SP 7: Unit Q Concept 2: Finding trig functions using identities

Original Problem: 

tan(x) = -1, cos(x)= 1

In this step above we have the substitution of cot that substitutes with 1/tan. 

In this next step we have the identity that includes both tan and sec. This identity allows us to simplify to a one term answer. 

 Next we come to another identity to where both cot and csc are found together. This identity allows the answer to come out to be one term and the steps include: adding, and square rooting. 

This following step shows the substitution of csc to sin. the substitution is to solve with the answer known for the answer we are in search of. 

These are the answers we came to finding with all the steps above. 

I/D #3: Unit Q: Pythagorean Identities

 Pythagorean Identities

1. sin2x=cos2x=1 comes from the the unit circle ratios. The sin2x symbolizes sin= y/r and the cos2x symbolizes cos= x/r. Combined together in an equation y/r turns into y, x/r turns into x, and added together it is x+y=r, or x2+y2=r2 which equals to one.

2. sin2x+cos2x=1

If we look back at our Unit Circle we can relate sin equaling to the ratio of y/r; leaving the answer to be y, since the r=1. To continue with the reference of the Unit Circle, cos has the ratio of x/r, leaving us with x. Plugging that into our original equation, we get y^2 + x^2 = 1. Looks similar, it should because it is just like the Pythagorean Theorem of a^2 + b^2 = c^2.

                                   

                                     

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

Please see my WPP13-14, made in collaboration with Anthony Lopez, 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 

The Law of Sines has great use when only given a few measurements of a not right triangle when needing all measurements. Figuring out the measurements of the angles and sides of a non right triangle is a bit more complex than with special right triangles because we cannot use SOH CAH TOA. Instead we use the law of sin to figure out one angle/side at a time. For example, in the picture below if we are only given Angle C, side b and Angle A, we can derive the law of sines to figure out the rest of the measurements. Given the measurements of two angles, we add them and divide by 180 to find the third angle. Once we find the third angle we match up sin (angle) B/ sin (side) b = sin (angle) of either A or C and continue that until all sides are found. 

(https://www.google.com/search?hl=en&site=imghp&tbm=isch&source=hp&biw=1680&bih=949&q=non+right+triangles&oq=non+right+triangles&gs_l=img.3..0.1425.6210.0.6666.23.21.2.0.0.0.150.1641.19j2.21.0....0...1ac.1.37.img..2.21.1450.ngevANF7olE#facrc=_&imgdii=_&imgrc=OMDCBtQcvuCROM%253A%3BibnpwnU4btR6lM%3Bhttp%253A%252F%252Fwww.drcruzan.com%252FImages%252FTrigNonRight%252FNonRightTriangleLabeling.png%3Bhttp%253A%252F%252Fwww.drcruzan.com%252FMathNonRightTrig.html%3B317%3B196) 

 4. Area formulas

Finding the area of oblique triangles all comes from the formula derived from the area of a triangle
A = 1/2bh. Using the picture below, we know that sinC= h/a, so to derive it to equal to the h we therefore get h = asinC. Next we substitute in our regular area equation for h and we get A= 1/2b(asinC). When solving for these triangles we are most likely going to be given two sides and one angle. The two sides are imputed into the equation as a and b. The angle that is given is always found with the use of sin. 

(https://www.google.com/search?hl=en&site=imghp&tbm=isch&source=hp&biw=1680&bih=949&q=non+right+triangles&oq=non+right+triangles&gs_l=img.3..0.1425.6210.0.6666.23.21.2.0.0.0.150.1641.19j2.21.0....0...1ac.1.37.img..2.21.1450.ngevANF7olE#hl=en&q=oblique%20triangle&tbm=isch&facrc=_&imgdii=_&imgrc=ffiBKZSnnuP3lM%253A%3B9k91py1yIcyOVM%3Bhttp%253A%252F%252Fwww.compuhigh.com%252Fdemo%252Flesson07_files%252Foblique.gif%3Bhttp%253A%252F%252Fwww.compuhigh.com%252Fdemo%252Ftriglesson07.html%3B780%3B355)

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

Problem: 

Solution:

I/D #2: Unit O Concept 7-8: How can we derive the patterns for our Special Right Triangles?

Inquiry Activity Summary 

30, 60, 90 Degree Triangle

In the 60, 60, 60 degrees triangle we use the rule that the triangles sides all equal one, and to get a  30, 60, 90 degrees triangle we must cut the original triangle in half, leaving us with one 30 degree angle, one 90 degree angle and last one remains 60 degrees. When dividing the triangle in half, we were left with one side equaling 1/2, the side across from the 30 degrees angle. The side across from the 90 degree angle is the hypotenuse which always equals 1. The next side is the one we search for. In searching for that side we use the Pythagorean Theorem (a^2 + b^2 = c^2) to give us the accurate answer, but to do that we use a variable that represents any number and  eliminates the fraction 1/2. plugging in the variable n, and equaling any number, such as 2, the 1/2 cancels to n, and the hypotenuse to a 2n. Next we plug those numbers and variables into the Pythagorean Theorem and solve. we are left with n radical 3; giving us our last side value.

45, 45, 90 Degree Triangle 

In a  90, 90, 90, 90 degrees square all sides are equaled to one; so in order to get a 45, 45, 90 degree triangle, we must divide the square into two triangles in half, diagonally. when dividing the square, we were left with two sides still equaling to 1, while the hypotenuse is undetermined so far. So we must use the Pythagorean Theorem (a^2 + b^2 = c^2) to solve for the "r" value; leaving us with radical 2. Once again the input of the "n" only means that it can be any number, as for the value does not always equal 1.

Inquiry Activity Reflection: 

1) Something I never noticed before about special right triangles is that the values that come with the sides and angles make sense in that they are not just random values, but are calculated to make a special right triangle.

2) Being able to derive these patterns myself aids in my learning because if i get lost and somehow forget my derivatives, i can use this method of finding the values of both the angles and the sides, by either using Pythagorean Theorem or simply remembering the steps and recalling the values.

ID #1: Unit N Concept 7: Knowing all Degrees and Radians Around the Unit Circle, Knowing all the Ordered Pairs Around the Unit Circle, Understanding and Applying ASTC to the Unit Circle

INQUIRY ACTIVITY SUMMARY

Many parts of this triangle have a specific label based on the Special Right Triangles rule. The first rule includes that the hypotenuse must equal 1 (ALWAYS). According to the 30, 60, 90 angles rules, the adjacent side to 30 (the side labeled x) has the value of x radical 3; the opposite side from the 30 degree angle (the side labeled y) has the value of 1/2; the hypotenuse of a 30 degree angle (the side labeled r) has the value of 2x . If this triangle were to be put in the unit circle across in the x- axis the origin (0,0) would be (radical 3/ 2, 0), and right above the previous point would be (radical 3/2, 1/2).

Along with the previous triangle, may rules apply to the sides and angles of the 45 degree triangle. The same rule about the hypotenuse applies to this triangle that it must equal 1 (labeled r) with the value of x radical 2. The adjacent side (labeled x) has the value of radical 2/2. the side opposite (labeled x) was valued the same of radical 2/2. The values are found when equaling the x's to the hypotenuse (x radical 2) leaving both of the sides equaling radical 2/2. If this triangle was inserted onto the unit circle, across from the origin (0,0) would lie the point (radical 2/2, 0), and above the previous point would be the point (radical 2/2, radical 2/2).

Just as the previous triangles, this 60 degree triangle includes rules to its sides and angles. The rule about the hypotenuse (labeled r) applies to this triangle with the value of 2x (the same as the 30 degree triangle). The side adjacent to the hypotenuse (labeled x) has the value of 1/2, and opposite from that side (labeled y) is valued radical 3/2. If this triangle were to be placed into the unit circle, across from the origin (0,0) on the x-axis would be point (1/2, 0), and right above would be point (1/2, radical 3/ 2).

                                                                                                               
All depending on where the ASTC is found on the circle, we will know which sin, cos, or cot is positive or negative. The ASTC starts from up right to the bottom from the left. Quadrant A (I) has ALL positive and NONE negative. Quadrant S (II), has csc/sin positive, and sec/cos, tan/cot negative. In quadrant T (III) tan/cot is positive, and csc/sin, cos/sec negative. Finally quadrant C (IV) has cos/sec positive and csc/sin, tan/cot negative. 

1. The coolest thing I learned from this activity was realizing that all along there were triangle involved that made the unit circle easier to understand. 
2. This activity will help me in this unit because it allowed me to find what i need to find using prior knowledge instead of just memorizing the entire unit circle. 
3. Something I never realized before about special right triangles and the unit circle is that they are connected through the point and radians values.