Reflection #1: Unit Q: Verifying Trig Identities

1. The meaning of verifying a trig identity is to prove that the equation is true and we do this by showing that both sides are equal to each other.

2. Tips and tricks that I found helpful are to try to always change the identities to sin(x) and cos(x) when possible. Another tip I would use is to try to convert the equation into a Pythagorean identity when it applies. Another important tip is to NEVER TOUCH THE RIGHT SIDE OF THE EQUATION! Always start with the left because you are verifying if it does equal to the right side. We love having our equations to have squares, so if your answers can not go further, you then can square both sides to get a Pythagorean identity out of it. A very important tip to always do when squaring both sides is to check your answer in the end to make sure it works, because the answer can be extraneous. A trick I would give you is to use the "u-substitution" because it is helpful when factoring trigonometric expressions. And it makes it easier during the factoring process because after you factor, you just plug everything back in! So easy!

3. My thought process through the steps when verifying trig identities is to look at the equation in general. I first check to see if we want the left and right side of the equation to equal to each other or if we are looking for a specific answer (degrees/radians). If we are looking for an answer, than the first thing I would do is to try to change the equation into sin(x) and cos(x). I would then look for a Pythagorean identity in order to have the same identity throughout the problem. I would also check to see what quadrants it lands on because it helps us determine where our answer will lay in. I would also see if I can use the "u-substitution" or "m-substitution" to make my life easier. If we were looking to see if the equations on both sides equal to each other, I would start with the left side and leave the right side alone and then bring it back down when in the end. I will also see if I can separate the equation, get a common denominator, FOIL, combine, cancel, replace, convert, or change identities. I need to have my options open to how I want to approach the problem. I also use my right side equation as guidance to where I want to be. My right side equation can tell me if a should combine and cancel or change and convert to a different Pythagorean identity. What I recommend to do is to look at the equation first and analyze the starting point and then see where we want to end. In my opinion, that is the key. Analyze the equation before making a move.

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.