# Trigonometric Proof Help

1. Jan 13, 2009

### Faint

1. The problem statement, all variables and given/known data

(tan2x) - (sin2x) = (tan2x)(sin2x)

Prove these two are equal.

2. Relevant equations

Various trig identities, mainly tan2x = sin2x / cos2x

3. The attempt at a solution

I tried putting in Sin2x / cos2x for tan, but I don't know where to go from there.

Anyone mind helping me out?

2. Jan 13, 2009

### jgens

This has a really simple solution. Factor out sin(x)^2 from the left hand side and then simplify.

3. Jan 13, 2009

### Faint

I don't how/where you can factor sin(x)^2 out in this case. This is what I have done to the left side thus far.

$$\frac{sin^{2}x}{cos^{2}x} - \frac{sin^{2}x}{1}$$

then

$$\frac{sin^{2}x}{cos^{2}x} - \frac{sin^{2}xcos^{2}x}{cos^{2}x}$$

which ends up as

$$\frac{sin^{2}x - sin^{2}xcos^{2}x}{cos^{2}x}}$$

4. Jan 13, 2009

### jgens

Well, using your steps, you're almost at the solution. Again, factor sin(x)^2 out of the numerator and simplify. :)

This is what I was talking about in my initial post: tan(x)^2 - sin(x)^2 = sin(x)^2 (sec(x)^2 - 1), which is easy to simplify.

5. Jan 13, 2009

### Faint

$$\frac{sin^{2}x (1 -cos^{2}x)}{cos^{2}x}}$$

I'm missing something, because I still don't get it. :grumpy: What am I overlooking? Sin/Cos is equal to Tan, but then I have (1 - cos2x)/(cos2x)

6. Jan 13, 2009

### jgens

Are you familiar with the trig. identity sin(x)^2 + cos(x)^2 = 1? If you are, the solution should seem fairly obvious in both instances.

7. Jan 13, 2009

### Faint

At this point I have:

$$\frac{sin^{2}x (1 -cos^{2}x)}{cos^{2}x}} = (\frac{sin^{2}x}{cos^{2}x})(\frac{sin^{2}x}{1}})$$

I know that identity, but I can't see how it can be used on either side.

I'm sorry for my complete lack of ability to grasp this.

8. Jan 13, 2009

### jgens

If you know that identity then you surely know 1 - cos(x)^2 = sin(x)^2 or similarly that sec(x)^2 - 1 = tan(x)^2.

9. Jan 13, 2009

### Faint

Just realized this:

$$\frac{sin^{2}x (1 -cos^{2}x)}{cos^{2}x}} = \frac{sin^{2}x(sin^{2}x)}{cos^{2}x}$$

which turns into

$$\frac{sin^{2}x (1 -cos^{2}x)}{cos^{2}x}} = \frac{sin^{2}x(1-cos^{2}x)}{cos^{2}x}$$

Did I do anything incorrect there? And thank you very much for your help.

10. Jan 13, 2009

### jgens

Nothing there is incorrect - in fact that result should tell you something - but you should have made a connection which it appears you have still failed to make. If 1 - cos(x)^2 = sin(x)^2 and we have (1 - cos(x)^2)/cos(x)^2, what substitution can I make to simplify the expression?

11. Jan 13, 2009

### Faint

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

Correct? I get confused when this happens though:

$$\frac{sin^{2}x (sin^{2}x)}{cos^{2}x}}$$

to

$$tan^{2}x\frac{sin^{2}x}{cos^{2}x}}$$

or does the cos(x)^2 drop?

12. Jan 13, 2009

### jgens

The cos(x)^2 term does not drop but is merely factored out essentially in the tan(x)^2 term; hence, your resulting equation should read tan(x)^2 sin(x)^2. Q.E.D.

13. Jan 13, 2009

### Faint

Okay, I understand. Thank you a lot for the great help. :)