Proving Trigonometric Identity: 1-cos2x+sin2x = tanx/1+cos2x+sin2x

[scorpa]

Hi everyone,

I'm sorry to bother you with such a trivial question but I really am stuck here. I don't know why I can't prove this one as I've managed to get harder ones than this, but as it is I need help with it.
This is the question:

Prove 1-cos2x+sin2x = tanx
1+cos2x+sin2x

I saw that there was there cos2x and sin 2x and substituted the double angle identities for them, but one I had substituted them and simplified them I was hopelessly stuck on them, no matter what I tried to do I couldn't get any farther with them. I've been doing nothing but math for two days now so I must have fried my brain or something. Any help you guys can give me I would really appreciate, I honestly don't know what I'm doing wrong. I'm sure it's going to end up being something terrible stupid but I thought I had better just ask. Oh and I have one more quick question!

If you have a question like:

(sin^4)x-(cos^4)x = 1
(sin^2)x-(cos^2)x

Can you just simplify by dividing the equation and getting (sin^2)x + (cos^2)x. And then that identity is one of the Pythagorean identities that end up equalling one, so that is the proof? Again sorry for the stupid question, I appreciate all the help anyone can give me on this.

 

[Data]

Try using \cos^2{x} = \frac{1}{2}(1 + \cos{2x}), \ \sin^2{x} = \frac{1}{2}(1 - \cos{2x}), and \sin{2x} = 2\sin{x} \cos{x}.

And for your second question, yes, that is perfectly fine.

 

[scorpa]

Try using \cos^2{x} = \frac{1}{2}(1 + \cos{2x}), \ \sin^2{x} = \frac{1}{2}(1 - \cos{2x}), and \sin{2x} = 2\sin{x} \cos{x}.

And for your second question, yes, that is perfectly fine.

Thanks very much for the help Data, but unfortunately I am still a bit confused

Why can you use identities you listed? I understand the sin2x=2sinxcosx, but I'm not sure why you can use the first two that you listed. It just doesn't seem to me like you have the right information for them.

The question was"

Prove 1-cos2x+sin2x =tanx
1+cos2x+sin2x

OK, now you said to use the first two identities that you listed, but I don't seem how they will work. I realize that I have the 1+cos2x, but I don't have the 1/2 needed to make it into the (cos^2)x, and the same with the 1-cos2x. Maybe I am just looking at it the wrong way. Could you maybe explain your choices to me so that I can understand why I should be using them? I don't want to just do the question without being able to understand why I am doing it. Thanks again you are being a great help!

 

[scorpa]

Oh and sorry my question keeps getting screwed up when I post it. It always looks right before I submit it, I just can't seem to be able to get the 1+cos2x=sin2x to go underneath the first expression.

 

[scorpa]

Oh and in the post above that was supposed to be 1+cos2x+sin2x not 1+cos2x=sin2x...sorry about that!

 

[seiferseph]

ok, for the first one, you know both cos2x and sin2x are identities, right? so put in the identity for sin2x,
and then find the identity for cos2x to eliminate the 1 that is in front(if you have only sin and cos, its easier to cancel or factor).
ex, if you have 1-cos2x, you would put in the identity that is 1-sin^2x, and the 1's would cancel.
after you do that for both top and bottom, try to factor out a term, and then cancel/solve. post if you have any more questions.

 

[cepheid]

Data's trig identities are correct.

We start with:

\cos 2x = \cos^2 x - \sin^2 x ...[1]

This double angle comes from the addition formula for cosine, in the special case in which the two angles being added together are just the same angle.

Now we can use the pythagorean identity:

\sin^2 x + \cos^2 x = 1

Rearrange it to solve for \cos^2 x:

\cos^2 x = 1 - \sin^2 x

Now substitute this in for the \cos^2 x in [1]:

\cos 2x = 1 - \sin^2 x - \sin^2 x = 1 - 2\sin^2 x

Rearrange to solve for \sin^2 x:

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

This is called the half angle identity. The second half angle identity that Data gave you (with an expression for \cos^2 x) is derived similarly. These identities are the starting point for solving your problem, since it involves the expressions 1 + cos2x and 1 - cos2x.

Don't question Data...what with his positronic brain and all, he's nearly infallible.

 

[scorpa]

OK I did try that, but I will try it again to see if I missed something along the way. The identities you listed were the exact ones I used, but as I said I will go give it another go!

 

[cepheid]

Post again with your progress if you still can't get it. Note that I made a mistake in my final step originally. It has been corrected.

 

[Curious3141]

Hi everyone,

I'm sorry to bother you with such a trivial question but I really am stuck here. I don't know why I can't prove this one as I've managed to get harder ones than this, but as it is I need help with it.
This is the question:

Prove 1-cos2x+sin2x = tanx
1+cos2x+sin2x

I saw that there was there cos2x and sin 2x and substituted the double angle identities for them, but one I had substituted them and simplified them I was hopelessly stuck on them, no matter what I tried to do I couldn't get any farther with them. I've been doing nothing but math for two days now so I must have fried my brain or something. Any help you guys can give me I would really appreciate, I honestly don't know what I'm doing wrong. I'm sure it's going to end up being something terrible stupid but I thought I had better just ask. Oh and I have one more quick question!

If you have a question like:

(sin^4)x-(cos^4)x = 1
(sin^2)x-(cos^2)x

Can you just simplify by dividing the equation and getting (sin^2)x + (cos^2)x. And then that identity is one of the Pythagorean identities that end up equalling one, so that is the proof? Again sorry for the stupid question, I appreciate all the help anyone can give me on this.

For the first one, use the double angle formulae for sin and cos :


\cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1

\sin 2x = 2\sin x \cos x

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


For the second one, as you said, just factorise and apply simple identities :

\frac{\sin^4 x - \cos^ 4 x}{\sin^2 x - \cos^2 x} = \frac{(\sin^2 x + \cos^2 x)(\sin^2 x - \cos^2 x)}{\sin^2 x - \cos^2 x} = 1

 

[cepheid]

Sure...just give him the answer why don't you?

 

[Curious3141]

Sure...just give him the answer why don't you?

Well, it looks like he's been trying for long enough, so why not ?

 

[Data]

Don't question Data...what with his positronic brain and all, he's nearly infallible.

NEARLY?

Edit: Now all I need to do is figure out how to get smilies in here! Oh, wait, what are all of those colorful symbols next to this white box I'm writing in?

 

[scorpa]

Ok guys thank you so much for the help, you have helped me out a lot here. After seeing the answer I've found what I was doing wrong. On the bottom half of the expression I was using the sine identity instead of the cosine, everything else I had right. I hate it when I am so close to getting an answer and then something stupid like that screws me up. Thanks again guys, I really appreciate it.

 

[scorpa]

Oh and yeah I feel very stupid right now because that was very easy :(