I have a Pre-Calc question

[thehstrybean]

OK, so my pre-calc teacher doesn't explain anything, and I don't understand "building up a denominator" on a trig verification. My exam's tomorrow, so any help is most appreciated.

EX: Verify 2tan(x)*sec(x)=(1/(1-sin(x)) - (1/(1+sin(x))
now I have written down (this was an example in class)
=1(1+sin(x))-1(1-sin(x))
------------------------------ (yeah, that's a fraction)
1(1+sin(x))*1(1-sin(x)

Now why does that happen?

 

[Zim Hosein]

What's the prize?

 

[thehstrybean]

That's a secret...but everyone loves it...espicially you, Zim...

 

[JayHu]

Originally posted by: thehstrybean
OK, so my pre-calc teacher doesn't explain anything, and I don't understand "building up a denominator" on a trig verification. My exam's tomorrow, so any help is most appreciated.

EX: Verify 2tan(x)*sec(x)=(1/1-sin(x))-(1/1+sin(x))
now I have written down (this was an example in class)
=1(1+sin(x))-1(1-sin(x))
------------------------------ (yeah, that's a fraction)
1(1+sin(x))*1(1-sin(x)

Now why does that happen?

getting a common denominator?
\frac{1}{1-sinx}-\frac{1}{1+sinx}

multiply the first one by \frac{1+sinx}{1+sinx} and the second one by \frac{1-sinx}{1-sinx}, and then subtract the tops.

 

[chuckywang]

Originally posted by: JayHu

Originally posted by: thehstrybean
OK, so my pre-calc teacher doesn't explain anything, and I don't understand "building up a denominator" on a trig verification. My exam's tomorrow, so any help is most appreciated.

EX: Verify 2tan(x)*sec(x)=(1/1-sin(x))-(1/1+sin(x))
now I have written down (this was an example in class)
=1(1+sin(x))-1(1-sin(x))
------------------------------ (yeah, that's a fraction)
1(1+sin(x))*1(1-sin(x)

Now why does that happen?

getting a common denominator?
\frac{1}{1-sinx}-\frac{1}{1+sinx}

multiply the first one by \frac{1+sinx}{1+sinx} and the second one by \frac{1-sinx}{1-sinx}, and then subtract the tops.

LaTex > Fusetalk

 

[JayHu]

Originally posted by: chuckywang

Originally posted by: JayHu

Originally posted by: thehstrybean
OK, so my pre-calc teacher doesn't explain anything, and I don't understand "building up a denominator" on a trig verification. My exam's tomorrow, so any help is most appreciated.

EX: Verify 2tan(x)*sec(x)=(1/1-sin(x))-(1/1+sin(x))
now I have written down (this was an example in class)
=1(1+sin(x))-1(1-sin(x))
------------------------------ (yeah, that's a fraction)
1(1+sin(x))*1(1-sin(x)

Now why does that happen?

getting a common denominator?
\frac{1}{1-sinx}-\frac{1}{1+sinx}

multiply the first one by \frac{1+sinx}{1+sinx} and the second one by \frac{1-sinx}{1-sinx}, and then subtract the tops.

LaTex > Fusetalk

I hope it's clearer than trying to draw a picture like he did.

 

[thehstrybean]

How come the minus stays at the top and you multiply at the bottom?

 

[chuckywang]

BTW, you should get 2*sin(x)/cos(x)^2 when you get a common denominator, which = 2*tan(x)*sec(x), since tan(x) = sin(x)/cos(x) and sec(x)=1/cos(x).

 

[ts3433]

Originally posted by: thehstrybean
How come the minus stays at the top and you multiply at the bottom?

That's how you merge fractions into one with a common denominator. All the merging is done in multiplication, so all addition operations that were between fractions stay because you now have all the fractions together (example: (2/4)+(7/4) = (2+7)/4).

Originally posted by: chuckywang
BTW, you should get 2*sin(x)/cos(x)^2 when you get a common denominator, which = 2*tan(x)*sec(x), since tan(x) = sin(x)/cos(x) and sec(x)=1/cos(x).

I got that from simply expressing the functions of the original in terms of sine and cosine (easier than manipulating that other side of the equation). Should be good.

 

[thehstrybean]

But what if they don't have common denominators? This one is two fractions, but one has a 1-sin(x) and the other has 1+sin(x)

 

[chuckywang]

Originally posted by: thehstrybean
But what if they don't have common denominators? This one is two fractions, but one has a 1-sin(x) and the other has 1+sin(x)

You make them have a common denominator.

 

[ts3433]

Originally posted by: thehstrybean
But what if they don't have common denominators? This one is two fractions, but one has a 1-sin(x) and the other has 1+sin(x)

As I speak, I'm banging my head against a wall. That's why you merge the fractions into one that has a common denominator.

I suppose I'll give you a more detailed procedure in case it wasn't clear earlier:
1. Multiply both denominators to get your common denominator: (1+sin x)(1-sin x) = 1-sin^2 x = cos^2 x.
2. The left original fraction doesn't have 1+sin x in it. Multiply that fraction by 1+sin x/1+sin x (yes, that is equal to one; that's how you're supposed to do it so that you don't change the value of the fraction).
3. Do something similar for the other one.
4. Combine the fractions and do the rest of your simplification.

 

[thehstrybean]

Ok, so you mulitply the bottom so they have common terms, then you multiply each side, so 1/1-sin(x) is multiplied by 1+sin(x), and the other side gets done with the other bottom term. I get that, but where do the 1s in 1(1+sin(x)) come in? They're on each side, but it doesn't make any sense as to why they're there.

 

[ts3433]

Forget about them--anything times 1 is itself. You should remember that from... third grade?

 

[thehstrybean]

Originally posted by: ts3433
Forget about them--anything times 1 is itself. You should remember that from... third grade?

Yeah, but I was wondering where it came from specifically. I think I get it...so, as promised, here is the suprise:























































Ready?







































































:beer: & for everyone! (and more :beer: for you, Zim...:heart: )

 

[blinky8225]

(1/(1-sin(x)) - (1/(1+sin(x))

They come in from the original problem. They are left in the example that you copied down in class for your understanding but are unnecessary as 1 * anything is itself.

 

[DrPizza]

I hope my eyes mis-read something above...
NO, in a trig identity, you cannot multiply both sides by something...

As they said above, multiply the top and bottom of a fraction by the same thing.
(I type and write "top" and "bottom", but I say "numerator" and "denominator")

 

[Zim Hosein]

Originally posted by: thehstrybean

:beer: & for everyone! (and more :beer: for you, Zim...:heart: )

:heart:

Cheers thehstrybean :beer:

 

[thehstrybean]

Originally posted by: DrPizza
I hope my eyes mis-read something above...
NO, in a trig identity, you cannot multiply both sides by something...

As they said above, multiply the top and bottom of a fraction by the same thing.
(I type and write "top" and "bottom", but I say "numerator" and "denominator")

No, not both SIDES, but both sides of one side...the LHS or RHS of the equal sign...