## Wednesday, March 26, 2014

### SP#7: Unit Q Concept 2: Finding All Trig Functions when given one Trig Function and Quadrant ( Using Identities)

This SP7 was made in collaboration with Daisy Larco. Please visit her awesome posts on her blog by going here.

1.) Finding all trigs by using identities.

In the first step the first thing we need to do is to see what quadrant the trig function will lie in.
So, in this case it lie in the second quadrant because that is the only place where tangent is negative and sine is positive.
The second step is to find all the trig functions by using the identities. So we know that tangent is -5/7, so we can solve for cotangent. Then we plug in the given trig for tangent and simplify to find cotangent. Once we got tangent we can solve for secant, by using 1+tan^2 theta = sec^2. Then plug in tangent and simplify to get secant. Next we solve for cosine, because we already have secant. We just need to plug in the formula cosine theta = 1/ secant theta. Then solve for cosecant, by plugging cotangent in the formula 1= cot^2 theta = csc^2 theta. That will get us cosecant. Then the only thing left we need to find is sine. Just by plugging in cosecant into the formula sin theta = 1/csc theta. This will help you find theta.

2.) Finding all trig by using SOHCAHTOA.

In order to find all the trig values, we must use SOHCAHTOA. Sine = opposite/hypotenuse, y/r. Cosine = adjacent/hypotenuse, x/r. Tangent = opposite/adjacent, y/x. Cosecant = hypotenuse/opposite, r/y. Secant = hypotenuse/adjacent, r/x. Cotangent = adjacent/opposite, x/y.

## Wednesday, March 19, 2014

### I/D #3: Unit Q - Pythagorean Identities

INQUIRY ACTIVITY SUMMARY:

1.) Where does sin^2s+cos^2x=1 come from to begin with? You should be referring to Unit Circle ratios and the Pythagorean Theorem in your explanation.

a. What is an "identity"? Why is the Pythagorean Theorem an "identity"?
An identity is a proven facts and formulas that are true. The Pythagorean Theorem is an identity because it is proven to always true.

b. Why is the Pythagorean Theorem using x, y, and r?
The Pythagorean Theorem uses x, y, and r because based on a right triangle in quadrant 1. It uses x, y, and r are like a, b, and c for the Pythagorean Theorem.

c. Perform an operation that makes the Pythagorean Theorem equal to 1.
We would have to divide r^2. In order to make the formula equal to 1.

d. What is the ratio for cosine on the unit circle?
The ratio for cosine on the unit circle is x/r. Based on SOHCAHTOA, cosine is adjacent over  hypotenuse.

e. What is the ratio for sine on the unit circle?
The same as the top one above. The ratio for sine is opposite over hypotenuse.

f. What do you notice about part (c) in relation to parts (d) and (a)? What can you conclude?

g. What is sin^2x+cos^2x=1 referred to as a Pythagorean Identity?
sin^2x+cos^2x=1 is referred as the Pythagorean Identity because in order to find it we use the Pythagorean Theorem.

h. Choose one of the "Magic 3" ordered pairs from the Unit Circle to show that this identity is true.

2.) Show and explain how to derive the two remaining Pythagorean Identities from sin^2x+cos^2x=1. Be sure to show step by step.
a. Perform a single operation fairly to derive the identity with Secant and Tangent.

b. Perform a single operation fairly to derive the identity with Cosecant and Cotangent.

INQUIRY ACTIVITY REFLECTION:

1.) "The connection that I see between Units N, O, P, and Q so far are..."that all of them refers back to the Unit Circle. Including the Pythagorean to find all the sides. As well as find the angles by using 180ยบ and subtracting the number we have to find the missing angles.
2.) "If I had to describe trigonometry in THREE words, they would be..." interesting but difficult.

## Monday, March 17, 2014

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

This WPP 13-14 was made in a collaboration with Katie. Please visit the other awesome posts on her blog by going here.

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## Sunday, March 16, 2014

### BQ#1: Unit P Concept 1 and 4: Law of Sines, Area of an Oblique Triangle, and their Derivations

1. Law of Sines

Why do we need it?
We need the Law of Sines in order to solve for a triangle that are not right triangle. While for a right triangle we can just use the Pythagorean Theorem to find missing angles and sides.

How is it derived and what we already know?

First lets look at a triangle, we know that it is not a right triangle because it does have a 90 degree angle label.

http://math.ucsd.edu/~wgarner/math4c/derivations/trigidentities/lawofsines.htm

To make a triangle a right triangle, we must draw a line from the angle C to line AB. The red line we call it perpendicular line and label it h, now we have 2 right triangles. We know that it is a right triangle because it has a square box, which means it is 90 degree angle.

Lets recall the area of the right triangle to be Area = 1/2 base times height. To find the area of the triangle, we know that sin A = h/c and that sin c = h/a. Then we divide them the first one by c and the second by a. To get csinA = h and asinC = h.

http://math.ucsd.edu/~wgarner/math4c/derivations/trigidentities/lawofsines.htm

To find angle B, we must make a perpendicular line from angle A to BC. Then use sinB = h/c and angle C is sinC = h/b.

That will give us the formula for Law for Sines, which is:
http://www.mathsisfun.com/algebra/trig-sine-law.ht

http://math.ucsd.edu/~wgarner/math4c/derivations/trigidentities/lawofsines.htm

4. Area of an Oblique Triangle

An oblique triangle is a triangle with all sides with different length.

How is the area of an oblique triangle derived?
Since we know that sinC = h/a, sinB = h/a, and sinA = h/c. We multiply them by their denominator and you will get:

http://wps.prenhall.com/wps/media/objects/551/564474/StudyGuide/7c1h7_1.gif

http://www.compuhigh.com/demo/lesson07_files/oblique.gif

How does it relate to the area formula that you are familiar with?
It basically uses the same formula, you just plug into the equation and find the area. We just need to use sine in order to find the height.

References:
http://www.mathsisfun.com/algebra/trig-sine-law.html
http://math.ucsd.edu/~wgarner/math4c/derivations/trigidentities/lawofsines.htm
http://wps.prenhall.com/wps/media/objects/551/564474/StudyGuide/7c1h7_1.gif
http://www.compuhigh.com/demo/lesson07_files/oblique.gif

## Wednesday, March 5, 2014

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

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## Tuesday, March 4, 2014

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

Inquiry Activities Summary

1.) 30-60-90 triangle

First of all, to derive this triangle, I should start with an equilateral triangle. Which means they have all the sides equal to each other, in this case all the sides equal to 1. Not only it is equilateral but it is also equiangular, meaning the angle are the same measure as well. So for an equilateral and equiangular triangle they must add up to 180°.

First we start with a equilateral triangle, which where are the sides equal to 1.  Then split the triangle in half, which will give me a 30-60-90 triangle. Which makes the base of the triangle equal to 1/2. So, what we have already know is "a" which equal to 1/2 and is the adjacent. I also know "c," which is 1, which is the hypotenuse. In order to find "b," I must use the Pythagorean Theorem. The Pythagorean Theorem is a^2 + b^2 = c^2. Then, I plug a, and c into the Pythagorean Theorem, a = 1/2, b = ?, and c = 1, and I got 1/4 + b^2 = 1. Then subtract 1/4 and got 3/4, and then take the square root of 3/4 and got radical 3/2.

Once, I got b, which equal to radical 3/2, then I multiply 1/2, radical 3/2, and 1 by 2 to get rid of fractions. So, I would have the hypotenuse equal to 2, the opposite equal to radical 3, and the adjacent equal to 1. Next, I take "n" multiply to 1, 2, and radical 3 because "n,"shows the relationships between the three sides of the triangle.  That will give me a the derive 30-60-90 triangle.

2.) 45-45-90 Triangle

In order to derive this square into a 45-45-90, we start off with a square with all equal sides. That means the square adds up to 360° In this case which also means that the sides equal to 1. Then split the square diagonally, so it will make it a 45-45-90 triangle. So the triangle have to add up to 180°.

Based on the triangle, I already know what "a" and "b" equals, we just need to find "c."
So, I know that a = 1 and b = 1, to find "c," we also use the Pythagorean Theorem just like I did to find  the 30-60-90 triangle. The Pythagorean Theorem is a^2 + b^2 = c^2. So, I plug it in the formula, 1^2 + 1^2 = c^2. Then, I got 2 equal c^2, so I have to take the square root and got radical 2 for c. Then I multiply n to 1, 1, and radical 2 and got n, n, and n radical 2. That will give me a derive 45-45-90 triangle.

Inquiry Activity Reflection:
1.) "Something I never notice before about special right triangle is..."the formula was found based and it is based on the Unit Circle.
2.) "Being able to derive these patterns myself aids in my learning because..."if I happen to forget how to solve for a special right triangle I can refer back to this.