I/D3: Unit Q - Pythagorean Identities

INQUIRY ACTIVITY SUMMARY

To find where sin^2x+cos^2x=1 we have to use the pythagorean theorem. X^2+Y^2=R^2. First we need to get R^2 to equal one. we do this by dividing R^2 by itself and divide everything else by R^2. we end up getting X^2/R^2+Y^2/R^2. After, we know that X^2/R^2= Cos^2x because we remember our unit circle. Y^2/R^2= Sin^2x. so we know that Cos^2x+Sin^2x=1.

It's good to know these, but these aren't the only trig functions we need to find. We also need the Tan^2x and the Cot^2x. We divide everything by Cos^2x. We come up with Sin^2x/Cos^2x+1=1/Cos^2x. Sin^2x/Cos^2x=Tan^2x since we remember the trig functions. The final answer will be Tan^2x+1=sec^2x.

To find Cot^2x we divide everything by Sin^2x this time.

We get 1+cos^2x+1/Sin^2x. from this we just simplify and we know that Cos/Sin = Cot because of our unit circle.  And 1/Sin is equal to Csc. so in the end we will get 1+Cot^2x=Csc^2x.

INQUIRY ACTIVITY REFLECTION

1. The connections that i see between Units N, O, P, and Q so far are the use of the trig functions throughout all of these. Also we have kept the unit circle through all of this.

2. If i had to describe trigonometry in THREE words, they would be eww (not sure if thats an acceptable word, if not then 'i don't like it'), difficult and helpful.

I.D#2. :Unit O :Concept.7-8. How can we derive the patterns for our special right triangles?

INQUIRY ACTIVITY SUMMARY:

In order to derive the pattern for a 45-45-90 triangle from from a square with a side length of 1, we first need to cut the square from corner to corner, as shown in the picture, to get the triangle. Since the square has a side length of 1, we know that the x and y both also equal one so we can label the sides 1.

Now you have the sides, so all you need no is the hypotenuse. To find it we will use the pythagorean theorem. a^2+b^2=c^2. plug in the numbers we have. Since we know a and b both equal 1 then we use 1+1=c^2 since 1^2 still equals 1. We end up having C^2=2. so we have to cancel he square by finding the square root of c^2 which will equal C... But what you do to one side, you do to the other so it'll end up being the square root of 2.

Now that we have the  numbers, we will now add the variable because the sides are not always gonna equal 1. You can use any letter for the variable, i used X. Since both a and b both equal 1 we can just use X for both of those sides. To find the hypotenuse we have to use the pythagorean theorem. we plug in X^2+X^2= 2X^2

Now that we're done with the 45-45-90, we can move on to the 30-60-90 from an equilateral triangle with a side length of 1 and angles of 60-60-60. We start by cutting it right down the middle.

 Since we cut it right down the middle. the top corner is also cut in half so instead of a 60, its now 2 30's. And also the bottom is cut in half so instead of being 1 is it 2 .5's or 1/2's.

We, again, have to use Pythagorean theorem to find the the Y or b. First we have to manipulate a (1/2)^2+b^2=1. Once we do, it will end up being b^2=1-(1/4). That means b will equal to radical3/2.

since these are just for the side, the letter variables you use are the exact same thing. Hypotenuse=n

x=n/2 y=nradical3/2. To get rid of the fractions all you do is multiply each side by 2. Hyp.=2n  x=n  y=n radical3. N is just what you use when The hypotenuse is unknown. so the completed triangle should look like this.

INQUIRY ACTIVITY REFLECTION

Something I never noticed before about special right triangles they are the exact same things as the Unit circle.

Being able to derive these patterns myself aids in my learning because it makes me independent and i don't have to rely on anything because i have all this in my head.

I/D #1: Unit:N Special Right Triangles and the Unit Circle. How do they relate?

Inquiry Activity Summary

To derive the unit circle values, in class we would use visuals to help us remember. We used the Special Right Triangles to find the dimensions of the UC.

We First have to label the triangle according to the rules of SRT. We will use the 30 degrees triangle for this example. Since we are using a unit circle we know that the radius is one. so we have to divide it by 2X to make it one. And what you do to one, you do to the others. then you just use algebra to solve it. Once you have the X and Y you plot it at the top right corner.

 The 30 degrees triangle has angles 30,60,90 and the sides are r=1, x=radical3/2, y=1/2.

 The 45 degree triangle has the angles of 45,45,90. The sides are r=1, x=radical2/2, y=radical2/2

The 60 degree triangle has the angles 60,30,90. the sides are r=1, x=1/2, y=radical3/2. The 60 is the same as the thirty but just switched t the Y and X.

The points the top right corner that you write are the points on the circle that fits the degrees. Each of these triangles lie in the 1st. quadrant.

You can use the SRT concept for every quadrant but you have to change the sign in either the X or Y or both depending on which quadrant you want it in.

“The coolest thing I learned from this activity was if you can't memorize the unit circle, you can just use the SRT to get the answers.”

“This activity will help me in this unit because it will make it easier to complete the unit circle.”

“Something I never realized before about special right triangles and the unit circle is how they really do relate to one another and if you cant remember one, use the other for help.”