# Very difficult trig identity

1. Nov 20, 2008

### soccertev

1. The problem statement, all variables and given/known data
I need help proving this identitiy.

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

2. Relevant equations

3. The attempt at a solution

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

2. Nov 20, 2008

### 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).

3. Nov 20, 2008

### soccertev

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

4. Nov 20, 2008

### Hootenanny

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

5. Nov 20, 2008

### HallsofIvy

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

6. Nov 20, 2008

### soccertev

ok so now i'm stuck on

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

7. Nov 20, 2008

### Chaos2009

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.

Last edited: Nov 20, 2008
8. Nov 20, 2008

### 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

9. Nov 20, 2008

### soccertev

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

10. Nov 20, 2008

### soccertev

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

11. Nov 20, 2008

### Chaos2009

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

12. Nov 20, 2008

### soccertev

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

13. Nov 20, 2008

### Chaos2009

Yes, yes it is.

14. Nov 20, 2008

### soccertev

so how do i solve this then :S

15. Nov 20, 2008

### 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.

16. Nov 20, 2008

### soccertev

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

17. Nov 20, 2008

### 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.

18. Nov 20, 2008

### soccertev

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

19. Nov 20, 2008

### 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.

20. Nov 20, 2008

### soccertev

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