Reflection #1: Unit Q - Verifying Trig Identities

1.) To verify a trig identity, it means both sides of an equation has to equal to each other. This requires simplification on the left side only, as the right side is not to be touched. In most or in all cases, the right side is already simplified to one or two trig identities, so rewriting it would create complications and more work. Throughout the verification and simplification, identities (definition from Mrs. Kirch: proven facts and formulas that are always true) are used for substitution, canceling, etc. The relations of the 11 identities all have some sort of pattern that go/relate with one another.

2.) There are a few tips and trick to verifying a trig identity. If the problem looks simple, start by checking if an identity would work. Usually from there, one identity will lead to another and trig functions will start canceling out perfectly. But if the problem is more complex, like those in Concept 5, the process is a bit different. For fractions, separation might work best but in some cases multiplying the conjugate works as well. In fractions, mainly, the problem looks most complicated, however one should not be afraid them as the fraction will simplify and decrease its size along the way. In other situations and problems, factoring a common multiple might help, while in the other hand powering up/down will also do the trick. Trig identities are also usually found where there is any addition/subtraction of number values. A main thing to remember is that one CANNOT divide by a trig function.   

3) When a trig identity is given to be verified, I feel less tense as the answer is technically already given but needs to be simplified/proven. I always look at the right side at the "answer" and start thinking how I would want to rearrange the problem to get there [to the answer]. I usually start simple by checking if an identity can be substituted in order to cancel out. I also search for some sort of relation that holds the problem together and that can be used to break it down. I prefer factoring since in many cases it would simplify to a trig identity. I try avoiding multiplying binomials/conjugates, squaring binomials, etc as it involves more work and can have several sections were mistakes can be made. 

SP #7: Unit Q Concept 2 - Finding Trig Functions Using Identities

Please see my SP #7, made in collaboration with Victoria Ventura, by visiting their blog here.  Also be sure to check out the other awesome posts on their blog.

I/D #3: Unit Q - Pythagorean Identities

INQUIRE ACTIVITY SUMMARY

1.) Where does sin^2x+cos^2x=1 come from to begin with? You should be referring to the Unit Circle ratios and the Pythagorean Theorem in your explanations. 

The Pythagorean Theorem is considered an "identity" (def: a proven fact or formula that is always true). The Theorem a^2+b^2=c^2 can also be written with different variable like x,y, and r to get: x^2+y^2=r^2. In order to get the equation equal 1, r^2 would need to be divided on both sides to get: (x/r)^2+(y/r)^2=1. From this new equation we can get information using our knowledge from the Unit Circle. If we remember, x/r is the trig ratio for cosines while y/r is the ratio for sines. The equation can be written as: cos^2(x)+sin^2(x)=1. This is referred to as a Pythagorean Identity since it was first derived from the Pythagorean Theorem, which is an identity that was simply rewritten but never lost its value. To prove that this equation is an identity and always true, we can take one of the "Magic 3" ordered pairs from the Unit Circle. For example, the ordered pair for a 45 degree angle is (radical2/2, radical2/2). When the ordered pair is plugged into the equation: (radical2)/2)^2 + (radical2)/2)^2, the answer WILL result to get 1. 

2.) Show and explain how to derive the two remaining Pythagorean Identities from cos^2(x)+sin^2(x)=1. Be sure to show step by step. 

     a) If the equation were to be divided by cos^2, the equation would look like:  cos^2(x)/cos^2(x)+sin^2(x)/cos^2(x)=1/cos^2(x). The first part (highlighted in blue) will cancel out to equal 1 since both numerator and denominator are the same. The second part (highlighted in green) can be substituted by a ratio identity of tan^2(x). The third part (highlighted in light blue) can be substituted by a reciprocal identity of sec^2(x). In the end, the derivation will result as 1+tan^2(x)=sec^2(x).

     b) If the equation were to be divided by sin^2, the equation would look like:  cos^2(x)/sin^2(x)+sin^2(x)/sin^2(x)=1/sin^2(x). The first part (highlighted in purple) can be substituted by a ratio identity of cot^2(x). The second part (highlighted in orange) cancels out to equal 1 since both numerator and denominator are the same. The third part (highlighted in red) can be substituted by a reciprocal identity of csc^2(x). In the end, the derivation will result as cot^2(x)+1=csc^2(x).

INQUIRE ACTIVITY REFLECTION

"The connections I see so far in Unit N, O, P, and Q so far are..." the repeating use of the trig functions of sine, cosine, tangent, cosine, cosecant, and cotangent from the Unit Circle AND the use of triangles.

"If I had to describe trigonometry in THREE words, they would be..." triangles, triangles, and triangles!...Just kidding; it would be complex, overwhelming, and triangles. 

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

Please see my WPP13-14, made in collaboration with Adrie Garcia, by visiting their blog here.  Also be sure to check out the other awesome posts on their blog.

BQ #1: Unit P Concept 1 & 4 - Law of Sines and Area Formula

1. Law of Sines - Why do we need it?  How is it derived from what we already know?

The law of sines is needed to figure out the missing parts (angles and sides) of a non-right triangle. Non-right triangles are more complex to solve for since they are not like special right triangles and have a special side relationships (ex: 45-45-90 triangle has a relationship of (x)-(x)-(x)radical 2).
The deviation to this law is shown through the process and pictures below:
First, there is the non-right triangle ABC. An imaginary perpendicular line is drawn to cut angle B. The imaginary line is labeled as h.

The perpendicular line created (2) right triangles. From there, the basic trig functions of sin (opposite/hypotenuse) can be used for angle A and C. The trig functions are then rearranged to equal to h.  

Since both functions are now equal to h, they can be equal to one another. Both are then over the corresponding sides used (sides a and c). Simplification can be done and the law is finished. (*Note: this implies the same for B if the perpendicular line was cut either from angle A or C)

4. Area formulas - How is the “area of an oblique” triangle derived?  How does it relate to the area formula that you are familiar with?

The area of an oblique triangle is derived from the triangle area equation and a trig function. The area of any triangle is (1a) A=1/2bh, where b is the base and h is the height. However, in an oblique triangle, the height is not given and the sin trig function must be used.
With the oblique triangle provided as an example, triangle ABC is cut perpendicularly in half from angle B to create 2 right triangles with the sharing side of h. Side h must be used with sin, in this example sin C is used with h/a (sin A could have been used as well). (1b)The sin trig function is rearranged to equal h.

Since the sin trig function now equals to h, it can be substituted into the (1a) area of triangles. (*Note: the perpendicular line from the beginning can cut any angle and create in total 3 ways to find the area with the same concept).

     It relates to the formula we are familiar with since it is mostly substitution, the new equation has parts from the old one. The triangle is still multiplied by 1/2 and consists of 2 sides. However in the new one, it also consists of an angle which is in between the two sides used. In the original one, meanwhile, uses the base and the perpendicular height. 

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

(www.acclaimimages.com)

Problems

1. Farco decided to put a flag vertically on top of his ice-cream building. After having it set up, he wonders how high up is the the tip of the flag from the ground. He stands 5 feet away from the building and at ground level the measure of elevation is 76 degrees. a) How high up is the tip of the flag in the nearest foot?
2. Farco decides to stand on top of the building, which is 15 ft high, and leans against the flag post. From that view, he sees an army of starving children on the other side of the street in the Pie Store. If the distance of Farco's position to the bottom of the Pie store is radical 62725 away, b) what is the angle of depression from the top of the building to the corner to the nearest hundreth?

Solutions

1.

2.

I/D #2: Unit O - How can we derive the patterns for our special right triangles?

INQUIRY ACTIVITY SUMMARY
1. 30-60-90 Triangle

     The 30-60-90 triangle comes from an equilateral triangle, which has equilateral sides with equiangles of 60 degree. The equilateral triangle is cut in half, making: (2) triangles, (2) 30 degree angles, and (2) 90 degree angle. To explain the relationship between the sides and angles, an equilateral triangle with the sides of 1 can be used. If the process would be applied to this example, and focusing on ONE of the two triangle, the angles would remain the same as mentioned. The side cut in half will equal to 1/2, while the other untouched side (the hypotenuse in this case) would remain as 1. However, one side remains missing with its  value, and in order to find it the Pythagorean Theorem (a^2 + b^2 = c^2) is used.

     Using the Pythagorean Theorem, c would be the hypotenuse while a and b would be the other two sides that are across the 30 and 60 degree angles. Side a would equal to 1/2 while c would equal to 1. Plug those into the equation to solve for b will result with b equaling to radical 3 over 2. Across the 30 degree angle, a= 1/2; across the 60 degree angle, b= radical 3/2; and across the 90 degree angle c= 1. The relationship for a 30-60-90 triangle is 1/2-radical 3/2-1. HOWEVER, not all triangles are the same and not all derive from an equilateral triangle with the side of 1. The variable n can be used on the relation to represent any value for a triangle, creating  (n/2)-n(radical 3)/2-(n). (And to make life simpler, the fraction can be removed my multiplying the relation by 2 to make (n)-n(radical 3)-2n..

2. 45-45-90 Triangle

     The 45-45-90 triangle comes from cutting a square directly in half diagonally. Cutting the square diagonally in half will create: (2) triangles, (4) 45 degree angles, and (2) 90 degree angles. To explain the relationship between the sides and angles, a square with the sides of 1 can be used. Cutting the square and focusing on ONE triangle, the right triangle created will have two sides with the length of 1 with a missing hypotenuse. The sides of 1 will be across the 45 degree angles, and the hypotenuse can be found using the Pythagorean Theorem.

     With the Pythagorean Theorem, side a will be 1 while side b will also be 1. After squaring the sides, adding them, and finding the square root, the hypotenuse will equal to radical 2. So a 45-45-90 triangle has a relationship of 1-1-radical 2. HOWEVER, not all triangles will derive from a square of 1. As in the 30-60-90 triangle, n can be used on the relation to represent any value of the triangle, creating (n)-(n)-n(radical 2).

INQUIRY ACTIVITY REFLECTION
“Something I never noticed before about special right triangles is…” that these special right triangles derived from equilateral triangles with equiangles. The 45-45-90 right triangle came from a square, a quadrilateral with the same length in sides and had (4) 90 right angles. The 30-60-90 right triangle derives from an equilateral triangle that had (3) 60 degree angles.
“Being able to derive these patterns myself aids in my learning because…” it gives me a better understanding where the numbers and relations actually come from and now know that a mathematician didn't randomly chose numbers. Also, in case I forget the relation during an important test, I could remember where and how these patterns are derived from.