SP#7: Unit Q Concept 3: Trigonometry substitution

Please see my SP7, made in collaboration with Kenia G., by visiting their blog here. Also be sure to check out the other awesome posts on their blog.

I/D# 3: Unit Q- Phytagorean Identities

INQUIRY ACTIVITY SUMMARY:

1. An identity is a proven fact that is always true. The Pythagorean Theorem is an identity because it can be proven that a squared plus b squared will equal c squared. In terms of x, y, and r as used in the Unit circle, the Pythagorean theorem is x^2+y^2=r^2. If we try to get the Pythagorean Theorem in terms of x,y, and r equal one, we must divide by r^2. The equation that results is (x/r)^2+(y/r)^2=1. The ratio for cosine on the unit circle is x/y and the ratio for sine on the Unit circle is y/r. If (y/r)^2 is replaced by sine is the equation it will be sin^2 and (x/r)^2 is replaced by cosine it will be cos^2. The equation will end up being sin^2x+cos^2=1. 

2. To derive the rest of the remaining identities we must divide different things to get them. We also must use our ratio identities and reciprocal identities to find them.

 INQUIRY ACTIVITY REFLECTION:
1.The connections that I see between Units N, O, P, Q are that they all use the Unit Circle as a basis to derive different equations. They also use the Magic Tree to  derive certain equations.
 

WPP# 13-14: Unit P Concept 6-7

                                      

                            Please see my WPP13-14, made in collaboration with Kenia G. Please visit the other awesome post on their blog by going here.
                                                           The Problem:

a) Danielle is due east from an oak tree. Danielle is looking at Andrew at a bearing of S32W. Andrew is looking at the same oak tree with a bearing of N15E. Andrew is also 52 feet away from the oak tree. How many feet apart are Andrew and Danielle?

b) Andrew and Danielle are now together and they go on a date. Sadly the time for them to be apart has come. They leave the same point. Their paths diverge at a bearing of 088 degrees. If Danielle walks 2.7 miles and Andrew walks 3 miles,how far apart are they at this time?


                                                                            The Solution:

BQ#1: Unit P Concept 1-5- Law of Sines AAS or ASA, Area of an oblique triangle.

1. Law Of Sines-

     The law of sines is needed when you are not working with a non-right triangles. This helps us with  the trig functions that are used for solving non-right triangles.
How is derived:
1.)

To create a right triangle we must drop a perpendicular line from angle B and it can be labeled as h.  Two right triangles are now created. To use the law of sines, two sides and two angles must be present.

2)

To find the missing parts of a triangle the following relationships are needed.  To find h you can use the transitive property and have cSineA=aSineC, since they both equal h.

3)

If you divide both side by ac then SinA/a= SinC/ c. This is one of the relationships used in the law of sines. The ratios can be used to find any parts of a non-right triangle.

4. Areas formulas: 

The area of an oblique triangle cannot be found without the value of h. The area of a right triangle is A= 1/2bh, where b is the base and h is the height of the triangle.

In this triangle sinC=h/a and we know that h=asinc. We substitute h for asinc in the are formula which is A=1/2bh. The new formula is A=1/2b(asinc) The area of an oblique triangle is one half of the product of two sides and the sine of the given angle. The angle that we are trying to find must be in between the two sides that were given. Two sides and a angle must always be present. The angle will always be sine of the given angle.

WPP#12: Unit O Concept 10

Create your own Playlist on LessonPaths!

1/D: #2 Unit O Concept 7-8: Derive the SRTs

Inquiry Summary Activity:

 To derive the Special Right Triangles, I started with a square with the side lengths of one and a equilateral triangle with the side lengths of 1. Theses two shapes help us derive the 45-45-90 Special Triangle and the 30-60-90 Special Triangle.The Pythagorean Theorem was also used to find missing parts of the triangles.

1) 30-60-90 Triangle:
To derive the 30-60-90 triangle,  I used a equilateral triangle with the side lengths of 1. An equilateral triangle has angles that are 60 degrees and all 3 angles are the same. Since a equilateral triangle has the same angles and the same sides, I sliced it down the middle. By slicing it down the middle a height and a 90 degree angle were created. A 30 degree angle was also created by splitting the 60 degree angle in half. Since the triangle was sliced now the base is 1/2. Since we know that on side is 1 and the base is now 1/2, using the Pythagorean Theorem we can figure what the height is. We use the a^2+b^2=c^2. a=One squared is a=1 and b=1/2 squared is b= 1/4. You add them and that gives radical 3 over 2.We multiply everything by 2 to get radical 3, 2 and 1. This translates into the normal n radical3, 2n, and n. The n is used as variable meaning that any number can be substituted in. N is used to expand the problem as needed.

2) 45-45-90 Triangle:
To derive the 45-45-90 triangle, I used a square with the sides of 1. The 4 angles of a square are 90 degrees.I sliced the square down its diagonal. By cutting it down its diagonal created  a hypotenuse and two 45 degree angles. Since we know that two sides are 1 we must now find the hypotenuse of the triangle created. I used the Pythagorean Theorem to find the hypotenuse. The equation is a^2+b^2=c^2. A=1 and B=1, which means c stays the same. 1 squared is 1 so that means a+b equals one. To get c we must get rid of the square root by squaring c and 2. That means c equals radical 2. The sides of the triangle are 1,1, and radical 2. That translates to the original pattern of a 45-45-90 triangle which is: n,n, n radical 2. The N is used a variable, which means any number can be substituted and that means that N is also used to expand the problem as needed.

Inquiry Activity Reflection:
1) Something I never noticed before about special right triangles is that they were created form other shapes.
2)Being able to derive these patterns myself aids my learning because if I need to do a problem that involves this and the triangles are not given I can do it on my own.