SP#7: Unit Q Concept 2: Finding All Trig Functions

 “This SP7 was made in collaboration with Mellanie T.  Please visit and check out her other cool blog posts by going here”.

First, determine the quadrant the problem will be in by looking at the signs. In his case, it will be in quadrant III. Then, it is best to use the reciprocal identities first because they are the easiest and fastest to find. However, you are not required to use them first. Afterwards, you can use the pythagorean, ratio, or reciprocal identities to find the remaining trig functions. You can also use SOHCAHTOA to solve this problem (as seen in the last image). Draw the triangle in quadrant III and label the sides. Then, use the trig ratios to solve for each trig function.

Please pay special attention to the final answers of the trig functions. There should never be a radical in the denominator. Remember to always rationalize! Furthermore,  it is important to pay attention to the signs in front of each answer. You should write the signs in at the beginning to make sure you do not forget. Good luck!

I/D#3: Unit Q - Pythagorean Identities

INQUIRY ACTIVITY SUMMARY
1. The Pythagorean identity, sin^2x+cos^2x=1, comes from the Pythagorean Theorem. If we arranged the Pythagorean theorem using x, y, and z, we would get x^2 + y^2 = r^2. And, if we made this equation equal one, we would have to divide everything by r^2. So, we would get: (x^2)/(r^2) + (y^2)/(r^2) = 1. We already know the trig ratio for cosine is x/r and the ratio for sine is y/r. So, (x^2)/(r^2)= cos^2 and (y^2)/(r^2) = sin^2. Therefore, for our final answer, we would get: cos^2 + sin^2 = 1, our Pythagorean theorem. Look at the pictures below for a visual.  Also, we can prove this by plugging in values from the unit circle. The example below uses the coordinates of the 30 degree reference angle: (rad 3/2, 1/2).

2. To derive the two remaining Pythagorean Identities from sin^2x+cos^2x=1, divide it by cos^2x and sin^2x. When you divide it by cos^2x, you shuold get 1 + tan^2x = sec^2x. Then, when you divide it by sin^2x, you should get 1 + cot^2x = csc^2x. Look at the pictures below to see, step by step, how I got them.

INQUIRY ACTIVITY REFLECTION
1. "The connections that I see between Units N, O, P, and Q so far are:" how the Pythagorean identities are derived from the Pythagorean Theorem (x^2 + y^2 = r^2) -> ( x^2/r^2 + y^2/r^2 = 1) ->(cos^2x + sin^2 = 1) and how the reciprocal and ratio identities are derived from the trig ratios (cotx= cosx/sinx) -> (x/y) / (y/r) = x/y
2. "If I had to describe trigonometry in THREE words, they would be:" triangles, identities, and ratios.

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

Please see my WPP13-14, made in collaboration with Mellanie T, by visiting her blog here.  And while you're there, be sure to check out her other cool posts as well. ^_^

BQ# 1: Unit P - Oblique Triangles, Law of Sines & Cosines, and Area

3. Law of Cosines

http://www.mathwarehouse.com/trigonometry/images/law-of-cosines/law-of-cosines-formula-and-picture2.gif

We need law of cosines to find certain sides or angles of an oblique triangle when we are given SSS (side-side-side) or SAS (side-angle-side). If we are given an oblique triangle and cut it down the middle of the base, we can create 2 right triangles (see pic below). 

Then, we can label the line h, for height. Now let's say we need to find the length of side a. We can use the distance formula, but to do that, we need the coordinates of angle B and angle C. If cosA=d/c, then d=ccosA. And, if sinA=h/c, then h=csinA. Therefore, angle B will be (ccosA, csinA). For angle C, b is the x value and 0 is the y value (b,0). Now that we have our two coordinates, we can plug these values in the  distance formula (see right): a^2=(ccosA-b^2)^2 + (csinA-0)^2. We need to FOIL (ccosA-b^2)^2 and simolify (csinA-0)^2. Then, we need to factor out c^2 from c^2sin^2A and c^cos^2A. (cos^2 + sin^2A) =1 (This is a trigonometric identity.) Thus, when we take the remaining numbers (highlighted in green), we obtain a^2=b^2+c^2-2bccosA, which is our Law of Cosines formula, which can be rearranged in three different ways (see 1st image).
Moreover, it is similar to the Pythagorean Theorem because the Pythagorean theorem is derived from the law of cosines.  If one of the angles is 90 degrees, then the cos(90)= 0. Therefore, we can leave out the -2bccosC, and it simplifies to a^2+b^2 = c^2.

5. Area Formulas

All three area formulas will work. They just use different values.

References:
http://www.mathwarehouse.com/trigonometry/images/law-of-cosines/law-of-cosines-formula-and-picture2.gif

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

I/D2: Unit O - How can we derive the patterns for our special right triangles?

INQUIRY ACTIVITY SUMMARY
1. We can derive the 30-60-90 triangle from an equilateral triangle with a side length of 1 by using the information we already know about equilateral triangles. An equilateral triangle has the same length on all sides (1) and has equal degrees as well (60 degrees). If we cut this triangle down the middle, the side length that is cut will become 1/2 for the 30 degree angle. And, the degree formed will become a 90 degree angle, which will have a opposite side length of 1 still. Since we know the hypotenuse and one leg, we can use the Pythagorean Theorem to solve for the second leg.  The work for this is shown below.  In the end, we get the leg to be rad 3/2. If we were to multiply all these side lengths by 2, to make the numbers easier to work with, we would get: 1 (30 degree), 2 (90 degree), and rad 3 (60 degree). If we were to replace these values with n, we would get: n, 2n, and n rad 3. The "n" is a variable that depends on the triangle and shows the relation between the sides, the ratio.

2. We can derive the 45-45-90 triangle from a square with a side length of 1 by using the information we already know about squares. All the sides are equal and all the angles are 90 degrees. Therefore, if we cut the square along its diagonal, the two legs of the triangle formed will remain 1. Also, the right angle will be cut in half, so the angles would be 45 degrees. In order to find the length of the hypotenuse, we can use the Pythagorean Theorem. The work shown below gives the answer of c= rad 2. Therefore, if we get our sides as: 1 (45 degree), 1 (45 degree), rad 2 (90 degree). If we replace the values with n, we get: n, n, and n rad 2. The "n" is a variable that depends on the triangle and displays the ratio of the sides, or, in other words, the relationship between them.

 INQUIRY ACTIVITY REFLECTION
1. "Something I never noticed before about special right triangles is:" how they are derived from basic shapes, follow the same rules, and make logical sense as well as mathematical.
2. "Being able to derive these patterns myself aids in my learning because:" it allows me to fully comprehend where these values come from, not just accept them blindly.