I/D #3 Unit Q Pythagorean Identities

Inquiry Activity Summary

1. So where do we get sin^2x + cos^2x =1? Well if we think back to our triangle, we can see in the picture below that x = cos,  y= sin and r = 1, and if we square this, we and up with our identity above.

   

   http://mathworld.wolfram.com/images/eps-gif/Trigonometry_1001.gif

   2. The two remaining identities can be derived from the identity above by dividing by cos^2x, so it would be sin^2x/cos^2x + 1 = 1/cos^2x we know from our reciprocal identity that 1/ cos^2x is equal to sec^2x and we know from our ratio identity that sin^2x/ cos^2x is equal to tan^2x so we are left with 1 + tan^2x = sec^2x To find the next one we dot he same steps except this time we divide by sin^2x which leaves us with 1 + cos^2x/ sin^2x = 1/ sin^2x after doing out identities, we are left with 1 + cot^2x = csc^2x 

Inquiry Activity Reflection

1. The connections that I have noticed between these previous units, is that we have had to use previous knowledge from our unit circle, the ordered pairs ect. Another thing I've noticed is that we have taken our roots of our ratios, from previous ones such as the pythagorean theorem and we've had to alter the because these are no longer 90 degree triangles. 
2. If I had to describe trigonometry in three words, I would say that it's puzzling, challenging and intriguing. 

SP #7 Unit Q Concept 2

This post was made in collaboration with Carlos Rendon, view other awesome math posts on his blog here

In our problem, we are given the information that sin<1 and also that cot is equal to -1/3
-Our first step was to find tan becasue we already have cot and we only have to do the reciprocal in order to get this. Our answer is -3 as shown in the picture below
-Now that we have tan, we can use the pythagorean identity 1 +tan ^2 = sec^2 so sec^2 is 1+ (9), radical 10
-We have sec so we can do the reciprocal in order to find cos, it is radical 10/ 10
-To find sin we can do 1- cos^2 and get 9/10, after we take the reciprocal and rationalize we should get 3 radical 10/ 10
-And to find csc we do the reciprocal of sin and get radical 10/ 3


                                                                     

In order to solve this using SohCahToa,
-We know sin is y/r which is 3/rad 10 and then it becomes 3rad 10/ 10 after we rationalize
-for csc it's r/y and it is rad10/ 3
- cos is x/r which is rad 10/10
-To get sec it's r/x and we get radical 10
-Next is tan, y/x which is -3
-And to get x/y (cot) it will be -1/3

                                                       

BQ #1 Unit P

Law of Sines- We need the law of sines whenever we only have two angles and a side of a triangle (ASA AAS). We use it by putting the sin of the angle divided by the side of that same letter, and then making it equal to the sin of an angle (if we have it) over the side of the same letter (we must have at least one of these in order for this to work)
                                                                             

http://www.mathwarehouse.com/trigonometry/law-of-sines/images/formula-picture-law-of-sines2.png

Area Formulas- The Area of an oblique triangle formula is A= .5* side *side times sin of angle. We must be given two sides and one angle of the triangle. The original area formula is A= .5 b *h. In this formula we substitute the sides in and the sin.

                                                                     

http://mathonweb.com/help_ebook/html/trig/trig27.gif

WPP # 13 & 14: Unit P Concepts 6 & 7

Please see my WPP-13-14 made in collaboration with Carlos Rendon by visiting their blog here. Also be sure to check out their other awesome blog posts.

WPP #12 Unit O Concept 10

http://piccolor.com/wp-content/uploads/2013/11/snow-mountain-wallpapers_22126_1920x1200.jpg

Erica looks to the top of a snow mountain, the mountain is 270 ft. tall, the angle from where Erica is stranding to the top of the mountain is 34 degrees. How far away is erica from the mountain (round to the nearest foot).
- tan 34 = 270/x
- x = 270/(tan 34)
- x = 401 feet

Erica decides that she wants to ski down the mountain. When she reaches the top, she looks back down at the cabin she was staying in. The mountain height is still 270 and the cabin is 750 feet away, what is the angle?
-tan x = 270/ 750
- Inverse tan = 270/ 750
- angle = 19.8 degrees

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

Inquiry Activity Summary

For the 45-45-90 triangle, we take a square and cut it in half diagonally. We cut diagonally in order to make two of the sides 45 degrees (cutting 90 in half). We can get the hypotenuse by using the pythagor theorem and solving for c. We use the variable 'n' in order to show that this is a ratio, and can be expanded or condensed.

1) In order to get a 30-60-90 triangle from an equilateral triangle, we have to cut it down the middle.This makes the hypotenuse equal to 1 and the bottom side equal to 1/2, using the pythagorean theorem we get the last side which is radical 3 over 2. The final step is to multiply all our sides by 2 so that we get 2, 1, and radical 3. In order to show that this is a ratio and can be used with larger or smaller numbers, we use the variable n.

http://www.freemathhelp.com/images/lessons/triangle306090-3.gif

2) In order to derive a 45-45-90 triangle from a square, we have to cut it in half diagonally, making two of the angles 45 (because we cut 90 in half). This gives us a hypotenuse of radical 2 ( from the original square side) And the other two side are both 1. If we use the pythagorean theorem, we can check this and it is true.

http://hotmath.com/hotmath_help/topics/45-45-90-triangles/triangle1.gif

3) It is important to note that we use variables in both of these to show that they can maintain the same ratio even if the numbers are altered, they all still have the same relationship.

Inquiry Activity Reflection

1) Something I never noticed before about special right triangles is how they both come from other shapes, such as the square and the equilateral triangle, and we just have to cut them in half.

2) Being able to derive these patterns myself aids in my learning because if I ever forget the relationship that these triangles have, I can always go back and solve it from memory.