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

Proof of Trigonometric Inequalities

After yesterday's inequality induction, it's time for some trigonometric induction. The proof is a little hard to understand at first, but it's definitely worth seeing.

Prove:
xsin(1/x) < 1 when x > 0

This question can definitely fit in the short question-long answer template. However, it doesn't necessarily fit short question-hard answer.

First, we know that x is positive, so we can divide by it without changing the inequality sign:
sin(1/x) < 1/x

Now the argument of the sine function and the right side of the inequality are the same, so let's call them y. All we have to prove now is that sin y < y for every positive y.

First, when y is greater than 1, this is obvious. A sine can never be greater than 1. When y is 1, there is no problem either, because sin 1 is about 0.84 (the argument is in radians, don't forget). The problem begins when 0 < y < 1.

To solve that, let's derivate the inequality we want to prove. We get cos y < 1. As you can see, the rate of growth of sin y is cos y, and the rate of growth of y is linear. When y goes up by 0.5, sin y goes by less than that. This means that sin y is always less than y when 0 < y < 1.

Since y = 1/x, we can conclude that:
sin(1/x) < 1/x (when 1/x > 0, which means x > 0)
xsin(1/x) < 1

Problem solved.
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

Trigonometric Derivation

Today I have a question about differentiation with the definition of differentiation, which means using the lim_{h -> 0} (f(x + h) - f(x)) / h. However, today it is going to be about trigonometric functions (it's always about trigonometry, isn't it?).

Differentiate cos(x) using the definition of differentiation.

First, the most important limit to remember is this: lim_{x -> 0} sin(x) / x = 1. When you have this in mind, you can solve anything (well, many things).

Let's write what we need to find:
lim_{h -> 0} (cos(x + h) - cos(x)) / h

Using the identity of the cosine of sums of angles, we can get:
lim_{h -> 0} (cos(x)cos(h) - sin(x)sin(h) - cos(x)) / h

Now follow closely:
lim_{h -> 0} ((cos(x)cos(h) - cos(x)) / h - sin(x)sin(h) / h)
lim_{h -> 0} (cos(x)cos(h) - cos(x)) / h - lim_{h -> 0} sin(x)sin(h) / h
lim_{h -> 0} cos(x)(cos(h) - 1) / h - sin(x) * lim_{h -> 0} sin(h) / h
lim_{h -> 0} cos(x)(cos(h) - 1)(cos(h) + 1)/h(cos(h) + 1) - sin(x) * lim_{h -> 0} sin(h) / h
lim_{h -> 0} cos(x)(cos²(h) - 1)/h(cos(h) + 1) - sin(x) * lim_{h -> 0} sin(h) / h
lim_{h -> 0} cos(x)(-sin²(h))/h(cos(h) + 1) - sin(x) * lim_{h -> 0} sin(h) / h

Now watch the beauty of mathematics reveal itself:
L'Hopitals rule says that to find lim_{x -> a} f(x)/g(x), you can also find lim_{x -> a} f'(x)/g'(x). Let's do that on the first limit:
lim_{h -> 0} -sin²(h) / h
= lim_{h -> 0} (-sin²(h))' / h'
= lim_{h -> 0} -2sin(h)cos(h) / 1
As h approaches 0, sin(h) also approaches zero, so the entire limit is zero, and thus the whole first part of the first big limit.

This leaves us with:
- sin(x) * lim_{h -> 0} sin(h) / h
As you can see, we have lim_{h -> 0} sin(h) / h = 1
That means:
-sin(x) * lim_{h -> 0} sin(h) / h = -sin(x) * 1 = -sin(x)

So the entire derivative equals -sin(x).

That's why super math tips exists. Or you can simply use Wikipedia's list of trigonometric derivatives (but where is the fun in that?).

Nadav

nadavs