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

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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 45º, 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.

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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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.

Resources:

SSS Packet

WPP#12: Unit O Concept 10 - Solving Angles of Elevation and Depression Word Problems

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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.

I/D #1 - Unit N Concept 7: How do SRTs & the UC relate?

Inquiry Activity Summary:
The unit circle activity is to help us develop our understanding of the special right triangles and how it correlates with the unit circle. There are three special right triangles: 30º, 45º, and 90º. The three sides of the triangle has to be simplified in such a way that the hypotenuse is equal to one because when inserting the triangle into the unit circle, the hypotenuse would be the radius, which always have to be one.

1. Rules for 30º triangle:

• Hypotenuse (r) is 2x *Remember to change it to 1 by dividing 2x*  
• Horizontal value (x) is x *When divided by 2x, it will make 1/2* 
• Vertical value (y) is x radical 3 *When divided by 2x, it will be radical 3/2*
• The 30º angle must be at the origin (0,0)

2. Rules for 45º triangle:

• Hypotenuse is radical 2 but remember to divide by radical 2 to make it 1
• Horizontal value is 1 over radical 2, when divided it'll be radical 2 /2
• Vertical value will be the same as horizontal value
• 45º angle must be at origin

3. Rules for 60º triangle:

• Hypotenuse is 2x, divide by 2x and it'll become 1
• Horizontal value is x radical 3, after dividing it'll be radical 3 / 2
• Vertical value is x, after dividing it'll be 1/2
• 60º must be at origin

4. The special right triangles activity helps us derive the unit circle because the points from the triangles will form a circle if all the points are connected. But of course, the 30º, 45º, & 60º are only in the first quadrant. So, you would have to find the other coordinate pairs, however, once you know quadrant 1, you basically know all the other quadrants.

5. The triangles drawn in this activity lies in quadrant 1. The values will change with positive or negative depending on which quadrant you're in. If you're in the 2nd quadrant, then your x value will be negative and your y value is positive. If it's in the 3rd, both the x and y value will be negative. If it's in the 4th, x value will be positive and y value negative.

In this 2nd quadrant, I showed the 30º reference angle,
the actual angle is 150º. As you can see, the coordinate pair
farthest to the right, is similar to the 30º angle in quadrant I.
The only difference is the x value is negative.

In this 3rd quadrant, I showed the 45º reference angle, the angle
is 225º. The coordinate pair closest to the bottom has both a negative
y and x value. It's also similar to 45º in quadrant I.

In quadrant IV, I showed the 60º reference angle and the angle
would be 300º. The coordinate pair closest to the bottom has a negative
y value and a positive x value. This angle is similar to the 60º angle
in quadrant I.

Inquiry Activity Reflection:

"The coolest thing I learned from this activity was..." The unit circle consists of special right triangles and it's not just a bunch of numbers you have to memorize; there's a pattern.

"This activity will help me in this unit because..." I'll be able to comprehend the unit circle without having to memorize random points.

"Something I never realized before about special right triangles and the unit circle is..." That just by knowing the first quadrant, you pretty much memorized the entire unit circle.