I'm a little confused. It could be Trig, Algebra, or both.

[mrtrixta]

Ok guys, please bare with me as this is my first post on here. I'm stuck on this problem. I'm not sure how to proceed to solving it. I'm in a Trig class but I'm sure this touches base on Algebra but it's been too long since I've taken it. Can't quite remember the steps. Guess I'm a bit too old and my memory isn't what it used to be. Thanks for the help.

cos ? = x / x+1

So far I used the Pythagoran Identity: sin² ? + cos² ? = 1
From there I got the alternative: sin² ? = 1 - cos² ?
Following steps are:
1). sin² ? = 1 - (x / x+1)²

2). sin² ? = 1 - x² / x²+2x+1


From there I'm stuck. Can't quite remember what that next step should be :?: . Please help. Also, let me know if I'm even on the right track. Any further clarification on anything stated please feel free to ask. I did my best to make this as clear as possible. Thanks!!

 

[stapel]

Ok guys, please bare with me...

I am not getting nekkid with you! :shock:

Or did you perhaps mean "bear" with you...?

I'm not sure how to proceed to solving it....

cos ? = x / x+1

Unless you were given a value for theta or for x (and I'm assuming you meant "x/(x + 1)", not "(x/x) + 1", as you posted), then there is no way to solve the posted equation. Sorry!

Eliz.

 

[mrtrixta]

Soooo sorry about that. You're right!! I did mean to say bear and not bare. I'm a little red. Also, I forgot to add another part to the original problem. With that equation, sin ? is supposed to be found. Also, about the "x/(x + 1)", not "(x/x) + 1", that's correct also. Like I stated it was my first time posting so I wasn't sure how to state the problem so that it could be understood clearly. Maybe with the info about finding sin ? with that given equation, the problem makes more sense. I hope!! :mrgreen: Maybe I should just re enter the problem so that it can be understood clearly.

 

[Loren]

You never did say what you were trying to do with the given equation. As was said in the previous answer, IF you are trying to solve for x, no can do. But maybe you were asked to express sin \(\displaystyle \theta\) in terms of x, or something else. If that is what you were asked the answer is \(\displaystyle \sin \theta = \frac{\sqrt{2x+1}}{x+1}\).

Typo now corrected. Sorry.

 

[galactus]

Given that \(\displaystyle cos{\theta}=\frac{x}{x+1}\), then the adjacent side is \(\displaystyle \sqrt{(x+1)^{2}-x^{2}}\)

That means \(\displaystyle sin{\theta}=\frac{\sqrt{(x+1)^{2}-x^{2}}}{x+1}=\frac{\sqrt{2x+1}}{x+1}\)

 

[mrtrixta]

Very interesting. Wow!!! Thanks a lot for all the help. I understand how the answer was found. So basically the way I was going is totally wrong. :lol: Oh well, at least I tried something. Thanks again for the help. I wonder though, was I on the right track or no??

 

[Loren]

You could have proceeded that way.
\(\displaystyle \sin^2\theta= 1- \cos^2\theta\)

\(\displaystyle \sin^2\theta =1-\frac{x^2}{(x+1)^2} = \frac{x^2+2x+1-x^2}{(x+1)^2}\)

\(\displaystyle \sin\theta=\pm\frac{\sqrt{2x+1}}{x+1}\)

Since your original equation had the cos \(\displaystyle \theta\) positive, theta must be in QI or QIV. The sine of an angle is negative in Quadrant IV, so our only possibility is its being in QI. Therefore the negative solution is discarded giving the final solution as

\(\displaystyle \sin\theta=\frac{\sqrt{2x+1}}{x+1}\)

 

[mrtrixta]

That is fantastic. Thanks so much for helping me out with this problem. I was just confused on how to proceed with the rest of the problem. I really appreciate it.

Quick question: How do you insert your text like that above. It just makes things so much clearer. I know that I'll be on using this site often so It'd be nice so that to make problems clear for all to understand. Thanks!!

 

[stapel]

How do you insert your text like that above. It just makes things so much clearer.

To learn how to use LaTeX, try following the links in the "Forum Help" pull-down menu at the very top of each forum page.

Eliz.

 

[mrtrixta]

Thanks, it's gonna be a huge help.