SP#7: Unit Q Concept 2 Finding trig functions with one trig function and quadrant

This SP7 was made in collaboration with Vanessa Topete. Please visit her other awesome posts on her blog by going here.

The problem gives us information about our cosine and our sine. Cosine is 6 over radical 61 and our sine is greater than 1. With our knowledge of ASTC we can say that cosine is positive in quadrants one and four and that sine is positive in quadrants one and two. Our triangle should be in quadrant one, even though we put the triangle in quadrant two, we messed up on the very first step. Since we have cosine of theta we can use the reciprocal identity of sec of theta=1/cos of theta. Our cosine is 6 radical 61 and to cancel out the denominator of 6 radical 61. The problem ends up being 1/6 radical 61 but we multiply by the reciprocal and secant should equal 6 radical 61/61. Next we solved for sin used the pythagorean identity of 1-cos^2theta=sin^2theta. 6/radical 61 was used for cos^2theta, it ended up becoming 36/61, next we subtracted it by 1 and ended up with radical 25/61, then we get 5/radical 61, finally we multiply the numerator and denominator by radical 61 and sin theta=5radical61/61. Since we have cosine and sin we used the ration identity of cot theta=cos theta/sin theta. With this ratio set up radical 61 over 61 canceled out in the numerator and the denominator and cot theta ends up equaling 6/5. Finally we use the reciprocal identity of tan theta=1/cot theta. The function should be 1/(6/5) and we multiply by the reciprocal and tan theta equals 5/6. 

We must now use SOHCAHTOA and we should start off by using the pythagorean theorem to find the hypotenuse. The hypotenuse ends up being radical 61 and now we find our trig functions. The ratio for sin theta is y/r and the ratio for cosecant theta is r/y. Sin theta is 5radical61/61 and cosecant theta is radical 61/5. The ratio for cosine theta is x/r and the ratio for secant is r/x. Cosine theta is -6radical61/61 and secant is -radical 61/6. The ratio for tangent theta is y/x and the ratio for cotangent is x/y. Tangent is -5/6 and cotangent is -6/5. 

ID#3: Unit Q Concept 1: Pythagorean Identities

The pythagorean Theorem is an identity because it is a proven fact or formula that is always true. The derivation actually comes from the unit circle. In quadrant 1 of the unit circle we get a triangle with all positive values and the x-axis is x, the hypotenuse is r, which equals 1, and the other side is y. The first thing we do is set up our equation, which is x^2+y^2=r^2. Next we divide both sides by r and the r will cancel, leaving us with 1, and our equation should be (x/r)^2+(y/r)^2=1. With this information we can tell that x/r is cosine, and y/r is sine, so now our equation should be cos^2x+sin^2x=1.

One of the next two identities is derived from the first identity. The first thing we do is divide all sides by cos^2x and the equation should be (cos^2x/cos^2)+(sin^2x/cos^2x)=(1/cos^2x). Cos^2x/cos^2x cancels and becomes 1, thus leaving us with 1+(sin^2/cos^2x)=(1/cos^2x). From our identities we know that sin^2x/cos^2x is the same as tanx and that 1/cos is the same as secx and we end the final equation of 1+tan^2x=sec^2x.

For the last identity we must divide everything by sin^2x and we should end up with (sin^2x/sin^2)+(cos^2x/sin^2x)=(1/sin^2x). (sin^2x/sin^2x) cancels out and leaves us with 1, so our equation should be 1+(cos^2x+sin^2x)=(1/sin^2x). From our identities we know that cosx/sinx is the same as cotx and that 1/sinx is the same as cscx. Our final identity should end up being 1+cot^2x=csc^2x.

The connections I see between units N, O, P, and Q are that most of the information comes from the unit circle, such as the trig functions. 

If I had to describe trigonometry in THREE words, they would be tricky, cool, and cornfusing. 

WPP #13 & #14: Unit P Concepts 6 & 7: Applications of Law of Sins and Cosines

This WPP was made in collaboration with Vanessa Topete. Please visit the other awesome blog posts on her blog by going here. 

Swanson and Robert both love pickles. They see a pickle tied to a balloon floating in the air and decide to go after it using their helicopters. Swanson's helicopter is 300 ft away from Robert's helicopter. Swanson flies N25*E and Robert flies N57*W. What is the distance that they will travel to reach the pickle?

Swanson is 13.2 feet away from Robert. He sees that Robert's helicopter hit a cliff and blew up. Robert parachutes out and lands safely on the ground. Swanson sees that the exploded helicopter is 17 ft away from the pickle. If Swanson goes straight for the pickle, how far will he have to travel?

After Swanson reached the pickle he found out that it was not a pickle, but really, IT WAS A CUCUMBER!

BQ #1: Unit P concepts 3 and 4: Law of Cosines and Area of an oblique triangle

3. Law of Cosines

(http://math.ucsd.edu/~wgarner/math4c/derivations/trigidentities/lawofcosines.htm)

Law of cosines is important because it helps us find the angles of a triangle, when doing SSS, or the value of two other sides in a triangle, when we have SAS, in this triangle angle C is the vertex, angle b is acosC and asinC, and angle A is at (b,0). 

(http://www.themathpage.com/atrig/Trig_IMG/cos6.gif)

To find our coordinates we must first draw a line going down the triangle and we end up with two right triangles. For this problem we will be using c^2=a^2+b^2-2abcosC. Since our angle B is now made up of two angles we solve for side b, which is made up of x and b-x. The coordinate for angle B is acosC and asinC. Once we foil them our result should be: 

(http://math.ucsd.edu/~wgarner/math4c/derivations/trigidentities/lawofcosines_files/eq0002M.gif)

4.Area Formulas 

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

For the area of an oblique triangle we have to use the formula of finding the area of a triangle, the area of finding a triangle is 1/2bh. As seen in the picture above we draw a line going down the triangle so we have the height of the triangle. Also, depending on what angle we are solving for we can have three different formulas to use: 1/2bcsinA, 1/2absinC, and 1/2acsinB. 

References:

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

http://www.themathpage.com/atrig/Trig_IMG/cos6.gif

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

(

BQ #1: Unit P Concept 3 and 4: Law of Cosines and Area of an oblique triangle

( (http://math.ucsd.edu/~wgarner/math4c/derivations/trigidentities/lawofcosines.htm)   

3. Law of Cosines

Law of cosines is important because it helps us find the angles of a triangle, when using SSS, or the value of the other two sides in a triangle, when we have an SAS triangle. In this picture angle C is the vertex, angle B is acosC and asinC, and angle A is at (b,0). 

( (http://www.themathpage.com/atrig/Trig_IMG/cos6.gif)    

To find our coordinates we must first draw a line going down our triangle and we end up with two right triangles. For this problem we will be using c^2=a^2+b^2-2abcosC. Since our angle B is now made up of two angles we solve for side b, which is made up of x and b-x. The coordinates for angle B is acosC and asinC and we foil them. Once we foil them our result should be: 

(http://math.ucsd.edu/~wgarner/math4c/derivations/trigidentities/lawofcosines.htm)

4. Area Formulas

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

For the area of an oblique triangle we have to use the formula of finding the area of a triangle, the formula is 1/2bh. As seen in the picture above we draw a line going down the triangle so we have the height of the triangle. Also, depending on what angle we are solving for we can have three different formulas to solve with. The formulas are 1/2bcsinA, 1/2acsinB, and 1/2absinC.

References: 

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

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

http://www.themathpage.com/atrig/Trig_IMG/cos6.gif