Cosine Simplification

Yahoo Answers is definitely the place everyone goes to find an answer for a math question. Otherwise, there is no explanation for the kinds of questions you can find there. Today's question is about simplification of a trigonometric expression:

Simplify: (cos 3x)(1 - cos 2x + cos 4x - cos 6x)

To solve that, we will need some major identities: the sum of cosines, the difference of cosines, the sum of sines, and the product of sines and cosines. With these formulas, this question can be easily solved.

Let's start with the formulas:
cos α + cos β = 2cos((α + β)/2)cos((α - β)/2)
cos α - cos β = -2sin((α + β)/2)sin((α - β)/2)
sin α + sin β = 2sin((α + β)/2)cos((α - β)/2)
sinαcosβ = 1/2(sin(α + β)sin(α - β))

Now, to avoid complications, let's aim for the average. We can turn 1 into cos 0, so we have the cosines of 0 and 6x and the cosines of 2x and 4x. In both formulas they are divided by two, so they should probably cancel each other.

cos 0 - cos 6x = -2sin((6x + 0) / 2)sin(6x - 0)/2 = -2sin² 3x
cos 4x - cos 2x = -2sin((4x + 2x)/2)sin((4x - 2x)/2) = -2sin 3x * sinx

In conclusion, the problem has just become:
(cos 3x)(-2sin² 3x - 2sin 3x * sin x)

Distribute:
-2sin 3x * sin 3x * cos 3x - 2sin 3x * cos 3x * sinx

Transform into a sum of a cosine and a sine (I picked α and β to be 3x, because there will be a sine of zero, which is zero):
-2 * 1/2 * (sin(3x + 3x) + sin(3x - 3x)) * sin 3x - 2 * 1/2 * (sin(3x + 3x) + sin(3x - 3x)) * sin x
-sin 6x * sin 3x - sin 6x * sin x

We are getting really close now, just hold on:
-(sin 6x)(sin 3x + sin x)
-(sin 6x)(2sin((3x + x)/2)cos((3x - x)/2)
-2sin 6x * sin 2x * cos x

And that's it.

Nadav

nadavs

Sinusoid Temperature

Temperatures are very hard to predict. Even weathermen get them wrong most of the times. However, there are some paradises which have a constant temperature range, such as Brazzaville (which is in Congo). Here is the question with the amazing details:

The temperature in Brazzaville is given by the following function:
C = 17 - 6cos(πt/12 + 5)
0 <= t < 24

1. What are the highest and lowest temperatures on Brazzaville?

2. What is the biggest rate of change (biggest drop/rise) in temperature?

To show how people complicate questions, I'm going to answer it using calculus, and then using a much simpler method.

The highest and lowest temperatures are achieved when the derivative of the function is 0, so let's do it:
C' = 6π/12 * sin(πt/12 + 5)
0 = πsin(πt/12 + 5) / 2

Divide by π / 2:
0 = sin(πt/12 + 5)
πt/12 + 5 = π*k (where k is an integer)
πt + 60 = 12πk
πt = 12πk - 60
t = 12k - 60/π

Now all we need to do is find the k values which give t's that match our requirement. Since 60/π is about 19, we need k's that are bigger than 1, so:
t₁ = 24 - 60/π = 4.901
t₂ = 36 - 60/π = 16.901

To verify which number is a maximum and which is a minimum, we can use the second derivative:
C'' = 6π²/144 * cos(πt/12 + 5)

Plug the t's:
t₁ makes it positive, so it's a minimum.
t₂ makes it negative, so it's a maximum.

Plug the two t's in the original equation:

C = 17 - 6cos(πt₁/12 + 5) = 17 - 6 = 11°C
C = 17 - 6cos(πt₂/12 + 5) = 17 - 6(-1) = 23°C

The temperature in Brazzaville ranges between 11°C - 23°C for the entire year. We can only be jealous of them...

To get the biggest rates of change, we need to find the maximum of the derivative, so we'll set the second derivative to zero and find the two points. One will give out a negative change and the other one will give a positive change. They are opposites of each other, since the cosine function is symmetrical on a vertical line going through a maximum or minimum.
C'' = 6π²/144 * cos(πt/12 + 5)
0 = 6π²/144 * cos(πt/12 + 5)
cos(πt/12 + 5) = 0
πt/12 + 5 = ±π/2 + 2πk (again, k is an integer)
πt + 60 = ±6π + 24πk
πt = ±6π + 24πk - 60
t = ±6 + 24k - 60/π

The t values in bounds are:
t₃ = 10.901
t₄ = 22.901

Plug them in C':
C' = π/2 * sin(πt₃/12 + 5) = π/2 = 1.57 °C/hour
C' = π/2 * sin(πt₄/12 + 5) = -π/2 = -1.57 °C/hour

So the sharpest temperature changes are 1.57 °C per hour.

This question can be solved using a much easier way: common sense.

The t inside the function is multiplied by π and divided by 12, so plugging in t values from 0 to 24 will give values from 0 to 2π inside the cosine function. This means it goes through an entire cycle from t = 0 to t = 24. With that, we can figure out that the maximum value for the cosine function is 1 and the minimum value is -1. Plug these two numbers and you get:
C = 17 - 1*6 = 11°C
C = 17 - (-1)*6 = 23°C


Also, the cosine function is the "steepest" when it goes through the x-axis, or more accurately, when the argument makes the cosine zero. All we need to do is find when the argument is zero and plug it in the derivative:
πt/12 + 5 = ±π/2 + 2πk

I already solved that, so I'm not going to do it again.

As you can see, thinking outside the box can really help when it comes to such long and tiring functions.

Yours,
Nadav

nadavs

Altitudes of Obtuse Triangles

Today's question is simple geometry. It deals with triangles, altitudes, sines, and other mathematical functions and objects. The question is quite long, but its answer is not that long. I guess it's another case for the "question theorem" - the length of the question and the length of the answer are inversely proportional.

In triangle ABC, A = 65 degrees, B = 13 degrees, C = 102 degrees. A line perpendicular to AC intersects the line defined by AC at point P. The perpendicular line also passes through point B. The length of PB is 17. Find the area of triangle ABC.

Get your ruler and draw this. If you read correctly, you should draw an obtuse triangle with an altitude outside the triangle. This altitude forms two right triangles: one is outside triangle ABC and one is including triangle ABC.

Since angle BCP is the supplement of angle C, it is 180 - 102 = 78 degrees. Since the outside triangle is a right triangle, we can calculate side BC of triangle ABC:
sin 78 = 17/BC
BC = 17/sin 78 = 17.379

Well, we have one side. To find the area of ABC, we need another side, and then we can use the area formula of 1/2 * a * b * sin C. It never fails.

Now that we have BC, we have two options. Use the law of sines or use a more creative way: find the hypotenuse of triangle APB. I always vote for creativity, so let's find that side.

Angle B is 13 degrees, as given. The angle adjacent to B, CBP, is 12 degrees (the complement of 78 degrees). This makes angle ABP a total of 25 degrees. Now that we have an angle and a side in a right triangle, we can find the other sides. Let's find side AB, because it uses the cosine function, which is... the cosine function:

cos 25 = 17 / AB
AB = 17 / cos 25
AB = 18.757

Now we have everything for the area formula: we have side c (AB), side a (BC), and angle B. Let's plug them all in:
S = 1/2 * c * a * sin B
S = 36.667 sq. units.

Problem solved. Wasn't that hard, was it?

Hope you enjoyed. There are more nice things like this at Super Math Tips.
Nadav

nadavs

Geometric Construction and the Sinus Function

Today I have a fantastic question from Yahoo Answers about geometric construction. The idea of geometric construction is to use a compass and a ruler (with no measurements) to construct different geometric shapes and figures.

Notice: to answer this question you need to know basic constructions: copying a segment, creating a segment X times larger than another, find a perpendicular bisector, and copy an angle. Without those you'll be lost.

Here is the question:

A triangle has sides a, b, and c. The ratio a/b = 7/4. You are given side c and the radius of the circumcircle, r. Construct the triangle.

This may seem simple, but it's a little more sophisticated than that. To construct that triangle, we need the law of sines.

The law of sines says that in a triangle, a/sin A = b/sin B = c/sin C = 2R (side a and angle A are opposite, R is the radius of the circumcircle). Let's play with this law a little:
a/sin A = b/sin B
That means:
a/b = sin A / sin B

In our triangle, a/b = 7/4. So all we need to do is find two angle whose ratio of sines is 7/4. So how do we do that?

The definition of sine A is the y-coordinate on the unit circle with the angle measure of A (read here about the unit circle). First, construct two segments, a' and b' in a way that a'/b' = 7/4. Now create a circle with a radius of more than a' and two perpendicular axes that intersect at the center of the circle. Position a' in the circle in such a way that it's perpendicular to the x-axis and touches the circle in one point (yet not tangent to it). Construct the line from the origin to the point of intersection of a' and the circle. Call the angle between the x-axis and that line angle A. Do the same thing for b' and angle B.

Since we used the definition of sine, we now have two angles with a sine ratio of 7/4. Now we need to create the triangle we want.

First, construct segment c. Since the circumcenter, the center of the circumcenter, is on the point of intersection of all perpendicular bisectors, construct the perpendicular bisector of segment c. Now, since the radius is given, use the end point of segment c and the perpendicular bisector to find the circumcenter and draw the circumcircle.

Now, when segment c is inside the circle, copy angle A to one of its sides and angle B to the other side. Complete the triangle, and you are done.

Feel free to send in more questions!
Nadav

nadavs

More Trigonometric Identities Revealed

For today there is another question from Yahoo Answers, also about trigonometric identities. This time it's not about simple ones like before, but a more complex one, like the ones I sometimes answer on super math tips. Here it is:

Prove:
(cos 4x + cos 2x) / (sin 4x + sin 2x) = cot 3x

At first glance this seems irritating and impossible, but with some creative thinking it's very possible.

Now, what do 2, 3, and 4 have in common? Of course, 3 is the average, and we're going to use that property heavily.

Since we want to have cot 3x at the end, let's change all sines and cosines to 3x + x or 3x - x and hope they cancel:
(cos (3x + x) + cos (3x - x)) / (sin (3x + x) + sin(3x - x)) = cot 3x.

Well, they don't cancel so fast, so we'll have to expand all expressions (remember what's sin (A + B)?) and then hope that we can cancel out some terms. Let's start (I put the arguments in parentheses so you can see them better):
(cos (3x + x) + cos (3x - x)) / (sin (3x + x) + sin(3x - x)) = cot 3x
Let's start from the numerator:
cos(3x)cos(x) - sin(3x)sin(x) + cos(3x)cos(x) + sin(3x)sin(x)
2cos(3x)cos(x)

Now, the denominator:
sin(3x)cos(x) + cos(3x)sin(x) + sin(3x)cos(x) - cos(3x)sin(x)
2sin(3x)cos(x)

Now divide those two terms and see what happens:
2cos(3x)cos(x) / 2sin(3x)cos(x)
The 2 and cos(x) cancel:
cos(3x) / sin(3x) = 1/(sin(3x) / cos(3x)) = 1/tan(3x) = cot (3x)

Job complete.

Besides solving these questions, here is a challenge for you: try to solve the previous question (with the towns) using calculus. I haven't solved it myself yet (except for the geometry part, which is easy).

The first one to solve (on comments or email) will be written in the hall of fame on the blog's list (which I'll create) with a link of his choice and the right to post here any question.

Go on to solving!
Nadav

nadavs

Trigonometry on the Table

Today I have a very simple question from Yahoo Answers. Hopefully tomorrow I'll be able to find something a little more interesting:

Prove that tan² x + 1 = sec² x

Every trigonometry student must remember this identity, but here is the reason why the trig students must know this identity:

As you all should know, sin² x + cos² x = 1 (if you don't know that, blame your math teacher). Now all we have to do is use some more trigonometric identities and rules of exponents, and we're all set:

tan x = sin x/cos x
sec x = 1/cos x
(a/b)ⁿ = aⁿ/bⁿ

Now we take the basic trigonometric identity and divide it by cos² x:
sin² x / cos² x + cos² x / cos² x = 1 / cos² x
(sin x / cos x)² + 1 = (1/cos x)²
tan² x + 1 = sec² x

Simple, nice, and easy.

Bring in more questions!
Nadav

nadavs