Unit Q Concept 2: Find all trig functions when given on trig function and quadrant using identities

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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.