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On The Shape Of Trig Graphs To A Precalc Student
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I want to know why 3 types of trig graphs have the shape that they do:
Secant graphs. Why would $dfrac 1cos theta$ result in a graph like this? I get the cosine graph - it makes sense when you compare it to a unit circle. But I don't understand it at the same level for secant graphs. Can someone please explain why the secant graph is shaped like that given it's equation?
Cosecant graphs. Same as above.
Cotangent graphs. Why would $dfrac 1tan theta$ cause the tangent graph to flip? Because when you look at a cotangent graph, it's basically a reflected tangent graph. Why would $dfrac 1cos theta$ create a graph that looks like that?
Can you please give the explanation for why the graphs are shaped like that not too rigorously, and at the level of a Precalculus student who hasn't learnt Calculus yet? Thank you.
algebra-precalculus trigonometry
edited Jul 26 at 15:56
bjcolby15
7921616
asked Jul 26 at 14:25
Ethan Chan
608322
add a comment |Â
up vote
1
down vote
favorite
I want to know why 3 types of trig graphs have the shape that they do:
Secant graphs. Why would $dfrac 1cos theta$ result in a graph like this? I get the cosine graph - it makes sense when you compare it to a unit circle. But I don't understand it at the same level for secant graphs. Can someone please explain why the secant graph is shaped like that given it's equation?
Cosecant graphs. Same as above.
Cotangent graphs. Why would $dfrac 1tan theta$ cause the tangent graph to flip? Because when you look at a cotangent graph, it's basically a reflected tangent graph. Why would $dfrac 1cos theta$ create a graph that looks like that?
Can you please give the explanation for why the graphs are shaped like that not too rigorously, and at the level of a Precalculus student who hasn't learnt Calculus yet? Thank you.
algebra-precalculus trigonometry
edited Jul 26 at 15:56
bjcolby15
7921616
asked Jul 26 at 14:25
Ethan Chan
608322
When $cos(theta) = pm 1$, then $sec(theta) = pm 1$. When $cos(theta) = 0$, then $sec(theta)$ is undefined---this implies the possible existence of an asymptote (you need limits to confirm this, and that is usually a topic for the end of pre-calc, or for calculus). Between the ones and the asymptotes, the cosine function either increases or decreases monotonically, so the secant function will have the same behaviour, only in the opposite direction. This can be made rigorous without calculus (though calculus makes it easier).
– Xander Henderson
Jul 26 at 14:31
That being said, I think that this is a question which might be more appropriate for the Mathematics Educators SE.
– Xander Henderson
Jul 26 at 14:32
@XanderHenderson I'm not an educator though; I'm a student. But yeah, thanks for the response. Last thing though, what about for cotangent graphs? Why would it be a perfect reversal of tangent graph?
– Ethan Chan
Jul 26 at 14:35
Because $tan(dfracpi2-x) = cot(x)$
– steven gregory
Jul 26 at 15:16
" Can someone please explain why the secant graph is shaped like that given it's equation? " Can you explain why you don't understand it? It's $frac 1cos x$ and as $cos x$ is periodic from 0 to 1 to 0 to -1 back to 0, It's reciprical will be periodic from $+infty$ to $1$ back to $infty$then an axis jump to $-infty$ up to $-1$ and back down. .... And that is exactly what it does do. So from my point of view $frac 1cos x$ exactly as I expect others would expect it to do. So ... I'm confused why you are confused.
– fleablood
Jul 26 at 16:03
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to know why 3 types of trig graphs have the shape that they do:
Secant graphs. Why would $dfrac 1cos theta$ result in a graph like this? I get the cosine graph - it makes sense when you compare it to a unit circle. But I don't understand it at the same level for secant graphs. Can someone please explain why the secant graph is shaped like that given it's equation?
Cosecant graphs. Same as above.
Cotangent graphs. Why would $dfrac 1tan theta$ cause the tangent graph to flip? Because when you look at a cotangent graph, it's basically a reflected tangent graph. Why would $dfrac 1cos theta$ create a graph that looks like that?
Can you please give the explanation for why the graphs are shaped like that not too rigorously, and at the level of a Precalculus student who hasn't learnt Calculus yet? Thank you.
algebra-precalculus trigonometry
edited Jul 26 at 15:56
bjcolby15
7921616
asked Jul 26 at 14:25
Ethan Chan
608322
I want to know why 3 types of trig graphs have the shape that they do:
Secant graphs. Why would $dfrac 1cos theta$ result in a graph like this? I get the cosine graph - it makes sense when you compare it to a unit circle. But I don't understand it at the same level for secant graphs. Can someone please explain why the secant graph is shaped like that given it's equation?
Cosecant graphs. Same as above.
Cotangent graphs. Why would $dfrac 1tan theta$ cause the tangent graph to flip? Because when you look at a cotangent graph, it's basically a reflected tangent graph. Why would $dfrac 1cos theta$ create a graph that looks like that?
Can you please give the explanation for why the graphs are shaped like that not too rigorously, and at the level of a Precalculus student who hasn't learnt Calculus yet? Thank you.
algebra-precalculus trigonometry
edited Jul 26 at 15:56
bjcolby15
7921616
asked Jul 26 at 14:25
Ethan Chan
608322
edited Jul 26 at 15:56
bjcolby15
7921616
edited Jul 26 at 15:56
bjcolby15
7921616
edited Jul 26 at 15:56
bjcolby15
7921616
7921616
asked Jul 26 at 14:25
Ethan Chan
608322
asked Jul 26 at 14:25
Ethan Chan
608322
asked Jul 26 at 14:25
Ethan Chan
608322
608322
When $cos(theta) = pm 1$, then $sec(theta) = pm 1$. When $cos(theta) = 0$, then $sec(theta)$ is undefined---this implies the possible existence of an asymptote (you need limits to confirm this, and that is usually a topic for the end of pre-calc, or for calculus). Between the ones and the asymptotes, the cosine function either increases or decreases monotonically, so the secant function will have the same behaviour, only in the opposite direction. This can be made rigorous without calculus (though calculus makes it easier).
– Xander Henderson
Jul 26 at 14:31
That being said, I think that this is a question which might be more appropriate for the Mathematics Educators SE.
– Xander Henderson
Jul 26 at 14:32
@XanderHenderson I'm not an educator though; I'm a student. But yeah, thanks for the response. Last thing though, what about for cotangent graphs? Why would it be a perfect reversal of tangent graph?
– Ethan Chan
Jul 26 at 14:35
Because $tan(dfracpi2-x) = cot(x)$
– steven gregory
Jul 26 at 15:16
" Can someone please explain why the secant graph is shaped like that given it's equation? " Can you explain why you don't understand it? It's $frac 1cos x$ and as $cos x$ is periodic from 0 to 1 to 0 to -1 back to 0, It's reciprical will be periodic from $+infty$ to $1$ back to $infty$then an axis jump to $-infty$ up to $-1$ and back down. .... And that is exactly what it does do. So from my point of view $frac 1cos x$ exactly as I expect others would expect it to do. So ... I'm confused why you are confused.
– fleablood
Jul 26 at 16:03
add a comment |Â
When $cos(theta) = pm 1$, then $sec(theta) = pm 1$. When $cos(theta) = 0$, then $sec(theta)$ is undefined---this implies the possible existence of an asymptote (you need limits to confirm this, and that is usually a topic for the end of pre-calc, or for calculus). Between the ones and the asymptotes, the cosine function either increases or decreases monotonically, so the secant function will have the same behaviour, only in the opposite direction. This can be made rigorous without calculus (though calculus makes it easier).
– Xander Henderson
Jul 26 at 14:31
That being said, I think that this is a question which might be more appropriate for the Mathematics Educators SE.
– Xander Henderson
Jul 26 at 14:32
@XanderHenderson I'm not an educator though; I'm a student. But yeah, thanks for the response. Last thing though, what about for cotangent graphs? Why would it be a perfect reversal of tangent graph?
– Ethan Chan
Jul 26 at 14:35
Because $tan(dfracpi2-x) = cot(x)$
– steven gregory
Jul 26 at 15:16
" Can someone please explain why the secant graph is shaped like that given it's equation? " Can you explain why you don't understand it? It's $frac 1cos x$ and as $cos x$ is periodic from 0 to 1 to 0 to -1 back to 0, It's reciprical will be periodic from $+infty$ to $1$ back to $infty$then an axis jump to $-infty$ up to $-1$ and back down. .... And that is exactly what it does do. So from my point of view $frac 1cos x$ exactly as I expect others would expect it to do. So ... I'm confused why you are confused.
– fleablood
Jul 26 at 16:03
When $cos(theta) = pm 1$, then $sec(theta) = pm 1$. When $cos(theta) = 0$, then $sec(theta)$ is undefined---this implies the possible existence of an asymptote (you need limits to confirm this, and that is usually a topic for the end of pre-calc, or for calculus). Between the ones and the asymptotes, the cosine function either increases or decreases monotonically, so the secant function will have the same behaviour, only in the opposite direction. This can be made rigorous without calculus (though calculus makes it easier).
– Xander Henderson
Jul 26 at 14:31
When $cos(theta) = pm 1$, then $sec(theta) = pm 1$. When $cos(theta) = 0$, then $sec(theta)$ is undefined---this implies the possible existence of an asymptote (you need limits to confirm this, and that is usually a topic for the end of pre-calc, or for calculus). Between the ones and the asymptotes, the cosine function either increases or decreases monotonically, so the secant function will have the same behaviour, only in the opposite direction. This can be made rigorous without calculus (though calculus makes it easier).
– Xander Henderson
Jul 26 at 14:31
That being said, I think that this is a question which might be more appropriate for the Mathematics Educators SE.
– Xander Henderson
Jul 26 at 14:32
That being said, I think that this is a question which might be more appropriate for the Mathematics Educators SE.
– Xander Henderson
Jul 26 at 14:32
@XanderHenderson I'm not an educator though; I'm a student. But yeah, thanks for the response. Last thing though, what about for cotangent graphs? Why would it be a perfect reversal of tangent graph?
– Ethan Chan
Jul 26 at 14:35
@XanderHenderson I'm not an educator though; I'm a student. But yeah, thanks for the response. Last thing though, what about for cotangent graphs? Why would it be a perfect reversal of tangent graph?
– Ethan Chan
Jul 26 at 14:35
Because $tan(dfracpi2-x) = cot(x)$
– steven gregory
Jul 26 at 15:16
Because $tan(dfracpi2-x) = cot(x)$
– steven gregory
Jul 26 at 15:16
" Can someone please explain why the secant graph is shaped like that given it's equation? " Can you explain why you don't understand it? It's $frac 1cos x$ and as $cos x$ is periodic from 0 to 1 to 0 to -1 back to 0, It's reciprical will be periodic from $+infty$ to $1$ back to $infty$then an axis jump to $-infty$ up to $-1$ and back down. .... And that is exactly what it does do. So from my point of view $frac 1cos x$ exactly as I expect others would expect it to do. So ... I'm confused why you are confused.
– fleablood
Jul 26 at 16:03
" Can someone please explain why the secant graph is shaped like that given it's equation? " Can you explain why you don't understand it? It's $frac 1cos x$ and as $cos x$ is periodic from 0 to 1 to 0 to -1 back to 0, It's reciprical will be periodic from $+infty$ to $1$ back to $infty$then an axis jump to $-infty$ up to $-1$ and back down. .... And that is exactly what it does do. So from my point of view $frac 1cos x$ exactly as I expect others would expect it to do. So ... I'm confused why you are confused.
– fleablood
Jul 26 at 16:03
add a comment |Â
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consider how behavior in a graph of $f(x)$ will affect behavior in a graph $frac 1 f(x)$. For values where $0 < f(x) < 1$ then we will have $infty > frac 1f(x)$, with little changes from $teensy$ to $frac 12 teensy$ resulting in huge stretches from $huge$ to $2 times huge$. And if $f(x)$ goes through from $teensy$ to $0$ to $-teensy$ then $frac 1f(x)$ will get asymptotic to infinity, be infinite(undefined) at a point, and jump to "negative infinity" and then imediate start reducing to more moderate negative values as $f(x)$ because more siginificant negative values.
If $1 < f(x)$ we will have $1 > frac 1f(x) > 0$ and if $f(x)$ grows blithely huge toward $huge$, $double-huge$ and $huge^2$ then $frac 1f(x)$ will tend to flatline toward $0$.
So bearing that in mind:
$cos x$ is periodic from $0$ to $1$ back down $0$ and and to $-1$ to $0$ to $1$ etc. and $frac 1cos x$ behaves exactly as we'd expect, for $+infty$ down to a minimum of $1$ (as $cos x$ goes from $0$ to $1$ and peaks) and then back to $infty$. Then as $cos x$ goes from tiny positive through $0$, to tiny negative. $frac 1cos x$ goess to $+infty$ and jumps to $-infty$ and starts surfaces up to $-1$ and then (as $cos x$ reaches a nadir of $-1$ and heads back to $0$) $frac 1cos x$ reaches max at $-1$ and starts plumeting back to $-infty.
As for $tan x$ and $frac 1tan x$. $tan x$ will go from $infty$ where $cos x = 1; sin x= 0$ to $1$ where $cos x = sin x = frac 1sqrt 2$. In this stage $sin x$ is increasing faster than $cos x$ is decreasing. So there will be a bulge to the right in the graph. Then as $cos xto 0$ and $sin x to 1$ we have $frac sin xcos x to infty$. and as $cos x$ is not decreasing faster than $sin x$ is increasing, the graph stretches to be long and skinny. Then as $cos x$ goes through $0$ to negative, we jump to negative infinity.
In the same interval $cot x = frac cos xsin x$ starts and $frac 10 = infty$ and decreases down to $1$ and then to $0$. The exact same but in reverse. Then as $cos x$ goes threough $0$ to negative $frac cos xsin x$ goes ther $0$ to negative.
Useful to realize $cot x =frac cos xsin x = frac sin(frac pi 2 - x)cos frac pi 2 - x) = tan (frac pi 2 - x)$ so that explains more rigororously why their graphs are symmetriic.
answered Jul 26 at 16:42
fleablood
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Pictures often help.
Here's a graphical depiction of $cos theta$ and $sec theta = frac1cos theta$ for $theta = fracpi3 = 60$ degrees.
The situation for $cos theta$ is represented by the right triangle $triangle ABC$. The radius is $CA = 1$, and the cosine of $theta$ is represented by the length of $CB = frac12$.
Now, extend $CA$ rightward to $A'$, which lies on the vertical line tangent to the unit circle. I submit that the secant of $theta$ is represented by the length $CA'$, on this basis:
- $triangle ABC$ and $triangle A'B'C$ are similar.
- Therefore $CA'$ is to $CB'$ as $CA$ is to $CB$.
- Since $CB' = CA = 1$, we have that $CA'$ is to $1$ as $1$ is to $CB$.
- Symbolically, $CA' = fracCA'1 = frac1CB$.
- And since $CB$ is $cos theta$, $CA' = frac1cos theta = sec theta$.
Finally, imagine increasing $theta$ toward $fracpi2 = 90$ degrees. It is evident that $A'$ must move vertically upward—without bound, in fact, and that is indeed what the plot of $sec theta$ does.
If you increase $theta$ beyond $fracpi2$, but continue to extend $AC$ rightward toward the same vertical tangent, you find that $sec theta$ is negative and initially large, but decreasing toward $0$ as you approach $theta = pi = 180$ degrees.
answered Jul 26 at 17:09
Brian Tung
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
consider how behavior in a graph of $f(x)$ will affect behavior in a graph $frac 1 f(x)$. For values where $0 < f(x) < 1$ then we will have $infty > frac 1f(x)$, with little changes from $teensy$ to $frac 12 teensy$ resulting in huge stretches from $huge$ to $2 times huge$. And if $f(x)$ goes through from $teensy$ to $0$ to $-teensy$ then $frac 1f(x)$ will get asymptotic to infinity, be infinite(undefined) at a point, and jump to "negative infinity" and then imediate start reducing to more moderate negative values as $f(x)$ because more siginificant negative values.
If $1 < f(x)$ we will have $1 > frac 1f(x) > 0$ and if $f(x)$ grows blithely huge toward $huge$, $double-huge$ and $huge^2$ then $frac 1f(x)$ will tend to flatline toward $0$.
So bearing that in mind:
$cos x$ is periodic from $0$ to $1$ back down $0$ and and to $-1$ to $0$ to $1$ etc. and $frac 1cos x$ behaves exactly as we'd expect, for $+infty$ down to a minimum of $1$ (as $cos x$ goes from $0$ to $1$ and peaks) and then back to $infty$. Then as $cos x$ goes from tiny positive through $0$, to tiny negative. $frac 1cos x$ goess to $+infty$ and jumps to $-infty$ and starts surfaces up to $-1$ and then (as $cos x$ reaches a nadir of $-1$ and heads back to $0$) $frac 1cos x$ reaches max at $-1$ and starts plumeting back to $-infty.
As for $tan x$ and $frac 1tan x$. $tan x$ will go from $infty$ where $cos x = 1; sin x= 0$ to $1$ where $cos x = sin x = frac 1sqrt 2$. In this stage $sin x$ is increasing faster than $cos x$ is decreasing. So there will be a bulge to the right in the graph. Then as $cos xto 0$ and $sin x to 1$ we have $frac sin xcos x to infty$. and as $cos x$ is not decreasing faster than $sin x$ is increasing, the graph stretches to be long and skinny. Then as $cos x$ goes through $0$ to negative, we jump to negative infinity.
In the same interval $cot x = frac cos xsin x$ starts and $frac 10 = infty$ and decreases down to $1$ and then to $0$. The exact same but in reverse. Then as $cos x$ goes threough $0$ to negative $frac cos xsin x$ goes ther $0$ to negative.
Useful to realize $cot x =frac cos xsin x = frac sin(frac pi 2 - x)cos frac pi 2 - x) = tan (frac pi 2 - x)$ so that explains more rigororously why their graphs are symmetriic.
answered Jul 26 at 16:42
fleablood
60.3k22575
add a comment |Â
up vote
0
down vote
consider how behavior in a graph of $f(x)$ will affect behavior in a graph $frac 1 f(x)$. For values where $0 < f(x) < 1$ then we will have $infty > frac 1f(x)$, with little changes from $teensy$ to $frac 12 teensy$ resulting in huge stretches from $huge$ to $2 times huge$. And if $f(x)$ goes through from $teensy$ to $0$ to $-teensy$ then $frac 1f(x)$ will get asymptotic to infinity, be infinite(undefined) at a point, and jump to "negative infinity" and then imediate start reducing to more moderate negative values as $f(x)$ because more siginificant negative values.
If $1 < f(x)$ we will have $1 > frac 1f(x) > 0$ and if $f(x)$ grows blithely huge toward $huge$, $double-huge$ and $huge^2$ then $frac 1f(x)$ will tend to flatline toward $0$.
So bearing that in mind:
$cos x$ is periodic from $0$ to $1$ back down $0$ and and to $-1$ to $0$ to $1$ etc. and $frac 1cos x$ behaves exactly as we'd expect, for $+infty$ down to a minimum of $1$ (as $cos x$ goes from $0$ to $1$ and peaks) and then back to $infty$. Then as $cos x$ goes from tiny positive through $0$, to tiny negative. $frac 1cos x$ goess to $+infty$ and jumps to $-infty$ and starts surfaces up to $-1$ and then (as $cos x$ reaches a nadir of $-1$ and heads back to $0$) $frac 1cos x$ reaches max at $-1$ and starts plumeting back to $-infty.
As for $tan x$ and $frac 1tan x$. $tan x$ will go from $infty$ where $cos x = 1; sin x= 0$ to $1$ where $cos x = sin x = frac 1sqrt 2$. In this stage $sin x$ is increasing faster than $cos x$ is decreasing. So there will be a bulge to the right in the graph. Then as $cos xto 0$ and $sin x to 1$ we have $frac sin xcos x to infty$. and as $cos x$ is not decreasing faster than $sin x$ is increasing, the graph stretches to be long and skinny. Then as $cos x$ goes through $0$ to negative, we jump to negative infinity.
In the same interval $cot x = frac cos xsin x$ starts and $frac 10 = infty$ and decreases down to $1$ and then to $0$. The exact same but in reverse. Then as $cos x$ goes threough $0$ to negative $frac cos xsin x$ goes ther $0$ to negative.
Useful to realize $cot x =frac cos xsin x = frac sin(frac pi 2 - x)cos frac pi 2 - x) = tan (frac pi 2 - x)$ so that explains more rigororously why their graphs are symmetriic.
answered Jul 26 at 16:42
fleablood
60.3k22575
add a comment |Â
up vote
0
down vote
up vote
0
down vote
consider how behavior in a graph of $f(x)$ will affect behavior in a graph $frac 1 f(x)$. For values where $0 < f(x) < 1$ then we will have $infty > frac 1f(x)$, with little changes from $teensy$ to $frac 12 teensy$ resulting in huge stretches from $huge$ to $2 times huge$. And if $f(x)$ goes through from $teensy$ to $0$ to $-teensy$ then $frac 1f(x)$ will get asymptotic to infinity, be infinite(undefined) at a point, and jump to "negative infinity" and then imediate start reducing to more moderate negative values as $f(x)$ because more siginificant negative values.
If $1 < f(x)$ we will have $1 > frac 1f(x) > 0$ and if $f(x)$ grows blithely huge toward $huge$, $double-huge$ and $huge^2$ then $frac 1f(x)$ will tend to flatline toward $0$.
So bearing that in mind:
$cos x$ is periodic from $0$ to $1$ back down $0$ and and to $-1$ to $0$ to $1$ etc. and $frac 1cos x$ behaves exactly as we'd expect, for $+infty$ down to a minimum of $1$ (as $cos x$ goes from $0$ to $1$ and peaks) and then back to $infty$. Then as $cos x$ goes from tiny positive through $0$, to tiny negative. $frac 1cos x$ goess to $+infty$ and jumps to $-infty$ and starts surfaces up to $-1$ and then (as $cos x$ reaches a nadir of $-1$ and heads back to $0$) $frac 1cos x$ reaches max at $-1$ and starts plumeting back to $-infty.
As for $tan x$ and $frac 1tan x$. $tan x$ will go from $infty$ where $cos x = 1; sin x= 0$ to $1$ where $cos x = sin x = frac 1sqrt 2$. In this stage $sin x$ is increasing faster than $cos x$ is decreasing. So there will be a bulge to the right in the graph. Then as $cos xto 0$ and $sin x to 1$ we have $frac sin xcos x to infty$. and as $cos x$ is not decreasing faster than $sin x$ is increasing, the graph stretches to be long and skinny. Then as $cos x$ goes through $0$ to negative, we jump to negative infinity.
In the same interval $cot x = frac cos xsin x$ starts and $frac 10 = infty$ and decreases down to $1$ and then to $0$. The exact same but in reverse. Then as $cos x$ goes threough $0$ to negative $frac cos xsin x$ goes ther $0$ to negative.
Useful to realize $cot x =frac cos xsin x = frac sin(frac pi 2 - x)cos frac pi 2 - x) = tan (frac pi 2 - x)$ so that explains more rigororously why their graphs are symmetriic.
answered Jul 26 at 16:42
fleablood
60.3k22575
consider how behavior in a graph of $f(x)$ will affect behavior in a graph $frac 1 f(x)$. For values where $0 < f(x) < 1$ then we will have $infty > frac 1f(x)$, with little changes from $teensy$ to $frac 12 teensy$ resulting in huge stretches from $huge$ to $2 times huge$. And if $f(x)$ goes through from $teensy$ to $0$ to $-teensy$ then $frac 1f(x)$ will get asymptotic to infinity, be infinite(undefined) at a point, and jump to "negative infinity" and then imediate start reducing to more moderate negative values as $f(x)$ because more siginificant negative values.
If $1 < f(x)$ we will have $1 > frac 1f(x) > 0$ and if $f(x)$ grows blithely huge toward $huge$, $double-huge$ and $huge^2$ then $frac 1f(x)$ will tend to flatline toward $0$.
So bearing that in mind:
$cos x$ is periodic from $0$ to $1$ back down $0$ and and to $-1$ to $0$ to $1$ etc. and $frac 1cos x$ behaves exactly as we'd expect, for $+infty$ down to a minimum of $1$ (as $cos x$ goes from $0$ to $1$ and peaks) and then back to $infty$. Then as $cos x$ goes from tiny positive through $0$, to tiny negative. $frac 1cos x$ goess to $+infty$ and jumps to $-infty$ and starts surfaces up to $-1$ and then (as $cos x$ reaches a nadir of $-1$ and heads back to $0$) $frac 1cos x$ reaches max at $-1$ and starts plumeting back to $-infty.
As for $tan x$ and $frac 1tan x$. $tan x$ will go from $infty$ where $cos x = 1; sin x= 0$ to $1$ where $cos x = sin x = frac 1sqrt 2$. In this stage $sin x$ is increasing faster than $cos x$ is decreasing. So there will be a bulge to the right in the graph. Then as $cos xto 0$ and $sin x to 1$ we have $frac sin xcos x to infty$. and as $cos x$ is not decreasing faster than $sin x$ is increasing, the graph stretches to be long and skinny. Then as $cos x$ goes through $0$ to negative, we jump to negative infinity.
In the same interval $cot x = frac cos xsin x$ starts and $frac 10 = infty$ and decreases down to $1$ and then to $0$. The exact same but in reverse. Then as $cos x$ goes threough $0$ to negative $frac cos xsin x$ goes ther $0$ to negative.
Useful to realize $cot x =frac cos xsin x = frac sin(frac pi 2 - x)cos frac pi 2 - x) = tan (frac pi 2 - x)$ so that explains more rigororously why their graphs are symmetriic.
answered Jul 26 at 16:42
fleablood
60.3k22575
answered Jul 26 at 16:42
fleablood
60.3k22575
answered Jul 26 at 16:42
fleablood
60.3k22575
answered Jul 26 at 16:42
fleablood
60.3k22575
60.3k22575
add a comment |Â
add a comment |Â
up vote
0
down vote
Pictures often help.
Here's a graphical depiction of $cos theta$ and $sec theta = frac1cos theta$ for $theta = fracpi3 = 60$ degrees.
The situation for $cos theta$ is represented by the right triangle $triangle ABC$. The radius is $CA = 1$, and the cosine of $theta$ is represented by the length of $CB = frac12$.
Now, extend $CA$ rightward to $A'$, which lies on the vertical line tangent to the unit circle. I submit that the secant of $theta$ is represented by the length $CA'$, on this basis:
- $triangle ABC$ and $triangle A'B'C$ are similar.
- Therefore $CA'$ is to $CB'$ as $CA$ is to $CB$.
- Since $CB' = CA = 1$, we have that $CA'$ is to $1$ as $1$ is to $CB$.
- Symbolically, $CA' = fracCA'1 = frac1CB$.
- And since $CB$ is $cos theta$, $CA' = frac1cos theta = sec theta$.
Finally, imagine increasing $theta$ toward $fracpi2 = 90$ degrees. It is evident that $A'$ must move vertically upward—without bound, in fact, and that is indeed what the plot of $sec theta$ does.
If you increase $theta$ beyond $fracpi2$, but continue to extend $AC$ rightward toward the same vertical tangent, you find that $sec theta$ is negative and initially large, but decreasing toward $0$ as you approach $theta = pi = 180$ degrees.
answered Jul 26 at 17:09
Brian Tung
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Pictures often help.
Here's a graphical depiction of $cos theta$ and $sec theta = frac1cos theta$ for $theta = fracpi3 = 60$ degrees.
The situation for $cos theta$ is represented by the right triangle $triangle ABC$. The radius is $CA = 1$, and the cosine of $theta$ is represented by the length of $CB = frac12$.
Now, extend $CA$ rightward to $A'$, which lies on the vertical line tangent to the unit circle. I submit that the secant of $theta$ is represented by the length $CA'$, on this basis:
- $triangle ABC$ and $triangle A'B'C$ are similar.
- Therefore $CA'$ is to $CB'$ as $CA$ is to $CB$.
- Since $CB' = CA = 1$, we have that $CA'$ is to $1$ as $1$ is to $CB$.
- Symbolically, $CA' = fracCA'1 = frac1CB$.
- And since $CB$ is $cos theta$, $CA' = frac1cos theta = sec theta$.
Finally, imagine increasing $theta$ toward $fracpi2 = 90$ degrees. It is evident that $A'$ must move vertically upward—without bound, in fact, and that is indeed what the plot of $sec theta$ does.
If you increase $theta$ beyond $fracpi2$, but continue to extend $AC$ rightward toward the same vertical tangent, you find that $sec theta$ is negative and initially large, but decreasing toward $0$ as you approach $theta = pi = 180$ degrees.
answered Jul 26 at 17:09
Brian Tung
25.2k32353
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Pictures often help.
Here's a graphical depiction of $cos theta$ and $sec theta = frac1cos theta$ for $theta = fracpi3 = 60$ degrees.
The situation for $cos theta$ is represented by the right triangle $triangle ABC$. The radius is $CA = 1$, and the cosine of $theta$ is represented by the length of $CB = frac12$.
Now, extend $CA$ rightward to $A'$, which lies on the vertical line tangent to the unit circle. I submit that the secant of $theta$ is represented by the length $CA'$, on this basis:
- $triangle ABC$ and $triangle A'B'C$ are similar.
- Therefore $CA'$ is to $CB'$ as $CA$ is to $CB$.
- Since $CB' = CA = 1$, we have that $CA'$ is to $1$ as $1$ is to $CB$.
- Symbolically, $CA' = fracCA'1 = frac1CB$.
- And since $CB$ is $cos theta$, $CA' = frac1cos theta = sec theta$.
Finally, imagine increasing $theta$ toward $fracpi2 = 90$ degrees. It is evident that $A'$ must move vertically upward—without bound, in fact, and that is indeed what the plot of $sec theta$ does.
If you increase $theta$ beyond $fracpi2$, but continue to extend $AC$ rightward toward the same vertical tangent, you find that $sec theta$ is negative and initially large, but decreasing toward $0$ as you approach $theta = pi = 180$ degrees.
answered Jul 26 at 17:09
Brian Tung
25.2k32353
Pictures often help.
Here's a graphical depiction of $cos theta$ and $sec theta = frac1cos theta$ for $theta = fracpi3 = 60$ degrees.
The situation for $cos theta$ is represented by the right triangle $triangle ABC$. The radius is $CA = 1$, and the cosine of $theta$ is represented by the length of $CB = frac12$.
Now, extend $CA$ rightward to $A'$, which lies on the vertical line tangent to the unit circle. I submit that the secant of $theta$ is represented by the length $CA'$, on this basis:
- $triangle ABC$ and $triangle A'B'C$ are similar.
- Therefore $CA'$ is to $CB'$ as $CA$ is to $CB$.
- Since $CB' = CA = 1$, we have that $CA'$ is to $1$ as $1$ is to $CB$.
- Symbolically, $CA' = fracCA'1 = frac1CB$.
- And since $CB$ is $cos theta$, $CA' = frac1cos theta = sec theta$.
Finally, imagine increasing $theta$ toward $fracpi2 = 90$ degrees. It is evident that $A'$ must move vertically upward—without bound, in fact, and that is indeed what the plot of $sec theta$ does.
If you increase $theta$ beyond $fracpi2$, but continue to extend $AC$ rightward toward the same vertical tangent, you find that $sec theta$ is negative and initially large, but decreasing toward $0$ as you approach $theta = pi = 180$ degrees.
answered Jul 26 at 17:09
Brian Tung
25.2k32353
answered Jul 26 at 17:09
Brian Tung
25.2k32353
answered Jul 26 at 17:09
Brian Tung
25.2k32353
answered Jul 26 at 17:09
Brian Tung
25.2k32353
25.2k32353
add a comment |Â
add a comment |Â
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When $cos(theta) = pm 1$, then $sec(theta) = pm 1$. When $cos(theta) = 0$, then $sec(theta)$ is undefined---this implies the possible existence of an asymptote (you need limits to confirm this, and that is usually a topic for the end of pre-calc, or for calculus). Between the ones and the asymptotes, the cosine function either increases or decreases monotonically, so the secant function will have the same behaviour, only in the opposite direction. This can be made rigorous without calculus (though calculus makes it easier).
– Xander Henderson
Jul 26 at 14:31
That being said, I think that this is a question which might be more appropriate for the Mathematics Educators SE.
– Xander Henderson
Jul 26 at 14:32
@XanderHenderson I'm not an educator though; I'm a student. But yeah, thanks for the response. Last thing though, what about for cotangent graphs? Why would it be a perfect reversal of tangent graph?
– Ethan Chan
Jul 26 at 14:35
Because $tan(dfracpi2-x) = cot(x)$
– steven gregory
Jul 26 at 15:16
" Can someone please explain why the secant graph is shaped like that given it's equation? " Can you explain why you don't understand it? It's $frac 1cos x$ and as $cos x$ is periodic from 0 to 1 to 0 to -1 back to 0, It's reciprical will be periodic from $+infty$ to $1$ back to $infty$then an axis jump to $-infty$ up to $-1$ and back down. .... And that is exactly what it does do. So from my point of view $frac 1cos x$ exactly as I expect others would expect it to do. So ... I'm confused why you are confused.
– fleablood
Jul 26 at 16:03