How to Prove a Difficult Trig Identity?

[soccertev]

Homework Statement

I need help proving this identitiy.

...cos2A-cos4A
tan3A = -------------- or tan3A = cos2A-cos4A/sin4A-sin2A
....sin4A-sin2A

there both the same, just different way of writing it. please help! :)

Homework Equations

The Attempt at a Solution

I honestly don't have any idea where to start :(

 

[Дьявол]

Hello! Do you feel comfortable to start with sin3A/cos3A ? My point is to prove the right side of the equation, so tg3A=sin3A/cos3A. My next step will be sin(2A+A)/cos(2A+A).
It will look like:
$$tan3A=\frac{sin3A}{cos3A}=\frac{sin(2A+A)}{cos(2A+A)}$$
Now use the trigonometric identities (product-to-sum formulas).
Consider the fact that this forum uses LaTeX. You can find the math expressions by clicking on the Σ button (on the right side of the tools).

 

[soccertev]

I honestly, don't understand what your doing here?

 

[Hootenanny]

I honestly, don't understand what your doing here?

What specifically don't you understand about the previous post?

 

[HallsofIvy]

He's using the sum formulas for sine and cosine. Do you know them?

 

[soccertev]

ok so now I'm stuck on

2sinAcosA+cos^2A-Sin^2ASinA / cos^2A-Sin^2AcosA+2sinAcosAsinA

 

[Chaos2009]

ok so now I'm stuck on

2sinAcosA+cos^2A-Sin^2ASinA / cos^2A-Sin^2AcosA+2sinAcosAsinA

I'm not quite sure where you get that expression. If you don't manipulating the tan3A side to prove the right side, you can try using the sum to product identities on the right side.
$$cos(a) - cos(b) = -2sin(\frac{a + b}{2})sin(\frac{a - b}{2})$$
$$sin(a) - sin(b) = 2cos(\frac{a + b}{2})sin(\frac{a - b}{2})$$

By the way, I like that Latex thing, now.

 

[soccertev]

Latex is very confusing for me, so sorry... ok well I'm at

sin2AcosA + cos2AsinA / cos2AcosA - sin2AsinA

i manipulated tan3A using the sum product identity

 

[soccertev]

actually i didn't use the product sum i used what i was told to use up a few posts

 

[soccertev]

chaos2009, how did they derive the product and sum formulas from the addition formulas?

 

[Chaos2009]

Hmm, well if your talking about the trigonometric identities, you just have to be really tricky about how you rewrite things.

 

[soccertev]

also, i think u made a mistake, in the cos- cos identity, isn't it = -2sin((a+b)/2)...

 

[Chaos2009]

Yes, yes it is.

 

[soccertev]

so how do i solve this then :S

 

[Chaos2009]

I'm not quite sure where the other people were trying to go with this problem. But, you can try simplifying the right side of your original equation using the sum to product identities. Your numerator is a cos x - cos y and the denominator is a sin x - sin y.

 

[soccertev]

i'm pretty sure this gives me -tan3A, does it not?

 

[Chaos2009]

After the sum to product identities, you should get something like:
$$\frac{-2sin(3A)sin(-A)}{2cos(3A)sin(A)}$$
The sine function is an odd function so:
$$sin(-x) = -sin(x)$$
Plug that back into what we got:
$$\frac{2sin(3A)sin(A)}{2cos(3A)sin(A)}$$
I don't think there should be a negative.

 

[soccertev]

ok, one last question, so to bug the hell outta u, can you explain how the sin function akes it back to positive

 

[Chaos2009]

It is that middle equation I have in my previous post. The sine function is an odd function which means:
$$sin(-x) = -sin(x)$$
Because we got a sin(-A) in the numerator, we replace it with -sin(A).
$$-2 * sin(3A) * sin(-A) = -2 * sin(3A) * -sin(A)$$
The two negative sign will kind of cancel each other out because a negative times a negative is a positive. Therefore, our new numerator is positive.

 

[soccertev]

why is sinA on the denominator positive, from what i understand sin((2A-4A)/2)= sin-A

 

[soccertev]

nvm :$

 

[icystrike]

$$-2(sin3A)\left[sin(-A)\right]/(cos3A)(sinA)$$
=$$2(sin3A)\left[sin(A)\right]/(cos3A)(sinA)$$
=$$\frac{sin3A}{cos3A}$$
=tan3A

I wondering if we can use LHS to prove RHS..
i can't do it, i tried.

 

[Дьявол]

$$-2(sin3A)\left[sin(-A)\right]/(cos3A)(sinA)$$
=$$2(sin3A)\left[sin(A)\right]/(cos3A)(sinA)$$
=$$\frac{sin3A}{cos3A}$$
=tan3A

I wondering if we can use LHS to prove RHS..
i can't do it, i tried.

Look at my second post. Maybe, it is a little bit messy to go with, but it is ok.

Starting with the right side it looks straight away.

 

[icystrike]

Look at my second post. Maybe, it is a little bit messy to go with, but it is ok.

Starting with the right side it looks straight away.

yes. of cos i know how to start wif the LHS.
but i can't derive with RHS, why not you try it too. (:
i also did tried with tan3A=$$\frac{tan2A+tanA}{1-(tan 2A)(tan A)}$$