# Is my reasoning correct?

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 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?

I can see that noone is interested in giving me an answer.

Oh well.

AKG
Homework Helper

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

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:
Thanks guys

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

More questions to come soon!