Graph of $r=3$

This is one of the simplest graphs to show how

]]>Graph of $r=3$

This is one of the simplest graphs to show how a polar equation works on a polar plane.

To create a line through the pole you can do this with the following function.

Graph of $y=x$ line using polar graph: $r=\beta$ where $\beta$ is just some constant.

To create a vertical line you can use the $cos(\theta)$ function as a fraction ratio.

Vertical Line: $r=\frac{2}{cos(\theta)} $

Horizontal Line: $r=\frac{2}{sin(\theta)} $

The next graph that's common is the trigonometric functions that you are used to working with. The following are the `cos(x)`

and `sin(x)`

.

Cosine Graph: $r=cos(\theta)$

Sine Graph: $r=sin(\theta)$

]]>`x`

and `y`

plane.
Before you start getting your graphing

]]>`x`

and `y`

plane.
Before you start getting your graphing paper out you are going to have to understand a couple of ideas first. Do you know the transformations that are happening to the base function? How are those changes affecting the given graph?

Here is a quick review of the basic transformations that a graph can go through.

1.How to find the amplitude of a trigonometric function?

2.How to find the vertical shift of a trigonometric function?

3.How do you find the frequency of a trigonometric function?

4.How do you find the period of a trigonometric function?

5.How do you find the phase shift of a trigonometric function?

If you break up the general form into smaller parts you will be left with the following review of graphing a function.

You can now use this understanding of the graph and some key points to graph any of the trigonometric functions.

]]>`x-plane`

of the graph.
$$Asin[B(x-\frac{C}{B})]+D$$

From the example above the phase shift of the graph would be.

$$\frac{C}{B}$$

Let's do a short

]]>`x-plane`

of the graph.
$$Asin[B(x-\frac{C}{B})]+D$$

From the example above the phase shift of the graph would be.

$$\frac{C}{B}$$

Let's do a short example of how the phase shifts would happen to a basic `sin(x)`

function.

$$sin(x)$$

If you take the function and shift it to the left $\frac{\pi}{2}$ length.

$$sin(x+\frac{\pi}{2})$$

You can also go the other way and shift it to the right $\frac{\pi}{2}$ length.

$$sin(x-\frac{\pi}{2})$$

Let's take a look at all three transformations.

]]>Period vs Frequency: The difference between the period and the frequency of a trigonometric function is that the frequency is how fast a period is happening

]]>Period vs Frequency: The difference between the period and the frequency of a trigonometric function is that the frequency is how fast a period is happening and a period defines how often a period happens.

$$f(s+p)=f(s)$$

For the smallest value of s, is what defines a set period of the function given.

Let's take a look at a simple `sin(x)`

graph between $-\pi$ and $\pi$.

You can take the same graph from a different perspective let's say from $0$ to $2\pi$.

As you can see it's the same graph. You can now define that this graph requires a full $2\pi$ length to complete a full period.

$$0\ to\ \frac { 2\pi}{ b } $$ where `b`

is always the absolute value.

Let's stretch the wave with a smaller frequency. For example `1/2`

.

$$sin(\frac{1}{2}x)$$

The graph would look like this.

If you work through the algebra.

$$y=Asin(Bx-C)+D$$

You will end up with a period of $4\pi$ and a frequency of $\frac {1}{2}$

Just to compare with a basic sin wave.

The red and black line defines a complete period for the graph of $sin(\frac{1}{2}x)$ and the purple and blue line define the period for just the standard $sin(x)$ function.

]]>This means for the base function to complete a full revolution around

]]>This means for the base function to complete a full revolution around the circle it's going to take $2\pi$ length. Below you will see the base function of $sin(x)$ with the length of $2\pi$.

$$sin(x)$$

The purple and black lines are $\pi$ length away from the origin and if you add up the values of both distances from the origin you will end up with $2\pi$ length.

If you continue with compressing the function based on the base function by a factor of 2 you will go twice as fast around the circle causing a compression in the $2\pi$ length.

$$sin(2x)$$

Same would apply for the stretching of the function. If you want to stretch the function you will have to multiply by a value that is less than one. In the example below, you can see that the function is getting delayed by half causing the function to be stretched in the $2\pi$ length.

$$sin(\frac{1}{2}x)$$

Now to wrap it all up let's see all three functions put together in the $2\pi$ length from the origin.

If you look at the graph it should make perfect sense.

]]>Just for fun and giggles keep this image in mind when you are thinking of frequency.

$$sin(x)$$

As you can see no shifts are happening to this function but

]]>$$sin(x)$$

As you can see no shifts are happening to this function but if we add to the function a value of +3 the function is going to get shifted 3 values up in the `y-axis`

.

$$sin(x)+3$$

The same concept applies if you use the same function and you shift it down -3 values in the `y-axis`

.

The values that you add or remove from the base function are not co-dependent in any way. If you want to shift the function in the vertical direction you just have to add or remove from the base function.

]]>`y-axis`

. Take for example the following function.
$$sin(x)$$

On this function, no compression or stretching on the `y-axis`

is happening but if you add an amplitude of 3

`y-axis`

. Take for example the following function.
$$sin(x)$$

On this function, no compression or stretching on the `y-axis`

is happening but if you add an amplitude of 3 the amplitude is going to stretch the function values up to the 3 mark on the `y-axis`

.

$$3sin(x)$$

The same concept applies to compressing the function for a value that is smaller than one.

$$\frac { 1 }{ 2 }sin(x)$$

If you now overlay all three functions together you will see that the function still have the same intercepts on the `x-axis`

just the values on the `y-axis`

have changed everything else has remained the same.

Remember the slope of the line is by how much a line is being tilted.

As you can see we can pick any

]]>Remember the slope of the line is by how much a line is being tilted.

As you can see we can pick any two points on any given line and find out what the slope of the line is at any given point using this concept.

All you need to know is two points on a given line to use this formula to figure out what the slope of any line is.

]]>

$Degrees$ | $Radians$ | $sin(x)$ | $cos(x)$ | $tan(x)$ |
---|---|---|---|---|

$0°$ | $0$ | $0$ | $1$ | $0$ |

$30°$ | $\frac { \pi }{ 6 }$ | $$\frac { 1 }{ 2 }$$ | $$\frac { \sqrt {3} }{ 2 }$$ | $$\frac { \sqrt {3} }{ 3 }$$ |

$45°$ | $\frac { \pi }{ 4 |

$Degrees$ | $Radians$ | $sin(x)$ | $cos(x)$ | $tan(x)$ |
---|---|---|---|---|

$0°$ | $0$ | $0$ | $1$ | $0$ |

$30°$ | $\frac { \pi }{ 6 }$ | $$\frac { 1 }{ 2 }$$ | $$\frac { \sqrt {3} }{ 2 }$$ | $$\frac { \sqrt {3} }{ 3 }$$ |

$45°$ | $\frac { \pi }{ 4 }$ | $$\frac { \sqrt {2} }{ 2 }$$ | $$\frac { \sqrt {2} }{ 2 }$$ | $$1$$ |

$60°$ | $\frac { \pi }{ 3 }$ | $$\frac { \sqrt {3} }{ 2 }$$ | $$\frac { 1 }{ 2 }$$ | $$1$$ |

$90°$ | $\frac { \pi }{ 2 }$ | $$1$$ | $$0$$ | $$undefined$$ |

$180°$ | $\pi$ | $$0$$ | $$-1$$ | $$0$$ |

$270°$ | $\frac { 3\pi }{ 2 }$ | $$-1$$ | $$0$$ | $$undefined$$ |

If you want to be able to memorize this entire table take a couple of minutes and review - How to memorize common trigonometric angles

]]>Instead of having your education based on if you could sing a song or not let's instead deduce this common angles based on what information you do

]]>Instead of having your education based on if you could sing a song or not let's instead deduce this common angles based on what information you do know. By the way, this is very much the same way that your memory stores away information.

All of this starts with a right angle that can be represented by the following triangle.

$${ a }^{ 2 }+{ b }^{ 2 }={ c }^{ 2 }$$

You can now find each side of the triangle just as long as you know two sides of the triangle based on this equation.

Do you remember the basic trig functions? If not here is a review of them sin(x) cos(x) tan(x).

If you are still having a hard time remembering each of this functions take a second and review: How to Remember Sin(x), Cos(x), Tan(x)?

Let's start with the most common angles.

**$0$ radians**. How would this triangle look like? This is a tricky question. If you think about it the triangle would end up being just a flat line across the $0$ mark on the plane.

I'm letting $a=1$ just to make the math easier for us. You will then need to find the $sin(0)$, $cos(0)$, and $tan(0)$.

$$sin(0)=0$$ $$cos(0)=1$$ $$tan(0)=0$$

*Here is where things get interesting*. If you let the angle be created from the left side of the line you would create a positive angle and you would be able to find $sin(0)$, $tan(0)$ but since no angle can be created at $0$ radians you are left with just $cos(0)$ which is just $1$.

**$\frac { \pi }{ 6 }$ radians**. How would this triangle look like? If you think about it the triangle would end up being just $\frac { \pi }{ 6 }$ radians away from the initial side.

Based on this information you can now find the rest of the parts of the triangle.

All of this is great! You now know all of the angles that compose this triangle. The problem becomes how do you find the $sin(\frac { \pi }{ 6 })$, $cos(\frac { \pi }{ 6 })$, and $tan(\frac { \pi }{ 6 })$ of this triangle?

Because you don't know any of the sides of this triangle and you can't assume any of the sides of the triangle let's start with a simpler isosceles triangle of the same length.

If you now take this very same triangle and cut it up in half you would end up with this triangle.

The only piece missing from this triangle is the rise of the triangle which you can find by using the Pythagorean theorem.

$$1^{ 2 }+{ b }^{ 2 }={ 2 }^{ 2 }$$ $$1+{ b }^{ 2 }=4$$ $${ b }^{ 2 }=4-1$$ $${ b }=\sqrt { 3 }$$

You now know all three pieces of the triangle you can complete the triangle with the following labels.

If you pay close attention to the triangle above you can now define the $sin$, $cos$, and $tan$ values for $\frac { \pi }{ 6 }$ and $\frac { \pi }{ 3 }$ radians.

Let's first find the values for $\frac { \pi }{ 6 }$.

$$sin(\frac { \pi }{ 6 })=\frac { 1 }{ 2 }$$ $$cos(\frac { \pi }{ 6 })=\frac { \sqrt { 3 } }{ 2 }$$ $$tan(\frac { \pi }{ 6 })=\frac { \sqrt { 3 } }{ 3 }$$

Let's find the values for $\frac { \pi }{ 3 }$.

$$sin(\frac { \pi }{ 3 })=\frac { \sqrt { 3 } }{ 2 }$$ $$cos(\frac { \pi }{ 3 })=\frac { 1 }{ 2 }$$ $$tan(\frac { \pi }{ 3 })=\sqrt { 3 }$$

The last two remaining angles to take a look at are $\frac { \pi }{ 2 }$ and $\frac { \pi }{ 4 }$ if I was to draw this triangle how would it look like?

Let's now find each of the sides in the triangle.

$${ a }^{ 2 }+{ b }^{ 2 }={ c }^{ 2 }$$ $${ 1 }^{ 2 }+{ 1 }^{ 2 }={ c }^{ 2 }$$ $$2={ c }^{ 2 }$$ $$c=\sqrt { 2 }$$

If you pay close attention to the triangle above you can now define the $sin$, $cos$, and $tan$ values for $\frac { \pi }{ 4 }$ and $\frac { \pi }{ 2 }$ radians.

Let's first find the values for $\frac { \pi }{ 4 }$.

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

Next up is $\frac { \pi }{ 2 }$

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

There you have it. You have completed 0, $\frac { \pi }{ 6 }$, $\frac { \pi }{ 4 }$, $\frac { \pi }{ 3 }$, and $\frac { \pi }{ 2 }$.

]]>$$right\ angle=90°$$ $$acute\ angle=x°<90°$$

]]>$$right\ angle=90°$$ $$acute\ angle=x°<90°$$

]]>**This problem is all sort's of FUN!** I was working on this problem trying to figure it out. Everything that I tried I kept always getting the wrong answer. Here is the big problem that I was facing.

I was trying to convert two revolutions into 1 revolution and cut the time in half. My thought here was since a circle is $2\pi$ is a complete revolution so should 1 revolution and 1 minute and 3.5 seconds. This is where my logic was WRONG. I cleaned this up by representing two revolutions in seconds instead of trying to convert them to a complete revolution around the circle, in other words, the circumference of the circle.

$$speed=\frac { 2 }{ 127 } (\frac { rev }{ sec } )$$

After doing this the math works out perfectly!

If you have a hard time reading the solution, it's probably a good idea to review how to find the solution

]]>You can also explain this relationship of the circumference of the circle with this.

$$c=

]]>You can also explain this relationship of the circumference of the circle with this.

$$c=2\pi r$$

This is going to apply to just about any other circle that you encounter. The $\pi$ value will always be a constant as the circle grows based on the radius.

]]>Imagine that you are out for a run, as you run around the circle you will create your linear speed as you go around

]]>Imagine that you are out for a run, as you run around the circle you will create your linear speed as you go around the circle.

You can define linear speed as such.

$$linear\ speed=\frac { arc }{ time } $$

To abstract this one more level you can use $s$ for the arc and $t$ for the given time to obtain the linear speed.

$$linear\ speed=\frac { s }{ t } $$

**REMEMBER**: The $s$ can be found by simply multiplying the radius by any given angle $\theta$. See How to find theta of any given angle?

$$s=r\theta$$

]]>To find the angle all you will need is to

]]>To find the angle all you will need is to divide the arc length by the radius.

$$\theta =\frac { s }{ r }$$

]]>