I/D3: Unit Q - Pythagorean Identities

Pythagorean Identities

We know that an identity is always true. So this makes the Pythagorean Theorem an identity, with a^2+b^2=c^2. We replace a,b, and c with x,y, and r making x^2+y^2=r^2. We need to make the theorem equal to 1. We do this by dividing everything by r. Our equation is now (x/r)^2+(y/r)^2=1. From our unit circle we know the ratio of cosine is x/r which means we can substitute cos for x/r. The ratio for sine from the unit circle is y/r and now we can plug in sin for y/r. Plugging both of these in gets us cos^2x+sin^2x=1 which is our main identity.

To derive the identity with secant and tangent from cos^2x+sin^2x=1 we must divide everything by cos^2. We now have sin^2x/cos^2x=1/cos^2x. From the Ratio identity we know tanx=sinx/cosx therefore we are able to plug in tan^2x for sin^2x/cos^2x. We are able to do this because the ratio identities are allowed to power up. As for the other side we know from the Recipricol identity that secx=1/cosx. The Reciprical identity is also allowed to power up, the only one not allowed to is the pythagorean identity. So now we are left with the final identity of 1+tan^2x=sec^2x.

To derive the identity with cosecant and cotangent we must divide everything by sin^2x. Giving us cos^2x/sin^2x+1=1/sin^2x. In the ratio identity cotx=cosx/sinx and even though the problem is squared the ratio identities are allowed to be powered up allowing it to be cot^2x to equal cos^2x/sin^2x. The reciprical identity says that cscx=1/sinx. This identity is also allowed to be powered up, so it equals csc^2x=1/sin^2x. The final equation is cot^2x+1=csc^2x. 

   INQUIRY ACTIVITY REFLECTION

The connections I see between units N, O, P, and Q are that they all use the trig functions in some way. They are also the same because it is all derived from the unit circle.

If I had to describe trigonometry in three words, they would be, simple once learned.

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

WPP 13 & 14 Unit P Concept 6&7

This WPP was made in collaboration with Daniel. Please visit the other awesome posts on their blog by going here.

Blake and Steve both see a ball. They get 3 miles apart east and west of each other to race to the ball. Blake runs at a bearing of 050* at 20 mph for 15 min to get to the ball. How far away is Steve away from the ball?

Law of Cosines

Once Blake gets to the ball he throws it at a bearing of N 25 W with a distance of .8 miles. When it lands Steve looks at the at a bearing of N 70 W. How far apart are Blake and Steve? 

BQ#1 - Unit P

BQ#1 - Unit P 

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

Here we have the non-right triangle and once we make a line straight down from B we get two right triangles and could use what we know with the trig functions.

To derive the Law of Sines we know that to get A we need to use Sine and get the opposite/hypotenuse giving us h/c. We multiply both sides by c to get h by itself. And now have cSinA=h. We must also do the same to get C. Taking the same steps as A, we will get aSinC=h. Since both equal h, this means they both equal each other, cSinA=aSinC. We now need to get SinA and SinC by themselves, so we divide both sides by ac. And c will cancel out on one side and a will cancel out on the other leaving us with SinA/a=SinC/c

1. 1. 
      
   
   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?
   
   We derive this formula because we first know the original equation is A=1/2bh. We dont know the h so to find this we must use the trig functions. We know that the SinC=h/a and to get h by itself we multiply each side by a. We now replace h with aSinC giving us A=1/2b(aSinC).
   
   
   This relates to the original area formula because we are using the exact same formula except replacing h because in order to find h we must use the trig function. And instead of just finding h and plugging it in, we plug in the trig function to find h in the original equation.

WPP #12: Unit O Concept 10

WPP #12: Unit O Concept 10

David is about to jog a a hill. The hill has a 25 degree elevations and is 46 feet high. What is the length he'd run to the top?

David gets to the top of the hill and takes a break. As he looks he sees the other side of the hill is much steeper with an angle depression of 32 degrees, he also sees a lone tree that is about 92 feet on the trail down. What is the length of the horizontal line from top of hill to the tree?

I/D #2: Unit O - Derive the SRTs

INQUIRY ACTIVITY SUMMARY

45-45-90

        

      To create a 45-45-90 triangle we must cut the square diagonally. Cutting the square diagonally will cut two of the 90 angles in half creating a 45 angle. To find the hypotenuse we must use the Pythagorean theorem. Since we know 'a' and 'b' we can plug that into the equation of a^2+b^2=c^2. Once we plug in 1 and 1 we get 2=c^2. We must get 'c' by itself and doing so by getting the square root of it canceling out the power of 2. What we do to one side we must do to the other and this gives us the 'square root of 2=c' and we now have the hypotenuse of 'square root of 2.'  We use 'n' on each side because the relationship between all 45-45-90 triangles would be the same. 

30-60-90

   

        To create the special right triangle of 30-60-90 we must cut an equilateral triangle right down the center. We do this to create two triangles and it divides the top 60 angle in half and creates a 30 angle in each triangle. Doing this also makes the 90 angles at the bottom of the triangle. Put it all together and we have the 30-60-90 triangle. Since the bottom length of the triangle was 1 cutting the triangle down the middle makes it 1/2 in both triangles. To get the height of the triangle we must use the Pythagorean theorem. a=1/2 and c=1 leaving us to find 'b', the height. Since a=1/2 we will multiply all sides by 2 to get rid of the 1/2. The equation will now be 1^2+b^2=2^2. After we solve for 'b' we will get the 'square root of 3' as the height. We again use 'n' to show the relation between all 30-60-90 triangles and how they each will have the same variables such as 2n,n, and n|3. 

 INQUIRY ACTIVITY REFLECTION

       Something I never noticed before about special right triangles is that every special right triangle will have the same relations no matter what the lengths of the triangle are.

Being able to derive these patterns myself aids in my learning because now I completely understand where the variables come from and it makes more sense of what I am doing when I am solving for problems that involve this.