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Wednesday, March 26, 2014

SP#7: Unit Q Concept 2 - Finding all Trig Function Values Using Identities

Please see my SP7, made in collaboration with Vanessa C. by visiting her blog HERE. Also, be sure to check out the other fabulous posts on her blog.

Wednesday, March 19, 2014

I/D#3: Unit Q - Pythagorean Identities

Inquiry Activity Summary:
1. The Pythagorean Theorem is an identity that can always be proven true no matter how much it is manipulated. It is evident in the Unit Circle when we take the sine and cosine of the 45ยบ, for instance. The ordered pair would be (√2/2, √2/2) and the sin45 is √2/2 and the cos45 is √2/2. If we plug it into the Pythagorean theorem, it would be one equals one, which is true. Sin2x+cos2x=1 comes from the pythagorean theorem, which is x2+y2=r2. Next, you have to manipulate the pythagorean theorem so that there is only one "r". This is because in the unit circle, r is equal to one. To do this, you divide everything by r2. Once you have done this, you will get (x2/r2)+(y2/r2)=r. However, you can arrange that to make it look simpler: (x/r)2+(y/r)2=1. Now, you know that x/r is equal to cosine and y/r is equal to sine. As a result, you just substitute or replace it with cos and sin. Then, it would look like cos2x+sin2x=1.

2. The second pythagorean identity is 1+tan2x=sec2x. To get this, you divide the first identity by cos2x so that it will cancel and is equal to one. It will look like this: 1+(sinx/cosx)2=(1/cosx)2. We know that sine is equal to y/r and cosine is equal to x/r, so you can cancel out the r's and end up with y/x. This would mean that it's tangent, so you substitute it in. Next, 1/cosx is secant because cosine is x/r but it's the denominator, so you take the reciprocal of it. The third pythagorean identity: 1+cot2x=csc2x. To get this, you divide the first identity by sin2x so that the sin2x can cancel and equal one. So far it will look like this: 1+(cosx/sinx)2=(1/sinx)2. We know that x/r is cosine and y/r is sine, so we can cancel out the r's and get x/y, which is cotangent. 1/sinx is (1/ y/r), to get rid of the fraction you do the reciprocal which is r/y, and that would be cosecant.

Inquiry Activity Reflection:
1. "The connections that I see between Units N, O, P, and Q so far are..." they all relate to the Unit Circle and they incorporate the Pythagorean Theorem one way or another

2. "If I had to describe trigonometry in three words, they would be..." Ratios, "Unit Circle", FUN.

Monday, March 17, 2014

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

This WPP 13-14 was made in collaboration with Diana Pham. Please visit the other awesome posts on her blog by going here.

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Saturday, March 15, 2014

BQ#1: Unit P Concept 2 & 5 - Law of Sines & Area of an Oblique Triangle

2. Law of Sines
SSA (Side Side Angle) is considered ambiguous because there is no one sure solution. The problem only gives us one angle of the triangle so there is some ambiguity of what the triangle looks like. There could be one possible solution, two possible solutions, or no solutions.
This connects to the Unit Circle trig function values because the Law of Sines can give us the measures for each line segment on the Unit Circle. In Unit N, we learned that the unit circle has a radius that is measured to be one, however, it is not always the case. As a result, the Law of Sines help us correlate the angles with the length of their sides.

4. Area Formulas
The "area of an oblique" triangle is derived from the regular area equation, A=1/2bh. In the triangle below, you can see that there is perpendicular line through the triangle, creating two triangles. If we are looking at angle C, we know from previous units that SinC=h/a , therefore, h=aSinC. Now, just substitute it into the regular area equation: A=1/2b(aSinC). The equation we just found, however, is only if the problem gives you angle C. The equation for angle A: 1/2bcSinA. Equation for Angle B: 1/2acSinB. We use the same process to get the area for angles A and B. This relates to the area we are familiar with because we have to use that equation in order to get the area for an oblique triangle. Like I said before, you plug what h (height) is into the normal area equation.
SSS Packet

Tuesday, March 4, 2014

I/D #2: Unit O Concept 7-8 - Deriving Patterns for the SRTs

Inquiry Activity Summary:
1. For a 30-60-90 triangle, we must cut it straight down because it was currently an equilateral triangle, meaning all the angles are the same. Since we cut it in half, the length will be cut in half as well, making it 1/2. Since it was given that the hypotenuse side is one and we found out that the length is 1/2, we just need to find the height. To find the height, you use the pythagorean theorem which is a^2+b^2=c^2 (look at picture for mathematical explanation). To get rid of the fraction, you multiply everything by two. We use "n" as if it were equal to one. As a result, the hypotenuse is 2n, the side across the 60 degrees angle is n radical 3 and the last side is n. We use n because it's a variable that represents 1 since it's in all the terms.

2.  We can derive a 45-45-90 triangle from a square by cutting it diagonally. Cutting it diagonally will split the square in half as well as the 90 degree angles, making it 45 degrees. We know that the sides opposite from the 45 degree angles are still 1 because we didn't cut any sides in half, just the angles. To get the hypotenuse, you use the pythagorean theorem. (1)^2 + (1)^2 = c^2, C should equal radical 2 according to the work shown below. Also, we use n as a constant or a variable which represents 1. Using "n" represents the relationship between the sides and angles.

Inquiry Activity Reflection:
1. "Something I never noticed before about special right triangles is..." That there is a reason why the sides are the way they are, it's not just memorization.
2. "Being able to derive these patterns myself aids my learning because..." There is logical reasoning in deriving the patterns and I won't have to memorize it like I did before.