circle: If $tan(x) = \frac{sin(x)}{cos(x})$, then: | $sin(0) = 0$ | $cos(0) = 1$ | $tan(0) = \frac{cos(0)}{sin(0)} = \frac{0}{1} =0$ | | ----------------------------------------- | ----------------------------------------- | ---------------------------------------------------------------- | | $sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ | $cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ | $tan(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}/\frac{\sqrt{2}}{2} = 1$ | | $sin(\frac{\pi}{2}) = 1$ | $cos(\frac{\pi}{2}) = 0$ | $tan(\frac{\pi}{2}) = \frac{1}{0} = DNF$ | Interpreting the above table: - When $x = 0$, $y = 0$ - When $x = \frac{\pi}{4}$, $y = 1$ - When $x = \frac{\pi}{2}$, there's an asymptote Without any transformations applied, the period of $tan(x) = \pi$. Because $tan$ is an odd function, $tan(-x) = -tan(x)$. # Cotangent $$ y = cot(x) $$ ![Graph of cotangent](assets/graphcot.svg) To find relative points to create the above graph, you can use the unit circle: If $cot(x) = \frac{cos(x)}{sin(x)}$, then: | $sin(0) = 0$ | $cos(0) = 1$ | $cot(0) = \frac{sin(0)}{cos(0)} = \frac{1}{0} = DNF$ | | ----------------------------------------- | ----------------------------------------- | ---------------------------------------------------------------- | | $sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ | $cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ | $cot(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}/\frac{\sqrt{2}}{2} = 1$ | | $sin(\frac{\pi}{2}) = 1$ | $cos(\frac{\pi}{2}) = 0$ | $tan(\frac{\pi}{2}) = \frac{1}{0} = DNF$ | Without any transformations applied, the period of $cot(x) = \pi$. Because $cot$ is an odd function, $cot(-x) = -cot(x)$. # Features of Tangent and Cotangent Given the form $y = A\tan(Bx - C) + D$ (the same applies for $\cot$) - The stretching factor is $|A|$ - The period is $\frac{\pi}{|B|}$ - 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) - 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) - The range of both is $(-\infty, \infty)$ - The phase shift is $\frac{C}{B}$ - The vertical shift is $D$ # Secant $$ y = \sec(x) $$ ![Graph of secant](assets/graphsec.jpg) $$ sec(x) = \frac{1}{\cos{x}} $$ 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. The general form of secant is: $$ y = A\sec(B{x} - C) + D $$ $A$, $B$, $C$, and $D$ will have similar meanings to the secant function as they did to the sine and cosine functions. # Cosecant $$ y = \csc(x) $$ ![Graph of cosecant](assets/graphsec.jpg) $$ \csc(x) = \frac{1}{\sin(x)} $$ 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. The general form of cosecant is: $$ y = A\csc(B{x} - C) + D $$ $A$, $B$, $C$, and $D$ will have similar meanings to the cosecant function as they did to the sine and cosine functions. # Features of Secant and Cosecant - The stretching factor is $|A|$ - The period is $\frac{2\pi}{|B|}$ - 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) - The domain of cosecant is all $x$, where $x \ne \frac{C}{B} + \frac{\pi}{|B|}k$, where $k$ is an integer. - The range is $(\infty, -|A| +D]\cup [|A| + D], \infty)$ - 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. - The vertical asymptotes of cosecant occur at $x = \frac{C}{B} + \frac{\pi}{|B|}k$, where $k$ is an integer. - The vertical shift is $D$. # Examples > Given $-2\tan(\pi*x + \pi) - 1$ $A = -2$, $B = \pi$, $C = -\pi$, $D = -1$ > Identify the vertical stretch/compress factor, period, phase shift, and vertical shift of the function $y = 4\sec(\frac{\pi}{3}x - \frac{\pi}{2}) + 1$ $A = 4$, $B = \frac{\pi}{3}$, $C = \frac{\pi}{2}$, $D = 4$ Vertical stretch: $|4| = 4$ Period: $\frac{2\pi}{\frac{\pi}} | Transformation | Equation | | -------------- | ------------------------- | | Stretch | $\|-2\| = 2$ | | Period | $\frac{\pi}{\|\pi\|} = 1$ | | Phase shift | $\frac{-\pi}{\pi} = -1$ | | Vertical shift | $-1$ |