SP# 7: Unit Q Concept 2: Find all Trig Functions when given one trig function and quadrant (using identities and SOH CAH TOA)

Finding All Trig Functions (with identities and SOH CAH TOA)

This SP7 was made in collaboration with Julian T. Please visit the other awesome posts on their blog by clicking here.

Given secx=-rad185/8 ; tanx= >0 . Find all the trigonometric functions.

Solved with Pythagorean/Ratio/Reciprocal Identities (Unit Q Concept 2): *note: The identity work is written in red letters.

First, we need to figure out in what quadrant our triangle is in given the clues. We must know this so we get the sign of our trigonometric functions correct. Based on our clues, our triangle lies in quadrant 2 meaning only tangent and cotangent will be positive, the rest will be negative.

Given secant, it is easy to find its inverse, cosine. What we do is use the reciprocal identity of secx=1/cosx.

Using the Pythagorean identity of 1+tan^2x=sec^2x we can find tangent.

Another Pythagorean identity which states that sin^2x+cos^2x=1 allows us to find cosine. We were then left with a value that could be positive or negative but from what we did in the beginning, we know sine must be negative.

Using the reciprocal identity of cscx=1/sinx we can find cosecant because we have already found sine.

We can then use the tangent which was already found in green writing to find cotangent using the reciprocal identity of cotx=1/tanx.

Our findings correspond with the prediction that the triangle lied in quadrant 2 and the signs correspond also.

Solved with right triangles [SOH CAH TOA] (Unit O Concept 5) *note: The SOHCAHTOA work is explained along with each picture.

Knowing Secant is -rad185/8 we can find cosine because we have both hypotenuse and adjacent. Our answer is negative, which corresponds with what we found in the beginning (that our triangle is in quadrant 2) since cosine is negative there.

From here we can use Pythagorean theorem to find our missing side, opposite. We find that it could be 11. 

Then we can find sin, now that we found opposite and were given hypotenuse.

Its inverse would be cosecant and is found by setting sine over one and rationalizing.

Tangent and cotangent are easily found because they are opposite over adjacent for tangent and adjacent over opposite for cotangent and we have already found the values for adjacent and opposite using what was given and the Pythagorean theorem.

From our answers and what we concluded given the clues we can conclude that our triangle looks something like this:

I/D# 3: Unit Q Concept 1: Pythagorean Identities

Pythagorean Identities

INQUIRY ACTIVITY SUMMARY:

Fig 1 (http://www.regentsprep.org/Regents/math/algtrig/ATT9/pythagoreanid.htm)

1. Where does sin2x+cos2x=1 come from?

a) An "identity" by definition according to Mrs. Kirch, is a "proven fact or formula that is always true". The Pythagorean theorem is an identity because it will work with any triangle that is legitimately a right triangle. Because right triangles reoccur in the unit circle (in 30, 60, and 45 degree angle smaller triangles), the Pythagorean theorem can be found all throughout.

b) The Pythagorean theorem is side x squared plus side y squared equals the hypotenuse (c) squared. These terms only work when we have a right angle, and luckily, we do in all 3 types of our unit circle triangles (30, 60, and 90).

Fig 2 (http://image.tutorvista.com/cms/images/38/pythagorus-theorem.JPG)

c) We can make the Pythagorean theorem equal to 1 simply by dividing both dies by c^2. This can be then simplified by writing the squared outside parenthesis, signifying everything inside is squared. What's left inside the parenthesis should look very familiar :)

What's left after simplifying is (x/r)^2+(y/r)^2=1

d) The ratio for cosine on the unit circle is (x/r),with x being "adjacent" and r being "radius". In the unit circle, r is always 1 so cosine can also be represented as simply "x" (x/1).

e) The ratio for sine on the unit circle is (y/r), with y being "opposite" and r being "radius". In the unit circle, the radius is always equal to 1 so sine can also be represented as simply "y" (y/1). 

f) The inside of the parenthesis in part c can be represented by the ratios we found.  (x/r)^2+(y/r)^2=1 becomes (cosine)^2+(sine)^2=1. I can conclude that the Pythagorean theorem is always true because it is a THEOREM, therefore the relationship I just came up with after simplifying is also true.

Fig 3(http://www.regentsprep.org/Regents/math/algtrig/ATT9/pythagoreanid.htm)

g) sin2x+cos2x=1 is referred to as a Pythagorean identity because when a circle is formed out of the unit circle with a radius of 1, the legs of that triangle can have ratios that are over 1 (cosine and sine) and thus represented simply as x and y. The picture above depicts this sort of triangle in the unit circle Because it is a right triangle, x^2 (leg 1 and also cosinex) + y^2 (leg 2 and also sinex) = 1 (radius in unit circle). 

h)Our "magic pair" of a 30 degree (rad3/2, 1/2) triangle can prove the validity of the Pythagorean identity.

2. You can derive the two remaining Pythagorean Identities from sin2x+cos2x=1 easily with both secant and tangent (a) and cosecant and cotangent (b).

a)

 Fig 4 (http://www.regentsprep.org/Regents/math/algtrig/ATT9/pythagoreanid.htm)

 Using our fundamental Pythagorean identity of sin2x+cos2x=1 we can derive a second equation that contains tangent and secant. Simply divide the formula by cos^2x. According to our ratio identities, sinx/cosx=tanx. We also know that the inverse of cosine (1/cosx) is secantx. By plugging in these new values we get tan^2x+1=sec^2x.

b)

Fig 5 (http://www.regentsprep.org/Regents/math/algtrig/ATT9/pythagoreanid.htm)

Starting off we our fundamental pythegorean identity of sin2x+cos2x=1, we can divide by sinx^2x. We know cosx/sinx=cotangentx because it is a ratio identity. We also know the inverse of sine (1/sinx) is cosecant x. Therefore, if we plug in those new values we know 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 same trigonometric ratios, the same validity tests that can be proven with our magic pair triangle coordinates, and the repeated use of the unit circle (with radius of 1 of course).

2. If I had to describe trigonometry in THREE words, they would be... ratios, identities, and derivations.

REFERENCES:

Fig 2: http://image.tutorvista.com/cms/images/38/pythagorus-theorem.JPG

Fig 1,3,4,5: http://www.regentsprep.org/Regents/math/algtrig/ATT9/pythagoreanid.htm

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

Heavenly Accident

This WPP 13-14 was made in collaboration with Julian T. Please visit the other awesome posts on their blog by clicking here.

Law of Sines Problem: 

Julian and Jeff were best of friends on the way to the BFC (Best Friend Convention) in Tijuana, Mexico when Mexican drug smugglers kidnapped Julian because he was an FBI agent. Jeff needed to rescue him. He planned and prepared, tracking down the smugglers so he could find Julian. Unfortunately, Julian and Jeff were both murdered. They became angels and both met up on a cloud that was 28 feet in width at horizontal locations opposite from one another. Julian was located at a bearing of 069 degrees and Jeff was located at bearing of 327 degrees. How many feet were they each from the gates of heaven?

After setting up the law of sines relationship, Jeff was 24 feet away and Jeff was 10.26 feet away from the gates of heaven.

Law of Cosines Problem:

After their fateful encounter, Julian and Jeff are left to explore heaven. Upon being distracted by different people and meeting up with past relatives, Jeff and Julian find themselves lost. They both started off at the gates of heaven. Jeff flew northeast 35 degrees (N35E) for 2.5 hours at a speed of 2 mph. Julian flew northwest 65 degrees (N65W) for 1.5 hours at a speed of 4 mph. How far are Julian and Jeff after they are done flying and have come to a stop?

In the end, the two best friends ended up in heaven and with the ability to fly could see each other whenever. 

BQ# 1: Unit P Concepts 1-5:Deriving Laws and Formulas

Big Question: Deriving oblique triangle laws and area formulas

PART 1 - Law of Sines:

Fig 1: http://hyperphysics.phy-astr.gsu.edu/hbase/imgmth/lsin.gif

The law of sines is needed when working with triangles that are not right triangles (those without a 90 degree angle measure). We need the law of sines to solve triangles (find all sides and angles), however, we can only use it after calculating the answers for corresponding sides and angles (i.e: knowing the values for angle B and side b). After having those values known, we can set up a proportion to solve for the rest. That proportion is derived from an oblique triangle and the result is what we know as the law of sines.

Deriving the Law of Sines: If we have an oblique triangle, we know that we can form a height from angle B that is perpendicular to the base to form two right triangles, like so:

If treated separately, those two triangles give different ratios for h. That is because in terms of sine, both triangles will use different values to form their ratios. In the end, the ratios are proportionate and h is the same value. With this relationship we can solve many triangles as long as we can calculate a relationship between a corresponding angle and side*.

*note: The law of sines will only work with AAS or ASA because these triangles are the only ones from which we can calculate a relationship corresponding with same angle and side. If it is not AAS or ASA the properties of the triangle would probably be found with law of cosines.  Examples of either are shown below:

Fig 2: http://www.mathwarehouse.com/trigonometry/law-of-sines/images/law-of-sines-and-cosines/law-of-sines-and-cosines-problem2.png

PART 2 - Area Formulas

The area of an oblique triangle is derived by substituting the value of h after forming two right triangles. We know that the traditional formula for the area of a triangle is:

Fig 3: http://www.calculateme.com/cArea/area-triangle-base-height.gif

If we do not know h, we can substitute it with a value of sine. The second cut triangle tells us h is equal to asinC.

The second cut triangle gives us a value for h:

This formula is the same as the formula for the area of a triangle, the value for h is just different, as it is written in terms of sine values. In the end, this formula works when the product of two sides and their included angle (i.e: SAS) is halved.

REFERENCES:
Fig 1: http://hyperphysics.phy-astr.gsu.edu/hbase/imgmth/lsin.gif
Fig 2: http://www.mathwarehouse.com/trigonometry/law-of-sines/images/law-of-sines-and-cosines/law-of-sines-and-cosines-problem2.png
Fig 3: http://www.calculateme.com/cArea/area-triangle-base-height.gif

WPP #12: Unit O Concept 10 - Elevation and depression word problems

This post will present, set up, and figure out a solution to a real life elevation/depression problem.

Introduction: Jeff's ski resort experience and findings

http://lodging4vacations.com/mammoth-holiday-home-rental/1-mammoth-mountain-ski-area.jpg

The Problems: Ascending up the gondola and observing the flag pole

Read the word problems carefully and answer the question. Show all work and steps neatly and correctly. Round answers tot he nearest tenth.

Jeff goes to Mammoth Lakes to enjoy his winter break. He it is at an elevation of 350 feet at a camping house in a city located on a hill. Jeff wants to see how much the ski gondola travels to take him to the top of the biggest mountain, white smoky. When he looks up, Jeff is looking at an angle of elevation of 13 degrees (He is looking at it while laying down on the floor, so no eye level must be found). He knows the mountain is 995 feet tall.

From that same lodging house in the city at an elevation of 350 feet on top of a hill, he makes eye with a flag pole that is planted at sea level and is 350 feet tall. He knows that flagpole is 300 feet away. Jeff wants to know what the angle of depression is when he observes where the flag pole is planted.

The Solutions: Jeff's ski calculations

These images show the solutions to the word problems. Make sure to work out your own before taking a look.

If he knows how elevated he is and the height of the mountain, Jeff can find the difference and set that as the value for opposite (645 ft). We are trying to find the ski gondola's traveled distance, which is the hypotenuse(x). We have the angle(13). We set sin13=645/x. It ends up being 2867.3 ft.

We know the adjacent distance (300) and the opposite (350) because the top of the flag pole and Jeff, which is elevated, are eye to eye. We don't have an angle measure. We set up tan x =350/300. To isolate x, we use arc tan and plug into the calculator. The angle is 49.4 degrees.