SP #7: Unit Q Concept 2: Finding all trig functions values when given one trig function and quadrant (using identities)

Please see my SP7, made in collaboration with Ashley V., by visiting their blog here. Also be sure to check out the other awesome posts on their blog.

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 with?

     An "identity" is proven facts and formulas that are always true. The Pythagorean Theorem is an identity because it has been proven to be always true. Since the Pythagorean Theorem consists of x, y, and r, it is also the same as a, b, and c.

    Since a^2 + b^2=1, we divide r^2 on both sides. After MEMORIZING the Unit Circle, we knew cosine is (x/r)^2 and (y/r)^2 is sine. The ratio for cosine on the unit circle is x/r and the ratio for sine is y/r.

Sin^2x + cos^2x=1 is referred to as a Pythagorean Identity because we used the Pythagorean Theorem to find it.

  I will choose one of the “Magic 3” ordered pairs (30, 45, or 60 degrees) from the Unit Circle to show that this identity is true; in this case, I chose 60 degrees.

2). Show and explain how to derive the two remaining Pythagorean Identities from sin^2x + cos^2x=1.

The picture below shows and explains how to get tan^2x + 1 = sec^2x. The identity with Secant and Tagent is tan^2x + 1 = sec^2x.

The picture below shows and explains how to get 1 + cot^2x = csc^2x. The identity for Cosecant and Cotangent is 1 + cot^2x = csc^2x.

INQUIRY ACTIVITY REFLECTION

1). The connections that I see between Units N, O, P, and Q so far are that the Unit Circle and its ratios play a big part in all these units so “MEMORIZE THEM” and they also involve triangles to use the Pythagorean Theorem to get the answer or the missing side/angle.

2). If I had to describe trigonometry in THREE words, they would be burdensome, exhausting, and painful.

BQ#1: Unit P Concepts 1 and 4: Law of Sines and Area of an Oblique Triangle

   1. Law of Sines:
                                                                           
                                         

   Why do we need it?

       We need the law of sines because not every triangle is a right triangle, so we use the law of sines to solve for non-right triangles. We cannot use the Pythagorean Theorem or the normal trig function to solve for non-right triangles like we do for right triangles, but the law of sines luckily works for any triangle.

     How is it derived from what we already know?

·         The first thing we do is draw and label our non-right triangle.

·         We then drop down a perpendicular from “angle B,” which makes two triangles, and we call it “h”.

·         Since we now have two triangles, we can use the normal trig functions like sin, cos, tan, csc, sec, and cot; so “we can drop perpendiculars from the other two vertices and get the other relationships.”

·         After getting the two sin ratios, we see that they have “h” in common, so after we simplify both of the equations by getting rid of the denominators, we equal both to each other since both of them equal to “h”. We then get cSinA=aSinC.

·         We then want to get Sin and its angle by itself so we divide by the coefficients, which are “a” and “c” in this case, and you then get “sinA/a = sinC/c”.

·         It all depends where you make your perpendicular line “h”. So we can find angle B if only the perpendicular line is from a different angle like A or C and you do the same process.

 4. Area of an Oblique Triangle:

1     Area formulas:

·   “The area of a triangle is A=1/2bh, where ‘b’ is the base and ‘h’ is the perpendicular height of the triangle”.

·    “The area of an oblique (all sides’ different lengths) triangle is one-half of the product of two sides and the sine of their included angle (the angle in between them)”.

·    Formula: It depends on what angles they give you, but the options you have are A=1/2bcsinA, A= 1/2acsinB, or A= 1/2absinC.

How is the “area of an oblique” triangle derived?

·    The area of an oblique triangle is derived by cutting the triangle into two, which makes the triangle into a right triangle; this is done by using “h” as a line. Because we now have two right triangles, we can use the normal trig functions. Depending where “h” is dropped down from a different angle, we can see which one we are using from the options of sinC=h/a, sinA=h/c, sinB=h/(a or c). Angle B has two options because “h” is dropped down so it can apply to both triangles.

·    Since we want to get “h” by itself, we multiply by the denominator to both sides to get rid of it.

·    We then plug each of the equations into the formula for the area of a triangle, which is “A=1/2bh”.

·    We then substitute in our regular area equation for “h,” and we get A=1/2b(aSinC) or A=1/2b(cSinA). This also applies to angle B when you do it in the other triangle.

How does it relate to the area formula that you are familiar with?

It relates to the area formula that we are familiar with by plugging in the values into “h,” which make new formulas, and it also uses the normal trig formula for the area of a triangle. The area of an oblique is conditional; it needs to obtain two side lengths, which cannot be the same letter, and their angles must be included. Or else it will not work.

References:

Unit P SSS Packet

WPP #13 and 14: Unit P Concept 6 and 7: Law of Sines and Law of Cosines

This WPP13-14 was made in collaboration with Ashley V. and Tina N.  Please visit the other awesome posts on their blog by going here and here.

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WPP #12: Unit O Concept 10: Angle Of Elevation And Depression

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I/D #2: Unit O Concept 7-8: Deriving Patterns for Special Right Triangles

Inquiry Activity Summary

The purpose of the activity was to help us apprehend the special right triangle and how its patterns gets what it has. With the special right triangle, it helps us understand where the patterns come from, and instead of memorizing the patterns for both the 45-45-90 degree triangle and 30-60-90 degree triangle, you see how we get each of their side lengths by deriving them.

1. 30-60-90 Triangle

An equilateral triangle makes up of 60 degrees for each of its 3 sides; and making each of its sides equal to one since all its sides and degrees are equal. But when cut in half, its degrees change, and so does its sides. Its degrees become 30-60-90 degrees; the number across 90 degrees stays the same since it side was not cut off, the number across 30 degrees is now “1/2” because 1 was cut in half, so it value was also cut in half, so know the only thing we know need to find is the number that is across 60 degrees.

We use the Pythagorean Theorem to solve for the number side across 60 degrees, which we then get √3/2 as the answer. We then see how each side effects and correlates to the other sides. To get rid of the fractions from each sides, we multiply each side by 2, which gets rid of the fractions; we then get n, n√3, and 2n. After doing that, we multiply each side by “n”. Since “n” is a ratio, it helps the sides get bigger but its value will always stay constant. 

2. 45-45-90 Triangle

When we cut a square that has its four sides equal to 90 degrees and making each of its sides equal to 1 will change when cutting it across diagonally. It cuts the 90 degree angles in half, making it into a 45-45-90 degree triangles. Since two sides of the now triangle stood the same, it is still equal to one, but we now need to find the other side that was cut in half, so we will now use the Pythagorean Theorem to find the hypotenuse, which is the missing side. After finding the hypotenuse side, √2, we put everything together. We then see how each sides correlate and effect each other.

Since there are two 45 degree angles, both of their sides now equal to 1; and the 90 degree angle equals to √2. We then multiply “n” to all three sides and we end up with n, n, and n√2. Since “n” is a ratio, it always stay constant, so even if the numbers you plug in to “n” make the sides bigger, its value will stay constant.

Inquiry Activity Reflection

1. Something I never noticed before about special right triangles is how even though you cut them down, its value will always stay consistent. 

2. Being able to derive these patterns myself aids in my learning because it helps me understand where everything comes from and how one side effects the other sides rather than just itself.

I/D#1: Unit N Concept 7: The Unit Circle

     The Special Right Triangle (SRT) and the Unit Circle relate to each other in that the SRT helps give you the 30, 45, and 60 degrees points in the UC, which will then help you find the rest of the points in the other three quadrants. But you have to remember the order the points have to be in for each quadrants since they do not go in order.

INQUIRY ACTIVITY SUMMARY

1. 30 Degree Triangle

The first thing I did was plug in the degrees, so I can be able to set it up correctly. I labeled the hypotenuse, vertical value, and horizontal value and it's value for each side so it would be easier for me to know where to put my answers in either "r", "x", and "y". 

Since this is a special right triangle, the hypotenuse or "r", will always be one. In order to turn 2x to 1, you divide 2x to it to get one. Since you did it to "2x", you must also do it to "x radical 3" and "x"; so "x radical 3" turns into "radical 3 over 2" and "x" turns into "1/2". So r=1, x=radical 3 over 2, and y=1/2. After doing all this, you get your 30 degree point, which is (radical 3 over 2, 1/2).

2. 45 Degree Triangle

The first thing I did was plug in the degrees, so I can be able to set it up correctly. I labeled the hypotenuse, vertical value, and horizontal value and it's value for each side so it would be easier for me to know where to put my answers in either "r", "x", and "y". Since this is a SRT, I equaled "r" to 1, and so far, left the rest blank.

In order to get "x radical 2" to equal one, I will have to divide it by "x radical 2". Since I did that to one side, I also do it to the other sides, so both the x's will then equal to "radical 2 over 2". r=1, x=radical 2 over 2, and y=radical 2 over 2. In the end, you get the 45 degree point which is (radical 2 over 2, radical 2 over 2).

3. 60 Degree Triangle

The first thing I did was plug in the degrees, so I can be able to set it up correctly. I labeled the hypotenuse (2x), vertical value (x radical 3), and horizontal value (x); and it's value for each side so it would be easier for me to know where to put my answers in either "r", "x", and "y". Since this is a SRT, the hypotenuse (r) will equal to one.

To be able to get "2x" to equal to one, you divide it by "2x". Since you do it to one side, you do it to the other sides. "x" when divided by "2x" will equal to "1/2". "x radical 3" when divided by "2x" will equal to "radical 3 over 2". So r=1, x=1/2, and y=radical 3 over 2; and the 60 degree point will be (1/2, radical 3/2).

4. This activity helps me derive the Unit Circle in the way that it gives me the points around the Unit Circle and it helps me understand where the numbers come from. It also helps you visualize the SRT in the Unit Circle and how each degree gets its point. You also see the correlation between SRT and UC and how the SRT draws you the details for the UC, but the UC gives you the whole picture and has everything put together.

5. The triangle that was drawn in the activity lies in the first quadrants, since everything is positive, and it is the base of the Unit Circle. To memorize the whole Unit Circle, you just need to memorize the five steps in the first quadrants, which are the 0, 30, 45, 60, 90 degrees, radiants, and points to it. After having those down, you just need to remember some patterns, but you get the concept. The values change when I draw the triangles in Quadrants II, III, and IV, because depending on what the ratios for the trig functions are, you'll know which ones are positive and negatives, but it is still the same.

(http://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/ebaa19ac-ff8b-43a6-a793-a00d9ac15e86.png)

This right here shows Quadrants II, III, and IV, which came from Quadrant I. As you can see, all three are the same in angles, but the only difference is their degree and the quadrants that they are in. All three have the same reference angles and all three share the same points but different connotations, depending if it's positive or negative.

(http://www.regentsprep.org/Regents/math/algtrig/ATT3/reftriex.gif)

This is a 45 degree triangle that is in Quadrant II, which has the same values as Quadrant I, but the only differences is its connotation, since the x-value is negative, while in Quadrant I it is all positive. 

(http://01.edu-cdn.com/files/static/learningexpressllc/9781576855966/The_Unit_Circle_21.gif)

The left picture shows a 60 degree triangle that is in Quadrant III, which also has the same values as Quadrant I, but in this case, because it is in the third quadrant, its x-value and y-value are both negative. On the right picture, it also shows a 60 degree triangle, but is in the fourth quadrant. Like Quadrant II, III, and IV, it has the same values as Quadrant I. In Quadrant IV, the y-value is negative and the x-value is positive.

HEADING FOR THIS SECTION: INQUIRY ACTIVITY REFLECTION

1. The coolest thing I learned from this activity was that everything connects with one another even if it is in different quadrants, its values are the same.

2. This activity will help me in this unit because I am able to memorize the Unit Circle because I am now able to visualize the SRT in the Unit Circle, so I am able to remember the degrees, radiants, and the points, so basically, now I know the whole Unit Circle.

3. Something I never realized before about special right triangles and the unit circle is that if you put them together, you can see the connection between them two and how it works and the details that make the Unit Circle up.

Citations:
(http://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/ebaa19ac-ff8b-43a6-a793-a00d9ac15e86.png)
(http://www.regentsprep.org/Regents/math/algtrig/ATT3/reftriex.gif)
(http://01.edu-cdn.com/files/static/learningexpressllc/9781576855966/The_Unit_Circle_21.gif)