Who here is good with the Precalc Trig?

[ZIO]

I'm coming asking for help. I know at my age, I should be past this, but I'm horrible with math.

I need help with something, and a real bad help. Since this is a social room, this topic isn't necessarily out of place.

I NEED A TUTOR!!!!

 

[nevershootme]

on all topics on trig?

 

[hungrybox]

I am a Pre-Calc wizard.

whazzup?

 

[ZIO]

I thought I'd just ask a broad question like that.

I specifically need help with some problems I want to answer from a chapter 7 in my book. The problems look and sound much like this -

Prove the identity -

1+tan(B)/1-tan(B)=cos2B/1-sin2B

I been told this stuff is easy, but I can't get my head around it for some reason.

 

[BBQ°]

for those kind of questions, I'm pretty sure you need side1 to look like side2.

start with the most complex side, and use your knowledge of identities to convert the complex trig stuff to the simple trig stuff on the other side.

Uhhh probably doesn't make sense.

 

[ZIO]

I mean, I have the rules right in front of me. It's a matter of observing what you got.

See, the (1+tanB)/(1-tanB) could be converted to a (1+{sinB/cosB})/(1-{sinB/cosB})

But I'm not sure about that as the other side

cos2B/(1-sin2B) could then be converted to (cos^2B-sin^2B)/(1-2sinBcosB)

If that makes any sense to you. But, see, now I need to know which side is right, and even then, what is it I am looking for? <_>

 

[hungrybox]

Hmmm

those involve trig identities...

don't know those yet.

=/

 

[dashdancedan]

cos2x also = 1 - 2sin²x

Ya, sometimes it's hard cause there are multiple ways to prove them and you don't know where to start. But just pick one side and change stuff to identities you know.

 

[Player-3]

or cosxsquared - 1

 

[dashdancedan]

Actually 2cos²x - 1

 

[NC-Echo]

Ill try to help later tonight. I'm pretty sure I remember how to do those.

 

[bladeofapollo]

cos2B/(1-sin2B) could then be converted to (cos^2B-sin^2B)/(1-2sinBcosB)

Start w/ that equation.

cos^2(B) - sin^2(B) (Eq 1)
-----------------------
1-2sin(B)cos(B)

We know:

sin^2(B) + cos^2(B) = 1 (Trig identity)

Therefore:

1 - 2sin(B)cos(B) = [sin^2(B) + cos^2(B)] - 2sin(B)cos(B) (Eq 2)

You can now take the square root of the side on the right.

sin^2(B) - 2sin(B)cos(B) + cos^2(B) = [sin(B) - cos(B)]^2 = [cos(B) - sin(B)]^2 (Binomial Square Theorem)

We now substitute our new Binomial Square into equation 1

cos^2(B) - sin^2(B) (Eq 3)
-----------------------
[cos(B) - sin(B)]^2


Now by Difference of Squares (a - b)*(a + b) = a^2 - b^2

cos^2(B) - sin^2(B) = [cos(B) + sin(B)]*[cos(B) - sin(B)] (Eq 4)

Sub'ing Eq 4 into Eq 3, we get:

[cos(B) + sin(B)]*[cos(B) - sin(B)] (Eq 5)
--------------------------------------
[cos(B) - sin(B)]^2

We see that one term of [cos(B) - sin(B)] cancels, and are left with

cos(B) + sin(B) (Eq 6)
-----------------
cos(B) - sin(B)

And I'm working on the rest.

 

[dashdancedan]

Also, you can multiply the 1+tan(B)/1-tan(B) * 1+tan(B)/1+tan(B)
and get 1+ 2tan(B) + tan²(B)/1-tan²B

2tan(B) can then be 2tan(B)/1-tan²(B) and 1+tan²(B) can be sec²(B) or 1/cos²(B)

I don't know if any of those help you, but those are options to consider.

 

[XIF]

just know that everything can be written in terms of sine and cosine.

tan= sin/cos
sec= 1/cos
csc= 1/sin
cot= cos/sin

iirc, may have secant and cosecant mixed up, should be in your book.

sin^2 + cos^2 = 1 is monumentally important as well.

there are I believe a couple more identities you may need to know, but that will get you very far.

Proofs in trig can either be fun or super frustrating. All you have to do is write both sides of the equation, and work on both sides in such a way that they are identical. I would start with the tangent side since the other side is already pretty simple, and if you get stuck on that side, leave it and fiddle with the other side to try and make them identical. At a glance, I would try to convert tan to sin/cos, and then change 1 to cos/cos and add the two, and see where that gets you.

 

[BBQ°]

to add on to xif:

just fyi, for your final answer, you can only work with one side. You may not work side1 halfway and then work side2 halfway until you get them to be equal. although if you are able to do that, it shouldn't be a problem for you to get the answer legally.

at least that's what I've been taught.

 

[bladeofapollo]

Finally

1 = cos(B)/cos(B) = sin(B)/sin(B)

So from the left side of the equation

1 + tan(B)
------------
1 - tan(B)

=

cos/cos + sin/cos
--------------------
cos/cos - sin/cos

=

cos + sin
-----------
cos - sin

 

[Masky]

..........

 

[Gerbil]

I wasn't online

*Pumps his Math Degree (in progress lawl)*

 

[Sub_Scorpion7]

I'll do the work for you:

First, convert the tan's to sin/cos and the number 1's to cosx/cosx

cosx/cosx + sinx/cosx
--------------------------- =
cosx/cosx - sinx/cosx

cosx + sinx
-------------- =
cosx - sinx

Now multiply the top and bottom by (cosx + sinx), which is the same as multiplying by 1

cos^2x + sin^2x + 2sinxcosx
------------------------------------ =...........Remember that cos^2x + sin^2x = 1
........cos^2x - sin^2x ........................and cos^2x = [1 + cos(2x)]/2 ; sin^2x = [1 - cos(2x)]/2

1 + 2sinxcosx..........cos(2x)
------------------ = ----------------...........Also remember that 2sinxcosx = sin(2x)
.....cos(2x)...........1 - sin (2x)

1 + sin(2x)............cos(2x)
------------------ = ----------------...........Cross multiply
.....cos(2x)...........1 - sin (2x)

1 - sin^2(2x) = cos^2(2x)

1 = sin^2(2x) + sin^2(2x)

1 = 1

Edit: I guess spaces don't work properly on Smashboards...

 

[Masky]

..........

 

[bladeofapollo]

Sub_Scorpion and I know what's up.

And by that I mean we got here first.

 

[Sub_Scorpion7]

I think I found your problem
-RockCrock

I think I know your problem.

You don't know math.

Edit: WRECKED.^2

 

[Masky]

..........

 

[Sub_Scorpion7]

It was indirectly implied.

 

[Hawks go Caw]

Ready to have your mind blown? Okay.
First, some identities:
tan(x) = sin(x)/cos(x)
cos(x)^2- sin(x)^2 = cos(2x)
cos^2(x) + sin^2(x) = 1
sin(2x) = 2cos(x)sin(x)

1. Starting with Left Hand Side, replace tan(x) with sin(x)/cos(x):
1+tan(x)/1-tan(x) = 1+[sin(x)/cos(x)]/1-[sin(x)/cos(x)]

2. Simplify numerator and denominator by finding a common denominator and adding:
1+[sin(x)/cos(x)]/1-[sin(x)/cos(x)] becomes:
[(sinx+cosx)/cosx]/[(cosx-sinx)/cosx]

3. The cosine x's cancel so you have:
(sinx+cosx)/(cosx-sinx)

Nothing tricky so far. Just used the first identity and simplified the expression. The trickiest part is the next step.

4. Multiply the top and bottom by (sinx-cosx). You'll be left with:
(cosx^2-sinx^2)/(cosx^2-2cosxsinx+sinx^2)

Notice that the top part can be simplified now to cos(2x) based on the second identity stated. The denominator can be rearraged to become:
(cosx^2+sinx^2)-2cosxsinx

As you know, the term in parenthesis equals 1 by Pythagorean Theorem. The second portion (2cosxsinx) is equal to sin2x from our last stated identity. So subbing this all back into the equation yields:

5. cos2x/(1-sin2x)

LHS = RHS

QED

 

[ZIO]

As you know, the term in parenthesis equals 1 by Pythagorean Theorem. The second portion (2cosxsinx) is equal to sin2x from our last stated identity. So subbing this all back into the equation yields:

5. cos2x/(1-sin2x)

WAIT! Why isn't the result for this 1/(sin2x)? EDIT - okay. I read through again and see where I was misunderstanding again.


Also, I appreciate the help. This was only one of many problems I was having, but like, I'll need someone to sit with me to go through it. I can't force you to step me through all of them. But showing me what goes on here may help guide me in my other problems.

You can now take the square root of the side on the right.

sin^2(B) - 2sin(B)cos(B) + cos^2(B) = [sin(B) - cos(B)]^2 = [cos(B) - sin(B)]^2 (Binomial Square Theorem)

We now substitute our new Binomial Square into equation 1

cos^2(B) - sin^2(B) (Eq 3)
-----------------------
[cos(B) - sin(B)]^2

You lost me at this step.

 

[Sandtru27]

I am a Pre-Calc wizard.

whazzup?

Hmmm

those involve trig identities...

don't know those yet.

=/

lol STOMP'd

 

[thegreatkazoo]

I'm coming asking for help. I know at my age, I should be past this, but I'm horrible with math.

I need help with something, and a real bad help. Since this is a social room, this topic isn't necessarily out of place.

I NEED A TUTOR!!!!

I used to tutor in this for a year. Get @ me on gchat or aim or send a VM if you want help.

NB I will be able to help you starting on the 11th, as dead week is kicking my @**.

 

[bladeofapollo]

You can now take the square root of the side on the right.

sin^2(B) - 2sin(B)cos(B) + cos^2(B) = [sin(B) - cos(B)]^2 = [cos(B) - sin(B)]^2 (Binomial Square Theorem)

We now substitute our new Binomial Square into equation 1

cos^2(B) - sin^2(B) (Eq 3)
-----------------------
[cos(B) - sin(B)]^2

Ok. Binomial Square is (a - b)*(a - b) = (a - b)^2 = a^2 - 2ab + b^2

We take
a = cos(b)
and
b = sin(b)

and you get
[cos(b) - sin(b)]^2 = cos^2(b) - 2cos(b)sin(b) + sin^2(b)

It's Binomial Square backwards. Is that any more enlightening?

 

[null55]

lol this **** was so fun, i remember.

 

[dashdancedan]

oh ya

10oh yas

 

[MarsFool!]

Actually (a+b)(a-b) = (a-b)² If I'm not mistaken.

a²-2ab +b²

bro, watch your brackets. The exponent outside of the brackets makes your equation

(a -b)(a -b)

L2FOIL
(a + -1b) * (a + -1b)
(a(a + -1b) + -1b * (a + -1b))
((a * a + -1b * a) + -1b * (a + -1b))


((-1ab + a2) + -1b * (a + -1b))
((-1ab + a2) + -1b * (a + -1b))
(-1ab + a2 + (a * -1b + -1b * -1b))
(-1ab + a2 + (-1ab + 1b2))


(-1ab + -1ab + a2 + 1b2)

-1ab + -1ab = -2ab
(-2ab + a2 + 1b2)

I think you meant (a²-b²)

 

[dashdancedan]

O ya, that's what I meant.

 

[XIF]

a²-2ab +b²

bro, watch your brackets. The exponent outside of the brackets makes your equation

(a -b)(a -b)

L2FOIL
(a + -1b) * (a + -1b)
(a(a + -1b) + -1b * (a + -1b))
((a * a + -1b * a) + -1b * (a + -1b))


((-1ab + a2) + -1b * (a + -1b))
((-1ab + a2) + -1b * (a + -1b))
(-1ab + a2 + (a * -1b + -1b * -1b))
(-1ab + a2 + (-1ab + 1b2))


(-1ab + -1ab + a2 + 1b2)

-1ab + -1ab = -2ab
(-2ab + a2 + 1b2)

I think you meant (a²-b²)

if you had an easy binomial product to solve, would we still call it Pracets?