Is this correct? Trig inequality

[dolina dahani]

Guys I know this is elementary but im really confused when can i multiply the bottom? So i got a question about trig equation and im confused i got the same answer but im not sure if its right? I didnt multiply the bottom but left it and subtracted both sides by sin(x/2) and changed sinx to 2sinxcosx and crossed the square of cos and upper cos(x/2) and just left it as (sin(x/2)(1-cos(x/2))/cos(x/2) So to not confuse you here's their answer... Again my question is why did they do that if im not mistaken its best to not cross multiply???

 

[dolina dahani]

I forgot to put my answer, I'm sorry

 

[Dr.Peterson]

I can't tell what is given or what the goal is, so I can't comment on whether any method is better. Please state the problem (in English).

 

[dolina dahani]

I can't tell what is given or what the goal is, so I can't comment on whether any method is better. Please state the problem (in English).

Oh I'm sorry man, so the question was just to solve the equality that I marked in blue .
(Sin(x/2))/(1+cosx)=sin(x/2).
Sorry if you cant see what I wrote but basically im not sure if it's incorrect if I dont do it like they did it I mean they didn't leave the cos^2(x/2) as divisor but multiplied it by itself to get rid of it. So is it incorrect if I don't multiply it I mean if I leave it as it is and just subtract sin(x/2) in both sides and then solve

 

[topsquark]

That should be
[math]\dfrac{sin(x)}{1 + cos(x)} = sin \left ( \dfrac{x}{2} \right )[/math]
I would attack it using your t = x/2 method, but I admit that the "book's" method is more elegant. In the long run it doesn't matter how you solve it so long as you get the correct answer in the end.

-Dan

 

[dolina dahani]

That should be
[math]\dfrac{sin(x)}{1 + cos(x)} = sin \left ( \dfrac{x}{2} \right )[/math]
I would attack it using your t = x/2 method, but I admit that the "book's" method is more elegant. In the long run it doesn't matter how you solve it so long as you get the correct answer in the end.

-Dan

Thank you, yeah maybe it was trivial question I was just wondering if it's not correct to not multiply the bottom in the end if I'm not mistaken I like my way and se it as a safer solution

 

[dolina dahani]

That should be
[math]\dfrac{sin(x)}{1 + cos(x)} = sin \left ( \dfrac{x}{2} \right )[/math]
I would attack it using your t = x/2 method, but I admit that the "book's" method is more elegant. In the long run it doesn't matter how you solve it so long as you get the correct answer in the end.

-Dan

I mean is it mistake if i leave cos(x/2) in divisor and not multiply by it to get rid of it?? Sorry guys i just got confused and scarred cuz i have a test lol.

 

[Dr.Peterson]

I mean is it mistake if i leave cos(x/2) in divisor and not multiply by it to get rid of it?? Sorry guys i just got confused and scarred cuz i have a test lol.

How could that be incorrect?

One benefit of your approach is that you avoid forgetting that the denominator can't be zero (which they appear to have taken note of early, but didn't explicitly mention at the end where it is needed).