Graphing Sine function

Sketch the graph y= 3sinΘ for 0 < x < 2π

In order to graph sine or cosine function, you need to determine:

·        -  Amplitude (a)

·         - Period (b)

General Form:
     y= asinb(Θ-c)+d

Determine the Amplitude of a sine Function
- Amplitude is the distance from the center of the axis to the highest or lowest point for a sinΘ or cosΘ function.

     Amplitude = Max- min
                                 2

     Amplitude= IaI

So

 Determine the period of sine function

- To find the period, you first need to determine b

Then plug it in the formula of a period which is    2π
                                                                                IbI

- the period of the sin function is 2π, which means the function repeats itself every 2π

You are now ready to graph a sine function!    

1.      Plot the period (make sure to have 5 coordinates)

2.      The amplitude will be the maximum and minimum y-coordinates

3.      If the equation is positive, after plotting zero the next point will be plot on the positive side. If the equation is negative, it will be on the negative side.

4.      then, the next point will be on the x-axis since the amplitude on the positive and negative side should equidistant, the graph should look like a wave

 

 

y= 3sinΘ

 

Solving Trigonometric Equations Part 1

        Hi Pre Calc. class! Last time you guys learned about the unit circle, and all of it's properties, special angles, etc. Well now you are ready to continue to the next step, which is solving Trigonometric Equations with a specified interval! How exciting right? I'll walk you through the the steps though, so don't be worried.

THINGS TO REMEMBER:
- Special triangles
-Quadrantal Angles
SOH CAH TOA
-CAST rule
- ΘR angles

Example 1
To solve for  Θ:
1) See if the interval is set in either radians or degrees. That will determine how you will write your final answer. 

2)Simplify the equation by isolating the unknown, which is the theta, and factor.

3)Find  ΘR 

and find the quadrants that the answer is in using the CAST rule.

4)Find values for  Θ in degrees or radians. 

Since the interval is in degrees, I will keep it my answer in degrees. 

You see how there are two answers? That's because there are two quadrants that my answer for  Θ could be in! 

* If there is a particular restriction, such as " Where sin  Θ> 0" in  a question such as " tan Θ= -1/√ 3 " tan would be possible in TWO quadrants, Q. 2 and 4. BUT because of restriction, sin cannot be in Q. 4, only Q. 2 therefore there will only be ONE answer.*

Here is the full question for you! 

So I hope that this gave you a little more understanding about solving trigonometric equations with a specific interval. Try to review your note as well, as they are very useful and important! 

Enjoy my fellow Pre calc mates!

Trigonometric Ratios

Hello class. I am going to be teaching you all about trigonometric ratios.. Just kidding Mr.P already taught us. I'm just going to be recapping what we may or may not already know. Bad joke? Yeah I know, anyways lets get started.

The first thing you'll need to remember are these: 

The 6 Trigonometric Ratios

Also SOH CAH TOA will come very handy in this unit. 

So for example, lets say the given points are A(-3/5, -4/5), we need to find 6 trigonometric ratios for theta(θ)

So first off we're going to need a diagram, 

Then you'll need to look at the given points again which is (-3/5,-4/5) and figure out which quadrant point A would go when using the CAST rule. You also need to remember that cos is x and sin is y. 

So since cos and sin are both negative, the point would go in quadrant 3. 

Cos is adjacent/hypotenuse which makes -3 the adjacent and sin is opposite/hypotenuse which makes -4 is opposite and the hypotenuse is 5

If you understand the CAST rule then you are able to plug the points in sin, cos and tan plus their reciprocals. 
So here are the following answers:

This is just one of many examples that are almost exactly the same as this. So now that you know this, the world is in your hands... haha just kidding but in all seriousness, you will be able to figure out any trigonometric ratio question that comes your way.

THE UNIT CIRCLE

Hey everyone hope you had a fun Halloween and enjoyed your weekend! But remember to keep up with your homework and don't forget about Pre-Calculus.

THE UNIT CIRCLE

The "Unit Circle" is a circle with a radius of 1.

Using the unit circle makes it easy to find the values of trigonometric functions at quadrantal angles.                                                                                                                    

 For example, a 90º rotation from the positive x-axis puts you on the positive y-axis, which intersects the unit circle at the point (0, 1). From this, you know that (cos 90º, sin 90º) = (0, 1).

This is a graph of the values of all three trigonometric functions at each quadrantal angle:

Important Angles 30º , 45º , 60º

Most angles on the unit circle are based on the reference angles of either 30º , 45º , 60º

You should remember the special triangles from grade 11 , If not here is a labeled diagrams to help you.

Finding the missing side of a right triangles

Finding the missing side of a right triangle is a pretty simple matter if two sides are known. One of the more famous mathematical formulas is a2+b2=c2, which is known as the Pythagorean Theorem. The theorem states that the hypotenuse of a right triangle can be easily calculated from the lengths of the sides. The hypotenuse is the longest side of a right triangle.

If you're given the lengths of the two sides it is easy to find the hypotenuse. Just square the sides, add them, and then take the square root. Here's an example:

Since we are given that the two legs of the triangle are 3 and 4, plug those into the Pythagorean equation and solve for the hypotenuse:

a2+b2=c2

32+42=c2

9+16=c2

c=5

If you are given the hypotenuse and one of the legs it's going to be slightly more complicated, but only because you have to do some algebra first. Suppose you know that one leg is 5 and the hypotenuse (longest side) is 13. Plug those into the appropriate places in the Pythagorean equation:

a2+b2=c2

52+b2=132

25+b2=169

b2=144

b=12

Thank you for reading my blog post, I hope this post helped you review/ learn about the unit circle and the special triangles involved. 

-Ravinder Sehira 

sources: 

Pre-calculus 12 McGraw-Hill Ryerson- Text Book

Pre-Calculus 40s Unit 4 Circular Functions- Notes

sites:

http://www.freemathhelp.com/q6-right-triangle.html

http://www.mathsisfun.com/geometry/unit-circle.html

http://www.wyzant.com/resources/lessons/math/trigonometry/unit-circle