Help solving problem-II

S

Scrutinize

Junior Member
Joined
Sep 16, 2019
Messages
52
So I got stuck on this other problem as well:

Given that (sin(x+y))/(sin(x−y)) = p/q . Express cot y in terms of x, p, q. I expanded the top and the bottom but I got stuck there.

I got to:

(sin(x) cos(y) + cos(x) sin(y) )/( sin(x) cos(y) - sin(y) cos(x)) = p/q

I can't figure out how to isolate the cos(y) or sin(y) so that I can substitute those parts into cot (y).
Or how to solve the problem in another way.

Any help would be appreciated, thank you!
 
Scrutinize said:
So I got stuck on this other problem as well:

Given that (sin(x+y))/(sin(x−y)) = p/q . Express cot y in terms of x, p, q. I expanded the top and the bottom but I got stuck there.

I got to:

(sin(x) cos(y) + cos(x) sin(y) )/( sin(x) cos(y) - sin(y) cos(x)) = p/q

I can't figure out how to isolate the cos(y) or sin(y) so that I can substitute those parts into cot (y).
Or how to solve the problem in another way.

Any help would be appreciated, thank you!
Click to expand...
What would you get for [p/q + 1]?

What would you get for [p/q - 1]?
 
Scrutinize said:
Given that (sin(x+y))/(sin(x−y)) = p/q . Express cot y in terms of x, p, q. I expanded the top and the bottom but I got stuck there.

I got to:

(sin(x) cos(y) + cos(x) sin(y) )/( sin(x) cos(y) - sin(y) cos(x)) = p/q
Click to expand...
Cross-multiply (that is, multiply both sides by the LCD) to eliminate fractions. Then do something that will change the sines and cosines to cotangents (hint: divide every term by something).
 
sk - p/q + 1 would be (sin(x+y))/(sin(x−y)) + 1 or (sin(x+y))/(sin(x−y)) + sin^2(x) + cos^2(x) but I don't understand how that helps even if its plus or minus 1?

Dr.Peterson - I did as you said and I got what cot(y) was because I was able to get cos(y) and sin(y) out by using common factors. Thanks.

Still though, I want to understand what you meant with your hint SK.

Attached is my work:
 

Attachments

  • Math Work.pdf
    1.4 MB · Views: 2
Scrutinize said:
So I got stuck on this other problem as well:

Given that (sin(x+y))/(sin(x−y)) = p/q . Express cot y in terms of x, p, q. I expanded the top and the bottom but I got stuck there.

I got to:

(sin(x) cos(y) + cos(x) sin(y) )/( sin(x) cos(y) - sin(y) cos(x)) = p/q

I can't figure out how to isolate the cos(y) or sin(y) so that I can substitute those parts into cot (y).
Or how to solve the problem in another way.

Any help would be appreciated, thank you!
Click to expand...
p/q + 1 = 2*sin(x)*cos(y)/sin(x-y) = (p+q)/q

p/q - 1 = 2*sin(y)*cos(x)/sin(x-y) = (p-q)/q

[(p+q)/q] / [(p-q)/q] = [2*sin(x)*cos(y)/sin(x-y)] / [2*sin(y)*cos(x)/sin(x-y)]

(p+q) / (p-q) = [2*sin(x)*cos(y) / [2*sin(y)*cos(x)] = cot(y) * tan(x)

cot(y) = (p+q)/(p-q) * cot(x)
 
Subhotosh Khan said:
p/q + 1 = 2*sin(x)*cos(y)/sin(x-y) = (p+q)/q

p/q - 1 = 2*sin(y)*cos(x)/sin(x-y) = (p-q)/q

[(p+q)/q] / [(p-q)/q] = [2*sin(x)*cos(y)/sin(x-y)] / [2*sin(y)*cos(x)/sin(x-y)]

(p+q) / (p-q) = [2*sin(x)*cos(y) / [2*sin(y)*cos(x)] = cot(y) * tan(x)

cot(y) = (p+q)/(p-q) * cot(x)
Click to expand...
I'm probably just being dumb but I still can't understand what you did.

p/q + 1 = 2*sin(x)*cos(y)/sin(x-y) = (p+q)/q

p/q - 1 = 2*sin(y)*cos(x)/sin(x-y) = (p-q)/q

[(p+q)/q] / [(p-q)/q] = [2*sin(x)*cos(y)/sin(x-y)] / [2*sin(y)*cos(x)/sin(x-y)]


Could you tell me how you did the first 2 steps and then what you did to get to the third step? From there I understand what you did, sorry.
 
Scrutinize said:
I'm probably just being dumb but I still can't understand what you did.

p/q + 1 = 2*sin(x)*cos(y)/sin(x-y) = (p+q)/q

p/q - 1 = 2*sin(y)*cos(x)/sin(x-y) = (p-q)/q

[(p+q)/q] / [(p-q)/q] = [2*sin(x)*cos(y)/sin(x-y)] / [2*sin(y)*cos(x)/sin(x-y)]


Could you tell me how you did the first 2 steps and then what you did to get to the third step? From there I understand what you did, sorry.
Click to expand...
What are you getting for p/q + 1?
 
I think you are failing to use your expanded form, (sin(x) cos(y) + cos(x) sin(y) )/( sin(x) cos(y) - sin(y) cos(x)) = p/q . Find what p/q + 1 is, starting there.
 
Ah okay, I see now. And then how did you go to?

[(p+q)/q] / [(p-q)/q] = [2*sin(x)*cos(y)/sin(x-y)] / [2*sin(y)*cos(x)/sin(x-y)]


Specifically the [(p+q)/q] / [(p-q)/q]? Like why did you do that?
 
Scrutinize said:
Ah okay, I see now. And then how did you go to?

[(p+q)/q] / [(p-q)/q] = [2*sin(x)*cos(y)/sin(x-y)] / [2*sin(y)*cos(x)/sin(x-y)]


Specifically the [(p+q)/q] / [(p-q)/q]? Like why did you do that?
Click to expand...
That eliminates the denominators on both sides!

On the LHS, we factor out (1/q) and for the RHS we factor out [1/sin(x-y)].

Work with pencil and paper (as opposed to staring at the screen) - you should see it clearly!
 
Subhotosh Khan said:
That eliminates the denominators on both sides!

On the LHS, we factor out (1/q) and for the RHS we factor out [1/sin(x-y)].

Work with pencil and paper (as opposed to staring at the screen) - you should see it clearly!
Click to expand...
I am working with pencil and paper just couldn't see it.
 
You must log in or register to reply here.
Top