Wednesday, March 26, 2014
SP#7: Unit Q Concept 2: Finding All Trig Functions When Given One Trig fFunction and Quadrant
Please see my SP7, made in collaboration with Elizabeth T., by visiting their blog here. Also be sure to check out the other awesome posts on their blog
Thursday, March 20, 2014
I/D #3: Unit Q Concept 1: Using Fundamental Identities to Simplify or Verify Expressions
INQUIRY SUMMARY ACTIVITY
1. Where does sin^2x + cos^2x=1 come from?
To begin this is a Pythagorean identity. An identity is a proven fact or formula that is always true. The first thing we did was write out the Pythagorean theorem as show below. Then we replaced a^2+b^2=C^2 with x^2+y^2=r^2 ( as shown above) We are able to replace it with these variables because it consists of the same characteristic as the unit circle, were using all the variables from the first quadrant. At this moment our formula is x^2 + y^2 = r^2 and we want to get to sin2x+cos2x=1. To do that, we divide everything by r^2. If we divide everything by r^2, we get this: x^2/r^2 + y^2/r^2 = 1 (r^2/r^2). So r^2/r^2 becomes 1 and we can put (x/r) ^2+ (y/r)^2=1. Then we should notice that these variables are the ratios of cosine and sine. Now that we know this we can replace then and get a final answer of cos^2x+sin^2x=1.
Does it really work?
1. Where does sin^2x + cos^2x=1 come from?
To begin this is a Pythagorean identity. An identity is a proven fact or formula that is always true. The first thing we did was write out the Pythagorean theorem as show below. Then we replaced a^2+b^2=C^2 with x^2+y^2=r^2 ( as shown above) We are able to replace it with these variables because it consists of the same characteristic as the unit circle, were using all the variables from the first quadrant. At this moment our formula is x^2 + y^2 = r^2 and we want to get to sin2x+cos2x=1. To do that, we divide everything by r^2. If we divide everything by r^2, we get this: x^2/r^2 + y^2/r^2 = 1 (r^2/r^2). So r^2/r^2 becomes 1 and we can put (x/r) ^2+ (y/r)^2=1. Then we should notice that these variables are the ratios of cosine and sine. Now that we know this we can replace then and get a final answer of cos^2x+sin^2x=1.
Does it really work?
In order to verify that it works we can plug in the 30,45 and 60 degree angles. ( All values from the unit circle) In this case I choose 60*. The 1/2^2 becomes 1/4 and the rad3/2 becomes 3/4 because the squared cancels the 3 and you just multiply 2 by 2. This gives us an answer of 1.
2. The other Pythagorean Identities.
The picture above shows the steps of how to get tan^2x + 1 = sec^2x. First, you will have to divide both sides by cos^2x. For this one, we want tangent and secant. We can look at that as y/r times r/x. So as a result we have y/x which is tangent. Cosine divided by cosine is simply 1. And 1 divided by cos2x which is sec2x because sec x = 1/cos x, we just powered up because everything is being multiplied in this reciprocal identity. (look at the picture above)
The other Pythagorean identity is 1+cot^2x=csc^2x. In this identity we have cot and csc. We divide everything by sin2x . We divide cos^2x by sin^2x and we get cot^2x. We divide 1 by sin^2x and we end up with csc2x. This is how we get the answer.
INQUIRY ACTIVITY REFLECTION
1.THE CONNECTIONS I SEE WITH UNIT N, O, P AND Q SO FAR ARE the trig functions and the ratios. Also the magic three from the unit circle to derive the things above.
2. IF I HAD TO DESCRIBE TRIGONOMETRY IN THREE WORDS, THEY WOULD BE sin, cos, and tan.
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Tuesday, March 18, 2014
WPP #13-14: Unit P Concept 6 & 7: Applications with Law of Sines and Cosines
Please see my WPP13-14, made in collaboration with Ismael R., by visiting their blog here. Also be sure to check out the other awesome posts on their blog”
Sunday, March 16, 2014
BQ #1: Unit P Concept 3 and 4: Law of Sines and Area of an Oblique Triangle
Law Of Sines:
1. Law of Sines - Why do we need it? How is it derived from what we already know?
We use the law of sines when we don't have a right triangle. In previous concepts we were able to use the Pythagorean Theorem but were not able to use it in this concept because we are not dealing with right triangles. .Also like in the unit circle the restriction for sine is still applied (it can't be bigger than one). It is very important that you know that this law can only be used when we have ASA or AAS. The Law of Sines can be used when it comes to non-right triangles with AAS and ASA. This law was derived from a non-right angle triangle. To begin we drew a perpendicular line from the top of the triangle to the bottom and labeled that "h." Now that we have the two triangles the next step is to take the sin of A and C ( opposite/ hypotenuse). The next thing that we have to do is get h by itself so what we do is multiply both sides by its denominators as shown in the picture. In addition since both equations equal h we can set them equal to each other and we try to get sin by itself on top. So we obtain an answer of sinA/a= sinC/c.
Area of an Oblique Triangle:
1. How is the "area of an oblique" triangle derived?
1. Law of Sines - Why do we need it? How is it derived from what we already know?
We use the law of sines when we don't have a right triangle. In previous concepts we were able to use the Pythagorean Theorem but were not able to use it in this concept because we are not dealing with right triangles. .Also like in the unit circle the restriction for sine is still applied (it can't be bigger than one). It is very important that you know that this law can only be used when we have ASA or AAS. The Law of Sines can be used when it comes to non-right triangles with AAS and ASA. This law was derived from a non-right angle triangle. To begin we drew a perpendicular line from the top of the triangle to the bottom and labeled that "h." Now that we have the two triangles the next step is to take the sin of A and C ( opposite/ hypotenuse). The next thing that we have to do is get h by itself so what we do is multiply both sides by its denominators as shown in the picture. In addition since both equations equal h we can set them equal to each other and we try to get sin by itself on top. So we obtain an answer of sinA/a= sinC/c.
Area of an Oblique Triangle:
1. How is the "area of an oblique" triangle derived?
The area of an oblique triangle is derived from the original area formula for triangles (A=1/2bh). Since the perpendicular line of the triangle is the height(h) and b is the base. While the oblique triangles area is one half of the product of two sides and the sine of their interior angle.
We know that sinC=h/a from previous concepts, so once you multiply 'a' on both sides h=a sin C, and plugging 'h' into that A=1/2bh formula we really have A=1/2b(a sin C).
How does it relate to the area formula you are familiar with?
It relates to the area formula were familiar with because there very similar the only difference is that h is replaced by the sin of the included angle. Another thing that we used that we had used in the past was SOHCAHTOA.
It relates to the area formula were familiar with because there very similar the only difference is that h is replaced by the sin of the included angle. Another thing that we used that we had used in the past was SOHCAHTOA.
Thursday, March 6, 2014
WPP #12: Unit O Concept 10: Solving angle of elevation and depression word problems
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Tuesday, March 4, 2014
I/D #2: Unit O - How can we derive the patterns for our special rights triangle?
INQUIRY ACTIVITY SUMMARY:
Then we cut straight down on the triangle and that gives us an angle of 30* on the top because since we divide it symmetrically the angle dives into two. If we were to add 30+30 we would get an answer of 60. Then we created a 90* angle on the bottom left because the line created a perpendicular angle with the bottom of the triangle. ( as shown in the first picture). Since its an equilateral angle we know that all sides are equal and in this case it has a measurement of 1 on all three sides. The bottom side is 1/2 because its half of one. To find the other side (height) we use the Pythagorean Theorem ( a^2+b^2=c^2). We plug in 1/2 for a, were looking for b, and we plug in for 1 for c. This gives us an answer of rad3/2 for b. However to make the sides easier do deal with we give all sides a variable, we will use n, and were going to give that n a value of 2n. We give it a value of two because we don't want to deal with fractions. (see picture 2) Once we have done that we get a hypotenuse of 2n a horizontal side of n, and a vertical side of n rad3. Remember n just represents any number and it keeps the relationship consistent.
2. 45-45-90 triangle
For the 45-45-90 triangle we begin with a square. A square has four equal sides and four 90* angles. (see the first picture)First we drew a diagonal which cut the 90* angle into two pieces and that left us with a 45 degree angle. Since all sides are equal the horizontal and the vertical sides are 1. Then we use the Pythagorean theorem to find c ( diagonal). We use 1 for a and b and that gives us an answer of rad2 (picture number 2) Then we add n because n represents any number and it remains the relationship consistent.
INQUIRY ACTIVITY REFLECTION
1. Something I never noticed before about right triangles is how we were able to find the ratios from the equilateral triangle and a square.
2. Being able to derive these triangles myself aids in my learning because I can understand how a special right triangle works and use this for concepts 7 and 8.
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