Can I Sketch the Graph of \sin\Theta = \frac{\pi}{2} - \Theta Symmetrically?

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The discussion centers on sketching the graph of the equation y = sin(π/2 - Θ) and determining its symmetry. The user initially misstates the problem but later clarifies that both y = sin(Θ + π/2) and y = cos(Θ) are equivalent. Participants confirm that the reasoning is correct, emphasizing that both equations represent the same function. Substitution is suggested as a method to verify their equivalence. Overall, the user’s approach to understanding the graph and its properties is validated.
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This is the problem:

Sketch the graph of:

\sin\Theta = \frac{\pi}{2} - \Theta


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

\Theta = \frac{\pi}{2}

So I can write the equation as

\sin\Theta = \frac{\pi}{2} + \Theta

This will be a graph of \sin\Theta pulled \frac{\pi}{2} 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!)
 
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What I am not understanding is how you sketch it on a coordinate plane if there is only one variable (theta).
 
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 = \sin(\frac{\pi}{2} - \Theta)

and using the same reasoning above I end up with

y = \sin(\Theta + \frac{\pi}{2})

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

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

But I want to 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?
 
I can see that no one is interested in giving me an answer.

Oh well.
 
Yes, your reasoning is fine.
 
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.
 
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Thanks guys

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

More questions to come soon!
 
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