• Support PF! Buy your school textbooks, materials and every day products Here!

Is my reasoning correct?

  • Thread starter Tau
  • Start date
  • #1
Tau
14
0
This is the problem:

Sketch the graph of:

[tex]\sin\Theta[/tex] = [tex]\frac{\pi}{2}[/tex] - [tex]\Theta[/tex]


Now the graph (between 0 and pi) is symmetrical about the line

[tex]\Theta[/tex] = [tex]\frac{\pi}{2}[/tex]

So I can write the equation as

[tex]\sin\Theta[/tex] = [tex]\frac{\pi}{2}[/tex] + [tex]\Theta[/tex]

This will be a graph of [tex]\sin\Theta[/tex] pulled [tex]\frac{\pi}{2}[/tex] units to the left.

Is this reasoning correct?

If it is, would you do it a different way?


Thanks
Jeremy

(Noone wants a gmail invite?lol!)
 

Answers and Replies

  • #2
57
0
What I am not understanding is how you sketch it on a coordinate plane if there is only one variable (theta).
 
  • #3
Tau
14
0
I am SUCH AN IDIOT!

I mistyped the entire question. You all now see what a dufus I am.

Anyway, here is the question (typed properly this time):

Sketch the graph of:

y = [tex]\sin(\frac{\pi}{2}[/tex] - [tex]\Theta[/tex])

and using the same reasoning above I end up with

y = [tex]\sin(\Theta[/tex] + [tex]\frac{\pi}{2}[/tex])

Which is a curve of [tex]sin\Theta[/tex] pulled half pi units to the left
Now- is this reasoning correct.

If it is, would any of you done it a different way?
 
  • #4
Both of the expressions are equal to cos theta.
 
  • #5
Tau
14
0
I know that both curves are the same.

But I wanna know if I reasoned it out correctly.

Were the actions I took (see first post) the correct ones. Was I thinking along the correct lines?
 
  • #6
Tau
14
0
I can see that noone is interested in giving me an answer.

Oh well.
 
  • #7
AKG
Science Advisor
Homework Helper
2,559
3
Yes, your reasoning is fine.
 
  • #8
280
1
Well, both equations do simplify to cos theta, so yeah they are the same function. If you meant, "how do I see for sure that they are the same function?" you can always try substitution.

sin theta when theta equals zero is zero. Substituting (using degrees because I don't have those pretty graphics), your first equation becomes:

y = sin (90 - 0) = sin (90) - sin (0) = 1 - 0 = 1

Then your next equation becomes:

y = sin (0 + 90) = sin (0) + sin (90) = 0 + 1 = 1

To check to see that both equations can be substituted for simply y = cos (theta).

y = cos (0) = 1

The math police would probably arrest me if they found me dumping in numbers like this when you are probably asking for a formal proof, but it's good enough for me so I hope it helped.
 
Last edited:
  • #9
Tau
14
0
Thanks guys

I thought for a moment that I was alone in the dark.

More questions to come soon!
 

Related Threads for: Is my reasoning correct?

Replies
4
Views
564
Replies
15
Views
631
Replies
2
Views
1K
Replies
1
Views
688
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
5
Views
1K
Replies
5
Views
9K
  • Last Post
Replies
4
Views
1K
Top