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

Exponential Equations Fury

Yesterday a friend called me and asked me to help her with a system of exponential equations. At first I thought the system is impossible to solve. But then, when I thought all is lost, I started over and used another property of exponentiation. Then it all became clear.

This system is special, so if you know how to solve exponential equations try to solve it first. It gives a great feeling of satisfaction when solved:

2^{x - 1} * 3^{y + 1} = 72 * 6^{y - x + 2}
3^{x - 1} * 2^{y + 1} = 3 * 6^{x - y - 2}

At first, like most people, you'll be tempted to get everything in terms of powers of 2 and powers of 3 (since 6^x = 2^x * 3^x). However, this will get you nowhere. Instead, notice that both equations have terms with equal exponents, so let's take advantage of that and divide the first equation by the second (I put the factors in an order that will allow me to manipulate them later):

2^{x - 1} / 3^{x - 1} * 3^{y + 1} / 2^{y + 1} = 72 / 3 * 6^{y - x + 2} / 6^{x - y - 2}

Now let's use the exponentation rule that says aⁿ / bⁿ = (a / b)ⁿ. We also need the rule which says aⁿ / a^m = a^{n - m}:
(2/3)^{x - 1} * (3/2)^{y + 1} = 24 * 6^{y - x + 2 - (x - y - 2)}
Using the exponentation rule of 1/aⁿ = a^-n, we can turn the 2/3 into 3/2 and take the opposite of the exponent. Also, let's get the final version of the exponent of 6:
(3/2)^{1 - x} * (3/2)^{y + 1} = 24 * 6^{2y - 2x + 4}

Using the rule that says aⁿ * a^m = a^{n + m}, we can add the exponents of 3/2:
(3/2)^{y - x + 2} = 24 * 6^{2y - 2x + 4}

Look at the exponent of 6, looks familiar, right? Using the rule of (aⁿ)^m = a^nm, we can say:
6^{2y - 2x + 4} = 6^{2(y - x + 2)} = (6²)^{y - x + 2} = 36 ^{y - x + 2}

Now we have a really nice equation with equal exponents on both sides:
(3/2)^{y - x + 2} = 24 * 36^{y - x + 2}

Divide by 36^{y - x + 2}. Using the rules of exponentations mentioned above, this equals:
((3/2) / 36)^{y - x + 2} = 24
(1/24)^{y - x + 2} = 24¹
24^{x - y - 2} = 24¹
x - y - 2 = 1
x - y = 3
x = y + 3

Finally, a connection between x and y. Now let's plug it in one of the original equations to get the real values of x and y:
3^{x - 1} * 2^{y + 1} = 3 * 6^{x - y - 2}
3^{y + 3 - 1} * 2^{y + 1} = 3 * 6^{y + 3 - y - 2}
3^{y + 2} * 2^{y + 1} = 3 * 6¹
3² * 3^y * 2¹ * 2^y = 18
18 * 3^y * 2^y = 18
6^y = 1 = 6⁰
y = 0

Finally, a value. Let's plug that value in the easy equation: x = y + 3. That means x = 3. Problem solved.

Final answer:
x = 3
y = 0

That was long, but it was worth it. Now it's your turn - solve the two towns question using calculus and you will be in the hall of fame. Go ahead, try it.

Nadav

nadavs