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

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Tangent Fractions Identities

Today's question comes from Yahoo Answers, and it includes trigonometric identities, fractions, and some hard work. Stay with me here, it will take some time.

Prove the identity:
(1 - tan x)/(1 + tan x) = cos 2x / (1 + sin 2x)

As you know, when working out an identity, you need to pick a side and work out until you reach the other side. Since the left side is more complex, I'll work from there to the simple right side. Here we go:

(1 - tan x)/(1 + tan x)

Using the tangent quotient (tan x = sin x / cos x), we can write the fraction as:
(1 - sin x / cos x)(1 + sin x / cos x)
(cos x / cos x - sin x / cos x)/(cos x / cos x + sin x / cos x)
((cos x - sin x)/cos x)/((cos x + sin x)/cos x)

Notice that both denominators are cos x, so they can be cancelled:
(cos x - sin x)/(cos x + sin x)

Multiply both parts of the fraction by (cos x + sin x)
(cos x - sin x)(cos x + sin x)/(cos x + sin x)²

Distribute the parentheses:
(cos² x - sin² x)/(cos²2 + 2sinxcosx + sin² x)

As you should know, cos² x - sin² x = cos 2x and cos² x + sin² x = 1, so:
cos 2x / (1 + 2sinxcosx)

As you should also know, 2sinxcosx = sin 2x, so:
cos 2x / (1 + sin 2x)

Q.E.D.

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

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

Trigonometric Identities in Fractions

Today I have another trigonometric identity. I also found a nice questions in modular arithmetic, but it will have to wait until tomorrow, since this trigonometry question is much better. The question is short, which means it has a long solution. Here it is, try before you peek:

A + B + C = pi (A, B, and C are in radians).
Prove:
(cot A + cot B) / (tan A + tan B) + (cot B + cot C) / (tan B + tan C) + (cot A + cot C) / (tan A + tan C) = 1

Yeah, that's all you need to prove. That given the first condition, this long sum equals 1. And no, you can't multiply everything by the denominators. That's not how you prove an identity (and besides that, I'm not sure you want to multiply three binomial terms).

All we're left to do is hard work. Lots of fun. Let's see what happens when you take each fraction and try to simplify it a little (I'll write cot A as cotA for simplicity, same for B and C):

First, let's start with the numerator:
cotA + cotB = cosA/sinA + cosB/sinB
Make a common denominator and add:
(cosAsinB + cosBsinA) / (sinAsinB)
The numerator here looks like a trigonometric identity which says:
sin(A + B) = sinAcosB + cosAsinB
So the numerator of the first fraction is:
sin(A + B) / (sinAsinB)

Now the denominator:
tanA + tanB = sinA/cosA + sinB/cosB
Again, make common denominator and add:
(sinAcosB + sinBcosA) / (cosAcosB)
Looks familiar, doesn't it?
sin(A + B) / (cosAcosB)

Now you have the numerator and the denominator of the first fraction. Let's see what we can get out of them:
sin(A + B)/(sinAsinB) / sin(A + B)/(cosAcosB)

As you can see, sin(A + B) cancels (thank god) and we're left with (cosAcosB)/(sinAsinB). As you should know, this equals to cotAcotB. Well, now we have something we can work with.

I'm not going to show work on the other two fractions, I just assume you'll understand how I make this transition now:
cotAcotB + cotBcotC + cotAcotC = 1

Now that's more like it. Let's take cotC out as a common factor. Considering that C = pi - A - B, this can make our life easier. Follow closely now:
cotAcotB + cotC(cotA + cotB)
cotAcotB + cot(pi - A - B)(cotA + cotB)
Since tan(pi - x) = -tan(x), cot(pi - x) = -cot(x), so:
cotAcotB + cot(A + B)(cotA + cotB)

tan(A + B) = (tanA + tanB) / (1 - tanAtanB), so cot(A + B) = (1 - tanAtanB)/(tanA + tanB).

cotAcotB - (1 - tanAtanB)/(tanA + tanB) * (1/tanA + 1/tanB)
Make a common denominator and add, again:
cotAcotB - (tanA + tanB)/(1 - tanAtanB) * (tanA + tanB)/(tanAtanB)

Luckily, we can cancel (tanA + tanB):
cotAcotB - (1 - tanAtanB)/(tanAtanB)
cotAcotB - 1/(tanAtanB) + (tanAtanB)/(tanAtanB)

The first fraction equals cotAcotB and the second one equals 1:
cotAcotB - cotAcotB + 1
= 1

Proved.

That was long, but worth it, wasn't it?

Tomorrow, if there's nothing better, we'll have some modular arithmetic.

Yours,
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