I/D#3- Unit Q- Pythagorean Identities

INQUIRY ACTIVITY SUMMARY:
1) sin^2x+cos^2x=1 is a Pythagorean Identity which means that it can be a proven fact and formula that is always true. So to start deriving it we use the Pythagorean Theorem. Pythagorean Theorem is always a^2+b^2=c^2 but we have learned different when dealing with unit circle. We know that the leg are the x-axis and the one going up and down is the y-axis and we know that the hypotenuse is r when it comes to the unit circle. So know we just switch it really instead of having a^2+b^2=c^2 we can a new one that is x^2+y^2=r^2 and we want to get to sin^2x+cos^2x=1. To do that we divide everything by r^2 and we should get x^2/r^2+y^2/r^2=1 it then turns to (x/r)^2+(y/r)^2=1. Right here we should notice something about the variables and the ratios. We know that cosine is x/r and we know that sin is y/r so, then we plug it into our formula and (y/r)^2 is sin^2x and (x/r)^2 is cos^2x so know we put it together and we should get that identity which should be sin^2x+cos^2x=1

I choose the 60º from the "Magic 3" in the Unit Circle to show that this identity is true. The picture is below to see that it can be proven.

2) OTHER PYTHAGOREAN IDENTITIES

    The first pythagorean identity contains Secant and Tangent. The only way we could get that is by dividing by either sine or cosine. In this case we divided by cosine so that we can get what we want which is tangent and secant. So we divide by cos^2x and for sin^2x/cos^2x you should know well you should memorize that it is always going to be tan^2x. for the second part IT SHOULD NOT BE ZERO IT SHOULD BE ONE.The third part we use the reciprocal identity and it should be sec^2x. So when you put it all together you get tan^2x+1=sec^2x

The other pythagorean identity contains Cosecant and Cotangent. Since we divided the last one by cosine lets divide this one by sine now. we notice for part 1 that IT IS NOT GOING TO BE ZERO IT IS GOING TO BE ONE DONT FORGET. The second part you should know by memory that it should be cot^2x. The last part is the reciprocal so the reciprocal of sin is csc so it is csc^2x so your final answer should be 1+cot^2x=csc^2x

INQUIRY ACTIVITY REFLECTION:

“The connections that I see between Units N, O, P, and Q so far are…” The connections are that we are still using the trig functions cosine sine tangent .etc. and the references of the unit circles.

“If I had to describe trigonometry in THREE words, they would be…” Stress, overwhelming RATIOS

WPP#13 & 14: Unit P Concept 6-7: Applications with Law of Sines & Law of Cosines

"This WPP13-14 was made in collaboration with Peter Nguyen.  Please visit the other awesome posts on their blog by going here"

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BQ#1: Unit P Concept 1 and 4: Law of Sines and Area Formula

1.) Law of Sines- Why do we need it? How is it derived from what we already know?
       The Law of Sines is needed to solve triangles that are not right triangles. The normal trig functions are defined for a right triangle and are not directly useful in solving non right triangles but we can use the trig function to determine the law of sines. So first you have a non right triangles labeled as ABC. Then you make a imaginary perpendicular line from angle B and we call that h. Now we have formed two triangles. Now we use the trig function of sin which is hypotenuse over adjacent and then on the other triangle we have the same as well but we don't know the numbers so for the triangle on the left side it is SinA=h/c and for the right it is SinC=h/a. We then want h by itself so we multiply by c and on the other by a and we get cSinA=h and the other aSinC=h then we equal them together cSinA=aSinC and you divide by ac and you get SinA/a=SinC/c and that is exactly what the law of sines is.

To get a picture in your head of how to do it instead of me telling you there is a video here.

4.) Area Formula- How is the "area of an oblique" triangle derived? How does it relate to the area formula that you are familiar with? 
       The area formula is derived from the area equation and the trig function. The area equation you should know form Geometry is A=1/2bh where b is the base and h is the height. We then draw a perpendicular line on triangle ABC in half from angle B to create two triangles with a sharing side of h. We then know that SinC=h/a and when you get h by itself you get h=aSinC. Then you substitute it in the area equation for h and we get A=1/2b(aSinC). The area of an oblique triangle is one half of the product of two sides and the Sine of their included angle.

To get a visual in your head of how it should look like and example problems as well watch this video here.
      It relates to the area formula by using 1/2bh as a start to find the area formula for an oblique triangle. We still use 1/2 and it still consists of two sides and the base for b but we have to find our own h by using the trig function of sine.

Works Cited
http://www.youtube.com/watch?v=zJsw_ltZxFo
http://www.youtube.com/watch?v=gAX_IleqeJQ

WPP#12: Unit O Concept 10: Solving angle of Elevation and Depression word problems

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I/D#2: Unit O - How can we derive the patterns for our special right triangle?

INQUIRY ACTIVITY SUMMARY:
1) 30º 60º 90º TRIANGLE

  The 30-60-90 triangle comes from an equilateral triangle, which has equilateral angles of 60 degrees. The equilateral triangle is cut in half, making: 2 triangles, (2) 30º degree angles, and (2) 90º degree angles.  Then we cut it in half and (b) will equal to 1/2 because its half of 1, while the other untouched side (c) the hypotenuse would remain as 1. However, one side remains missing with its value (a), and in order to find it we use the  Pythagorean Theorem (a^2 + b^2 = c^2).

As you can see from the pic above to the left we use it and it should 1/2 squared should be 1/4 and now we subtract it from 1 and it should be radical 3/4 but 4 is a perfect square so we just have a equaling radical 3/2 so now we know that (a) is radical 3/2 (b) is 1/2 and (c) is 1. So now were not done because we dont want a fraction so all we have to do is multiply by 2 on each side and by looking at the pic above on the right side you see that when we multiply by 2 on (a) it should now be radical 3 and (b) should be 1 lastly (c) should be 2. Then we use use "n" to represent any value for the triangle creating n radical 3 on (a) 1n or just n on(b) and 2n on (c). 

2) 45º 45º 90º TRIANGLE

The 45-45-90 triangle comes from cutting a square directly in half diagonally. Cutting the square diagonally in half will create: 2 triangles, (4) 45º degree angles, and (2) 90º degree angles. We just have to focus on one since we know that its a 45º 45º 90º triangle we know that (a) and (b) have to be the same but by looking at the pic above we know that they are both 1. To look for (c) we need to use the Pythagorean Theorem (a^2 + b^2 = c^2).

To find (c) we do 1 squared plus 1 squared equals c squared. (c) equals radical 2. Then we add "n" to represent any value for the triangle to create n radical 2 that is (c) and n or 1n for (a) and n or in as well for (b) because remember its a 45º 45º 90º triangle and you know that (a) and (b) have to be the same.

INQUIRY ACTIVITY REFLECTION:

  “Something I never noticed before about special right triangles is…” how they got those numbers as in the sides like radical 2 and what the n had to do with all of these in the first place. How similar it is for deriving a special right triangle is.

“Being able to derive these patterns myself aids in my learning because…” it gives me a better understanding on hoe they got the numbers and what n really meant instead instead of just knowing that it just had to be there and why someone chose these numbers for a reason and if i forget i can always remember where these patterns were derived from.

I/D#1: Unit N Concept 7: How do SRT and UC relate?

INQUIRY ACTIVITY SUMMARY
1.) 30º TRIANGLE

Well to start we have to label all the sides. The hypotenuse has to equal to 1, and due to the rules of the 30º,60º,90º right triangle the side that is adjacent to 30º (x) has a value of x radical 3, the side opposite (y) has a value of x and the hypotenuse (r) as 2x. Then we divided by 2x in the hypotenuse,  since we did that we must do the same and then simplify the other sides by 2x . So x-radical 3/2x turns to radical 3/2 which is your x value. Then we do the same for our y value and x/2x turns to 1/2 and that is your y value. Then we draw the coordinate plane on the triangle.The origin is (0,0), while  going across the x axis is (radical 3/2,0) and then going directly up from the previous coordinate is (radical 3/2,1/2).

2.)45º TRIANGLE

For the 45º triangle you have to label it accordingly by the special rule for the 45º.  (x) is the side adjacent to the given degree with value of x, the (y) is opposite to the angle but similar to the adjacent side too so it has the same value of x and the hypotenuse, (r) is equaled to 1 still and is x radical 2. Then you are suppose to divide the other two sides by x radical 2 so x/x radical 2 is 1/radical 2 but we could NEVER leave a radical in the denominator so we multiply it by radical 2 and it should be radical 2/ 2. Since the x and y have the same value y would have to be also radical 2/2. Then draw the coordinate plane of the triangle. You start at the origin (0,0) then go to the right and that is (radical 2/2,0) then go up from the previous coordinate and it would be (radical 2/2,radical2/2).

3.) 60º TRIANGLE

The 60º triangle is exactly the same as the 30º triangle. (x) is adjacent but has a value of x, (y) would be the opposite angle with a value of  x radical 3 and the hypotenuse (r) still equal to 1 and is 2x. Then divide the other two sides by 2x like the 30º triangle and x/2x would be 1/2 the value of x, for the value of y it is x radical 3/2x and that would be radical 3/2. When drawing it in the coordinate plane you start at the origin (0,0) then move alongside to the x and it is (1/2,0) then go up with the previous coordinate plane and that is (1/2, radical 3/2).

4.) HOW DOES THIS ACTIVITY HELP YOU TO DERIVE THE UNIT CIRCLE?

This activity helps me derive the unit circle by figuring out the certain points and if you know these three you can know the rest of the other points but you need to know there signs if its going to be negative or positive.

5.)

Depending on which quadrant the SRT is found, the value will change by its sign positive or negative. If the triangle is found on Quadrant II, then x from the coordinate pair will be negative. In Quadrant III, both x and y will be negative. Finally in Quadrant IV, only y will be negative. In Quadrant I, all trig functions are positive, while in Quadrant II only sine and cosecant are positive and the rest negative. Meanwhile in Quadrant III, tangent and cotangent are the only positive. Finally in the Quadrant IV, only cosine and secant are positive. you can also memorize it by All Students Take Calculus.

6.) INQUIRY ACTIVITY REFLECTION

"THE COOLEST THING I LEARNED FROM THIS ACTIVITY WAS..." well i took cp Algebra ll so when i saw this i didnt know what this was at all as in the Unit Circle but the activity was Geometry so i knew that and learned that this activity was actually the first three in the unit circle in quadrant l.

" THIS ACTIVITY WILL HELP ME IN THIS UNIT BECAUSE..." I noticed that if we know the first quadrant then we can fill out the whole unit circle that is amazing and to help you more you can sing the song to get the ordered pairs and other methods as well.

"SOMETHING I NEVER REALIZED BEFORE ABOUT SPECIAL RIGHT TRIANGLES AND THE UNIT CIRCLE IS..." That i never knew that they were so related to one another. i didnt know that the coordinates can be different shapes as well.

RWA#1: Unit M Concpet 4-6- Conic Sections in Real LIfe

1. Parabola- Set of all points equal distance from a focus and the directrix
2.Algebraically

Graphically










                                                                                                 (image)
(image)

The parabola has FOUR key features: vertex, directrix, focus, and the axis of symmetry. The vertex is the point at which the parabola starts. The directrix is a line outside the parabola that is perpendicular to the axis of symmetry. The axis of symmetry is a line that breaks the graph into two equal sections. The focus is the point inside the parabola. Lastly the distance from the focus to the vertex and the vertex to the directrix is the same and known as "p". 
For graphing the Parabola you need to make sure the formula is in standard form. After you have done that you can find your parts you know what the vertex is it is (h,k). Then plot the focus which you should know is p by now and you plot it away from the vertex meaning it should be inside the parabola. To know if p is going up, down, left or right you need to know that when "p" is positive and the equation is x^2 it goes up, if "p" is positive and equation is y^2 it goes right, if "p" is negative and equation is x^2 it opens down, if "p" is negative and the equation is y^2 it opens left. the effect the focus does to the shape is the distance from the vertex to the Focus which determines how skinny or fat the parabola is going to be. The closer the focus is to the vertex the fatter the graph will be, the further away the focus is from the vertex the skinnier the graph will be.The eccentricity of the parabola is equal to 1 all the time.You then put in the directrix which is a line p units away from the vertex in the opposite direction of the focus. Then put in the axis of symmetry which is perpendicular to the directrix. Notice that the vertex and the focus are both going to be ordered pairs while the directrix and the axis of symmetry are going to be lines.
3. RWA

http://upload.wikimedia.org/wikipedia/commons/4/48/Headlight_lens_optics_schematic.png The cars headlights can be a great example of having a parabolic shape."The light is placed in the focus of a parabolic mirror, as the light travels and meets the mirror, it is reflected in lines parallel to the axis . This is why the light beam from the headlights of cars are so strong."( http://www3.ul.ie/~rynnet/swconics/UP.htm) source should be the vertex that bounces all of the dispersed lights from the reflector in a straight line straight forward.

  "Parabolic reflector has the property that if a light source is placed at the focus of the reflector then the light rays will reflect from the reflector as rays that is parallel to the axis. To have a safe night drive you don't want all the light rays to be parallel to the axis. Some of the light must be aimed far down the road to the side upward for signs or bridges."(http://www.pleacher.com/mp/mlessons/calculus/appparab.html)

Here it talks more about the headlight and how you have to find the right vertex so that when it bounces all the rays stay straight.

4. Works Cited Page
 http://upload.wikimedia.org/wikipedia/commons/4/48/Headlight_lens_optics_schematic.png
(image)
(image)
Car Headlights

Headlights