Math problems

[silvercomet1525]

Originally Posted by lord_carbo

0.99999999999 repeating = 1.

Yup. This website may help for anyone who thinks otherwise. Nine different proofs on there.

[og4lif]

Originally Posted by og4lif

X = variable

(1/sin^2 X) + (1/cos^2 X) = (tan X + 1/tan X)^2
no calculators 4 this 1

We already figured that out but can any1 solve this?

[vifs]

Originally Posted by og4lif

We already figured that out but can any1 solve this?

umm... X = 2 dozen boxes of Pi

lol i could most likely solve it if i was in school, im pretty much shut down even though school started in 4 days

[MrRubix]

I'm a little rusty when it comes to trig identities but here goes:

(1/sin^2X) + (1/cos^2X) = (tan X + 1/tanX)^2

First I'd combine the first two terms:

(cos^2X+sin^2X)/(sin^2X*cos^2X) = (tan X + 1/tanX)^2

The numerator, cosX^2 + sin^2X, is equal to 1:

1/(sin^2X*cos^2X) = (tan X + 1/tanX)^2

This is the same as:

(1/(sinX*cosX))^2 = (tan X + 1/tanX)^2

So now we are basically trying to equate the things within the second-power-raise, so we aim to equate 1/(sinX*cosX) and tan X + 1/tanX.

Since tan = sin/cos and 1/tan or cot = cos/sin, we see that the right hand side is basically sin/cos + cos/sin, so we have two instances of each. Currently, the left hand side has half as many instances. Therefore we should try to prove this by simplifying the right-hand side:

tanX + 1/tanX
We then change tanX into tan^2X/tanX so we can combine like terms:
(1 + tan^2X)/tanX
And change our tangents into their sine and cosine equivalents:
(1 + (sinX/cosX)^2)/(sinX/cosX)
We then change the 1 into cos^2X/cos^2X so we can combine it as a like term later:
((cos^2X/cos^2X) + (sin^2X/cos^2X))/(sinX/cosX)
Combining terms:
(((cos^2X + sin^2X)/cos^2X))/(sinX/cosX)
cos^2X + sin^2X is equal to 1:
(1/cos^2X)/(sinX/cosX)
Then we simplify things a bit:
cosX/(sinX*cos^2X)
And finally drop the common cosX from the numerator and denominator:
1/(sinX*cosX)

Since we just proved 1/(sinX*cosX) = tanX + 1/tanX, we therefore prove:

(1/(sinX*cosX))^2 = (tan X + 1/tanX)^2

And since we showed that (1/sin^2X) + (1/cos^2X) = (1/(sinX*cosX))^2, we therefore prove:

(1/sin^2X) + (1/cos^2X) = (tan X + 1/tanX)^2

Edit: Oh yeah, QED

[og4lif]

good job rubix you don't look that rusty too me
must have taken a while too get that all down

[Relambrien]

For anyone too lazy to understand the (1/3) + (1/3) + (1/3) = 1 thing, I'll write an explanation that hopefully most can understand.

First, 1/3 = 0.333...

The "..." means that the sequence repeats infinitely, to the point where the last digit is a "3" and you cannot add a zero after that. It may sound weird, but trust me, that's the simplest explanation.

Now with simple algebra, we can understand more things.

0.333... = 1/3
0.999... = 1 (Multiply each side by 3)

But how can this be true? This is how:

Oh, and the space filter makes this a little hard to read. Sorry.

Let c = 0.999...

c = 0.999...
10c = 9.999... (Multiply each side by 10)
9c = 9.000... (Subtract 1c from each side)
c = 1.000... (Divide each side by 9)
c = 1 (Simplify)
0.999... = 1 (Substitution property of equality)

Yes, I stole this from Wikipedia.

And about the second problem, it just goes to show you that simplifying after every step is key . If you do that, you find out that you get 0=0 early on.