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November 01, 2025, 03:07:12 pm
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Author Topic: Circular functions help  (Read 2709 times)  Share 

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Juddinator

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Re: Circular functions help
« Reply #15 on: April 12, 2010, 09:14:19 pm »
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Quote from: fady_22 on April 12, 2010, 09:09:21 pm
Just change it into a familiar form: e.g.
Yes however if this is we eventually get
« Last Edit: April 12, 2010, 09:17:55 pm by Juddinator »
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fady_22

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Re: Circular functions help
« Reply #16 on: April 12, 2010, 09:19:42 pm »
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?

What's wrong with that?
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m@tty

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Re: Circular functions help
« Reply #17 on: April 12, 2010, 09:24:05 pm »
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Is it that it is >1 that worries you? Both cosec and sec have a range of . Because their respective reciprocals, sin and cosine, have a range of .
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Juddinator

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Re: Circular functions help
« Reply #18 on: April 12, 2010, 09:28:06 pm »
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yeah... that's what I had but I was trying to make it a whole number lol... my bad sorry :S

How do you go about finding your asymptotes then? I get confused when you have to adapt them...
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Juddinator

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Re: Circular functions help
« Reply #19 on: April 12, 2010, 09:29:26 pm »
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Quote from: m@tty on April 12, 2010, 09:24:05 pm
Is it that it is >1 that worries you? Both cosec and sec have a range of . Because their respective reciprocals, sin and cosine, have a range of .
No I understand that, it's just the asymptotes that annoy me the most...
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Yitzi_K

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Re: Circular functions help
« Reply #20 on: April 12, 2010, 09:31:35 pm »
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Quote from: Juddinator on April 12, 2010, 09:29:26 pm
Quote from: m@tty on April 12, 2010, 09:24:05 pm
Is it that it is >1 that worries you? Both cosec and sec have a range of . Because their respective reciprocals, sin and cosine, have a range of .
No I understand that, it's just the asymptotes that annoy me the most...

Asymptotes of reciprocal graphs are wherever there was an x-intercept on the original graph
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fady_22

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Re: Circular functions help
« Reply #21 on: April 12, 2010, 09:37:38 pm »
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Well, the asymptotes occur when the reciprocal function is undefined, this occurs when the function on the bottom is equal to zero. So just find the values for when this function equals to zero, and you'll have the equations of your asymptotes.
e.g.
Let


And those are your asymptotes.
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m@tty

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Re: Circular functions help
« Reply #22 on: April 12, 2010, 09:42:33 pm »
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Quote from: Yitzi_K on April 12, 2010, 09:31:35 pm
Quote from: Juddinator on April 12, 2010, 09:29:26 pm
Quote from: m@tty on April 12, 2010, 09:24:05 pm
Is it that it is >1 that worries you? Both cosec and sec have a range of . Because their respective reciprocals, sin and cosine, have a range of .
No I understand that, it's just the asymptotes that annoy me the most...

Asymptotes of reciprocal graphs are wherever there was an x-intercept on the original graph

The basic logic behind this is undefined and cannot be represented on a graph, hence an asymptote is placed there to show this, similarly the vertical asymptotes of a graph become the zero's of the reciprocal as or .
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Juddinator

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Re: Circular functions help
« Reply #23 on: April 12, 2010, 10:02:07 pm »
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Quote from: m@tty on April 12, 2010, 09:42:33 pm
Quote from: Yitzi_K on April 12, 2010, 09:31:35 pm
Quote from: Juddinator on April 12, 2010, 09:29:26 pm
Quote from: m@tty on April 12, 2010, 09:24:05 pm
Is it that it is >1 that worries you? Both cosec and sec have a range of . Because their respective reciprocals, sin and cosine, have a range of .
No I understand that, it's just the asymptotes that annoy me the most...

Asymptotes of reciprocal graphs are wherever there was an x-intercept on the original graph

The basic logic behind this is undefined and cannot be represented on a graph, hence an asymptote is placed there to show this, similarly the vertical asymptotes of a graph become the zero's of the reciprocal as or .
Yeah! I'm with you guys now! :D :D  Thanks!
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