I/D3: Unit Q - Pythagorean Identities

Inquiry activity Summary

1.      Where does sin^2x+cos^2x=1 come from to begin with?

a.       To explain this we need to think back and remember basic trig functions. We know that sin=y/r and cos=x/r. The unit circle gives us a value for r which is r=1. We also need to recall that the Pythagorean Theorem is a^2+b^2=c^2. Now we have everything we need to prove this. First we choose an ordered pair form the unit circle, like (1/2,3/2). We can now plug this in to (y/r) ^2+(x/r) ^2=1. The equation we should plug into our calculator is to (1/2)^2+ (3/2)^2=1. This shows how we can use the unit circle and the Pythagorean Theorem to get sin^2x+cos^2x=1.

2.      Show and explain how to derive the two remaining Pythagorean Identities from sin^2x+cos^2x=1

a.      Once you read this paragraph you will definatly realize how easy it is to derive the rest of the Pythagorean Identitites. First, you have your original identity: sin^2x+cos^2x=1 you will divide this by cos^2. Once you have done this your new identity will be 1+tan^2x=sec^2x. To get the next identity you dived the original by sin^2, you will get 1+cot^2=csc^2.

Inquiry Activity Reflection

1. The connection that I see between Units N,O, and P are trig functions, trig ratios, and pythagorean theorem.

2. If I had to describe trigonometry in 3 words, they would be challenging, a little frustrating and rewarding. 

WPP 13-14: Unit P Concepts 6-7: Application of the Law of Sines and the Law of Cosines

Please see my WPP13-14, made in collaboration with Sarah Ngo, by visiting their blog here.  Also be sure to check out the other awesome posts on their blog.

BQ#1 Unit P Concept 1 and 4: Law of Sines and Area Formula

    1.  Law of Sines

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

Why do we need it?

• We need the law of sines because sometimes we are not dealing with right triangles. We don’t have the normal trig functions to solve non-right triangles. However, if we drop a perpendicular line h we can use the trig functions to help us solve the non-right triangles.

How is it derived?

• First, we cut the triangle in half. This reveals two right triangles within the non-right triangle. The law of sine’s are derived by taking the sine of angle A and angle C. we should end up with sinA= h/c & sinC= h/a Then we solve for h and simplify. This gives us c x sinA =h & a x sinC= h. If both equations are equal to h then they are both equal to each other. Now, to simplify this set a denominator of ac and cancel. You should end up with. SinA/a = SinC/c

2.   Area Formulas

• How is it related to the previously known formula? This formula is related to it because it is basically the same formula. The only difference is that we don’t know the value of h. to find the value of h we substitute (a x Sin C). In the next paragraph we will learn why and how we substitute this value.
• How is it derived?  The area formula is derived by finding the value of h. we know the are of a triangle is A= ½ bh. We know that SinC=h/a we solve for h, h= a x SinC. Plug this into the formula, A= ½ b( a x SinC ).

WPP #12: Unit O Concept 10- PICTURE

ANGLE OF ELEVATION

A man is on top of a mountain, the mountain is 200 feet tall the angle of depression is 35°. How far away ( laterally ) is he from the village at the bottom?

ANGLE OF DEPRESSION

The Los Angeles police department is preparing to invade a house. The angle of elevation is 45°, if the house is 50 feet away from the helicopter what is the distance that the officers have to drop from the helicopter to the ground?

I/D #2 Unit O- How Can We Derive the Patterns of Our Special Right Triangles?

INQUIRY ACTIVITY SUMMARY

          30 60 90 Triangles

·        The first thing you want to do is cut the equilateral triangle in half. To find the degrees we know that the bottom right corner is equal to 90, to bottom left corner is equal to 60 and the top since it was split in half is 30. After this, label the sides (a, b, c) from this we learn that a=1/2 b= unknown, and c=1. To find b we use the Pythagorean Theorem.

After this you should get b to equal (√3/2).now, we multiply everything by 2. The values should equal 1n, n√3, 2n. We use N because it is a variable, that shows that the ratio can be used in other multiples of the triangle.

45 45 90 Triangles

·        For this triangle we also start by cutting it in half diagonally. The angles are much easier to find in the triangle. Label all sides a, b, & c. we should find that a=1 and b=1.

 We plug these values in the Pythagorean Theorem and find that c=√2. We multiply all these values by “N” because it is a variable and it shows that this ratio works with all multiples of this triangle. Our final ratio should be a=n b=n & c=√2

INQUIRY ACTIVITY REFLECTION

·        Something I never noticed before about special right triangles is… you can derive the ratios for these triangles by setting the sides to 1 and using the square and the equilateral triangle.

·        Being able to derive these patterns myself aids in my learning because… when I forget the ratios I can still find them by using what I learned.