Proving (tan2x) - (sin2x) = (tan2x)(sin2x): Trigonometric Proof Help

[Faint]

Homework Statement

(tan²x) - (sin²x) = (tan²x)(sin²x)

Prove these two are equal.

Homework Equations

Various trig identities, mainly tan²x = sin²x / cos²x

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?

 

[jgens]

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

 

[Faint]

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

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

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

 

[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.

 

[Faint]

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.

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

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

 

[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.

 

[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.

 

[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.

 

[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.

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.

 

[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?

 

[Faint]

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?

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?

 

[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.

 

[Faint]

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.

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