Trig Symmetry: show y = 2csc(x/2)-2 symm. abt x = pi

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Nar
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Trig Symmetry: show y = 2csc(x/2)-2 symm. abt x = pi

Show that $y=2cosec(x/2)-2$ is symmetrical about $x=\pi$.
Use $f(a+x) = f(a-x)$ for symmetry about $x=a$

I know that I need to substitute $\pi+x$ and $\pi-x$ in and show they are the same but I'm not sure how to start and how to simplify

Thanks!

buddy
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Re: Trig Symmetry: show y = 2csc(x/2)-2 symm. abt x = pi

Show that $y=2cosec(x/2)-2$ is symmetrical about $x=\pi$.

I know that I need to substitute $\pi+x$ and $\pi-x$ in and show they are the same but I'm not sure how to start
Start by doing the substituting and use 1/sine instead of cosecant. Remember that sin(x + pi) = -sin(x) because of how the sine wave works.

Nar
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Re: Trig Symmetry: show y = 2csc(x/2)-2 symm. abt x = pi

Not sure how that works with the /2 in there. So would I have
$f(\pi+x) = 2cosec((\pi+x)/2)-2$
$= 2(1/sin((\pi+x)/2) - 2$

But I can't use $sin(\pi+x) = -sinx$ there with it all being /2 can I?

maggiemagnet
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Re: Trig Symmetry: show y = 2csc(x/2)-2 symm. abt x = pi

Do x/2 = @ (theta) and use the formula they show at the link.

Nar
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Re: Trig Symmetry: show y = 2csc(x/2)-2 symm. abt x = pi

Sorry, but I'm not sure what you mean by
Do x/2 = @ (theta) and use the formula they show at the link.
I'm really struggling with this and yet it seems like it is simple but I'm really missing the point I think.

stapel_eliz
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Sorry, but I'm not sure what you mean by
Do x/2 = @ (theta) and use the formula they show at the link.
I'm really struggling with this and yet it seems like it is simple but I'm really missing the point I think.
What are you getting when you use the indicated substitution?