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Can someone please explain to me why cotangent graphs look the way they do?
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Can someone please explain to me why cotangent graphs look the way they do? I want to know why they basically look like mirror reflected tangent graphs. I get that if $tantheta=y/x$, then $cottheta=x/y$. But why would this lead to a graph that looks like a tangent graph that was reflected over?
Can you please try to keep the answers at the level of a high school pre-calc student?
algebra-precalculus trigonometry
edited Aug 3 at 3:58
N. F. Taussig
37.9k93053
asked Aug 3 at 3:53
Ethan Chan
591322
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up vote
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down vote
favorite
2
Can someone please explain to me why cotangent graphs look the way they do? I want to know why they basically look like mirror reflected tangent graphs. I get that if $tantheta=y/x$, then $cottheta=x/y$. But why would this lead to a graph that looks like a tangent graph that was reflected over?
Can you please try to keep the answers at the level of a high school pre-calc student?
algebra-precalculus trigonometry
edited Aug 3 at 3:58
N. F. Taussig
37.9k93053
asked Aug 3 at 3:53
Ethan Chan
591322
1
Not sure how helpful this is (especially because of how it's inverted) but the cotangent function can be thought of as recording the x value of the intersection of the line $y=1$ and the line that travels through origin to a point on the unit circle. Here is a visual of this: desmos.com/calculator/qbdj61u2t9
– Mason
Aug 3 at 4:16
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up vote
3
down vote
favorite
2
up vote
3
down vote
favorite
2
2
Can someone please explain to me why cotangent graphs look the way they do? I want to know why they basically look like mirror reflected tangent graphs. I get that if $tantheta=y/x$, then $cottheta=x/y$. But why would this lead to a graph that looks like a tangent graph that was reflected over?
Can you please try to keep the answers at the level of a high school pre-calc student?
algebra-precalculus trigonometry
edited Aug 3 at 3:58
N. F. Taussig
37.9k93053
asked Aug 3 at 3:53
Ethan Chan
591322
Can someone please explain to me why cotangent graphs look the way they do? I want to know why they basically look like mirror reflected tangent graphs. I get that if $tantheta=y/x$, then $cottheta=x/y$. But why would this lead to a graph that looks like a tangent graph that was reflected over?
Can you please try to keep the answers at the level of a high school pre-calc student?
algebra-precalculus trigonometry
edited Aug 3 at 3:58
N. F. Taussig
37.9k93053
asked Aug 3 at 3:53
Ethan Chan
591322
edited Aug 3 at 3:58
N. F. Taussig
37.9k93053
edited Aug 3 at 3:58
N. F. Taussig
37.9k93053
edited Aug 3 at 3:58
N. F. Taussig
37.9k93053
37.9k93053
asked Aug 3 at 3:53
Ethan Chan
591322
asked Aug 3 at 3:53
Ethan Chan
591322
asked Aug 3 at 3:53
Ethan Chan
591322
591322
1
Not sure how helpful this is (especially because of how it's inverted) but the cotangent function can be thought of as recording the x value of the intersection of the line $y=1$ and the line that travels through origin to a point on the unit circle. Here is a visual of this: desmos.com/calculator/qbdj61u2t9
– Mason
Aug 3 at 4:16
add a comment |Â
1
Not sure how helpful this is (especially because of how it's inverted) but the cotangent function can be thought of as recording the x value of the intersection of the line $y=1$ and the line that travels through origin to a point on the unit circle. Here is a visual of this: desmos.com/calculator/qbdj61u2t9
– Mason
Aug 3 at 4:16
1
1
Not sure how helpful this is (especially because of how it's inverted) but the cotangent function can be thought of as recording the x value of the intersection of the line $y=1$ and the line that travels through origin to a point on the unit circle. Here is a visual of this: desmos.com/calculator/qbdj61u2t9
– Mason
Aug 3 at 4:16
Not sure how helpful this is (especially because of how it's inverted) but the cotangent function can be thought of as recording the x value of the intersection of the line $y=1$ and the line that travels through origin to a point on the unit circle. Here is a visual of this: desmos.com/calculator/qbdj61u2t9
– Mason
Aug 3 at 4:16
add a comment |Â
3 Answers
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We have $$ cot(x) = fraccos(x)sin(x) = fracsin(pi/2-x)cos(pi/2-x) = tan(pi/2-x)$$
so it is reflected and shifted by $pi/2.$
answered Aug 3 at 3:57
spaceisdarkgreen
27.1k21546
But how do we know this till holds when theta, or x in this case, is greater than pi/2? For instance, what if x was 3pi/4? How would this still hold? Can you prove it me please?
– Ethan Chan
Aug 3 at 4:07
6
@EthanChan Do you know what the graphs of $sin$ and $cos$ look like for all $x$?
– spaceisdarkgreen
Aug 3 at 4:17
8
Note: "it is reflected and shifted by $À/2$" holds for every trigonometric function when you change function to cofunction or the other way around. Quite useful. The answer uses this formula with $cos$ and $sin$ to obtain it for $cot$.
– Kamil Maciorowski
Aug 3 at 6:50
3
@EthanChan: In case you were taught the definition of $sin,cos$ in terms of the right-angle triangle, here is the correct way to extend it to other angles. Consider the unit circle $C$ in the $x,y$-plane (centred at origin with radius $1$). For angle $t$, draw a ray $R$ through the origin that is the result of rotating the $x$-axis by $t$ anti-clockwise. Let $P$ be the intersection of $R$ and $C$. Then define $cos(x),sin(x)$ as the $x,y$ coordinates of $P$. Equivalently, $P$ is the result of starting from $(1,0)$ and moving anti-clockwise along $C$ by a distance (arc-length) of $t$.
– user21820
Aug 3 at 11:59
2
@EthanChan: Then observe that if you change the angle from $t$ to $À/2-t$ it is equivalent to reflecting $P$ across the line $x=y$. This is the geometric reason why $cos(t) = sin(À/2-t)$ and $sin(t) = cos(À/2-t)$, because the $x,y$-coordinates are swapped in such a reflection.
– user21820
Aug 3 at 12:02
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$$cottheta = fraccosthetasintheta = fracsin(fracpi2-theta)cos(fracpi2-theta) = tanBig(fracpi2-thetaBig)$$
answered Aug 3 at 3:57
mr_e_man
843119
Okay. Last thing, how does this work for values of theta greater than 90 degrees?
– Ethan Chan
Aug 3 at 3:58
These identities are always true, as long as $sinthetaneq0$.
– mr_e_man
Aug 3 at 3:59
But can you prove why they will always be true?
– Ethan Chan
Aug 3 at 4:08
1
That depends on your definitions of the trig functions. Cotangent is always cosine over sine, and tangent is always sine over cosine. All that remains is whether $costheta = sin(pi/2-theta)$ (which is equivalent to the same with $cos$ and $sin$ swapped; try $theta=pi/2-phi$).
– mr_e_man
Aug 3 at 4:10
If you're using the unit circle, defining $costheta=x$ and $sintheta=y$, then swapping $sintheta$ and $costheta$ is the same as reflecting across the line $y=x$. This reflects the angle relative to $pi/4$. $$theta'-fracpi4=-Big(theta-fracpi4Big)$$ $$theta'=fracpi2-theta$$ $$costheta=sintheta'=sin(pi/2-theta)$$
– mr_e_man
Aug 3 at 4:23
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up vote
4
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Just as a visual complement for the answers you already have, notice the following animation.
Green angle: $theta$, yellow angle: $pi/2 - theta$ and $cot(x)$ is the red curve.
For more details regarding these types of constructions, see this question.
answered Aug 3 at 14:59
A. Arredondo
656
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
19
down vote
accepted
We have $$ cot(x) = fraccos(x)sin(x) = fracsin(pi/2-x)cos(pi/2-x) = tan(pi/2-x)$$
so it is reflected and shifted by $pi/2.$
answered Aug 3 at 3:57
spaceisdarkgreen
27.1k21546
But how do we know this till holds when theta, or x in this case, is greater than pi/2? For instance, what if x was 3pi/4? How would this still hold? Can you prove it me please?
– Ethan Chan
Aug 3 at 4:07
6
@EthanChan Do you know what the graphs of $sin$ and $cos$ look like for all $x$?
– spaceisdarkgreen
Aug 3 at 4:17
8
Note: "it is reflected and shifted by $À/2$" holds for every trigonometric function when you change function to cofunction or the other way around. Quite useful. The answer uses this formula with $cos$ and $sin$ to obtain it for $cot$.
– Kamil Maciorowski
Aug 3 at 6:50
3
@EthanChan: In case you were taught the definition of $sin,cos$ in terms of the right-angle triangle, here is the correct way to extend it to other angles. Consider the unit circle $C$ in the $x,y$-plane (centred at origin with radius $1$). For angle $t$, draw a ray $R$ through the origin that is the result of rotating the $x$-axis by $t$ anti-clockwise. Let $P$ be the intersection of $R$ and $C$. Then define $cos(x),sin(x)$ as the $x,y$ coordinates of $P$. Equivalently, $P$ is the result of starting from $(1,0)$ and moving anti-clockwise along $C$ by a distance (arc-length) of $t$.
– user21820
Aug 3 at 11:59
2
@EthanChan: Then observe that if you change the angle from $t$ to $À/2-t$ it is equivalent to reflecting $P$ across the line $x=y$. This is the geometric reason why $cos(t) = sin(À/2-t)$ and $sin(t) = cos(À/2-t)$, because the $x,y$-coordinates are swapped in such a reflection.
– user21820
Aug 3 at 12:02
add a comment |Â
up vote
19
down vote
accepted
We have $$ cot(x) = fraccos(x)sin(x) = fracsin(pi/2-x)cos(pi/2-x) = tan(pi/2-x)$$
so it is reflected and shifted by $pi/2.$
answered Aug 3 at 3:57
spaceisdarkgreen
27.1k21546
But how do we know this till holds when theta, or x in this case, is greater than pi/2? For instance, what if x was 3pi/4? How would this still hold? Can you prove it me please?
– Ethan Chan
Aug 3 at 4:07
6
@EthanChan Do you know what the graphs of $sin$ and $cos$ look like for all $x$?
– spaceisdarkgreen
Aug 3 at 4:17
8
Note: "it is reflected and shifted by $À/2$" holds for every trigonometric function when you change function to cofunction or the other way around. Quite useful. The answer uses this formula with $cos$ and $sin$ to obtain it for $cot$.
– Kamil Maciorowski
Aug 3 at 6:50
3
@EthanChan: In case you were taught the definition of $sin,cos$ in terms of the right-angle triangle, here is the correct way to extend it to other angles. Consider the unit circle $C$ in the $x,y$-plane (centred at origin with radius $1$). For angle $t$, draw a ray $R$ through the origin that is the result of rotating the $x$-axis by $t$ anti-clockwise. Let $P$ be the intersection of $R$ and $C$. Then define $cos(x),sin(x)$ as the $x,y$ coordinates of $P$. Equivalently, $P$ is the result of starting from $(1,0)$ and moving anti-clockwise along $C$ by a distance (arc-length) of $t$.
– user21820
Aug 3 at 11:59
2
@EthanChan: Then observe that if you change the angle from $t$ to $À/2-t$ it is equivalent to reflecting $P$ across the line $x=y$. This is the geometric reason why $cos(t) = sin(À/2-t)$ and $sin(t) = cos(À/2-t)$, because the $x,y$-coordinates are swapped in such a reflection.
– user21820
Aug 3 at 12:02
add a comment |Â
up vote
19
down vote
accepted
up vote
19
down vote
accepted
We have $$ cot(x) = fraccos(x)sin(x) = fracsin(pi/2-x)cos(pi/2-x) = tan(pi/2-x)$$
so it is reflected and shifted by $pi/2.$
answered Aug 3 at 3:57
spaceisdarkgreen
27.1k21546
We have $$ cot(x) = fraccos(x)sin(x) = fracsin(pi/2-x)cos(pi/2-x) = tan(pi/2-x)$$
so it is reflected and shifted by $pi/2.$
answered Aug 3 at 3:57
spaceisdarkgreen
27.1k21546
answered Aug 3 at 3:57
spaceisdarkgreen
27.1k21546
answered Aug 3 at 3:57
spaceisdarkgreen
27.1k21546
answered Aug 3 at 3:57
spaceisdarkgreen
27.1k21546
27.1k21546
But how do we know this till holds when theta, or x in this case, is greater than pi/2? For instance, what if x was 3pi/4? How would this still hold? Can you prove it me please?
– Ethan Chan
Aug 3 at 4:07
6
@EthanChan Do you know what the graphs of $sin$ and $cos$ look like for all $x$?
– spaceisdarkgreen
Aug 3 at 4:17
8
Note: "it is reflected and shifted by $À/2$" holds for every trigonometric function when you change function to cofunction or the other way around. Quite useful. The answer uses this formula with $cos$ and $sin$ to obtain it for $cot$.
– Kamil Maciorowski
Aug 3 at 6:50
3
@EthanChan: In case you were taught the definition of $sin,cos$ in terms of the right-angle triangle, here is the correct way to extend it to other angles. Consider the unit circle $C$ in the $x,y$-plane (centred at origin with radius $1$). For angle $t$, draw a ray $R$ through the origin that is the result of rotating the $x$-axis by $t$ anti-clockwise. Let $P$ be the intersection of $R$ and $C$. Then define $cos(x),sin(x)$ as the $x,y$ coordinates of $P$. Equivalently, $P$ is the result of starting from $(1,0)$ and moving anti-clockwise along $C$ by a distance (arc-length) of $t$.
– user21820
Aug 3 at 11:59
2
@EthanChan: Then observe that if you change the angle from $t$ to $À/2-t$ it is equivalent to reflecting $P$ across the line $x=y$. This is the geometric reason why $cos(t) = sin(À/2-t)$ and $sin(t) = cos(À/2-t)$, because the $x,y$-coordinates are swapped in such a reflection.
– user21820
Aug 3 at 12:02
add a comment |Â
But how do we know this till holds when theta, or x in this case, is greater than pi/2? For instance, what if x was 3pi/4? How would this still hold? Can you prove it me please?
– Ethan Chan
Aug 3 at 4:07
6
@EthanChan Do you know what the graphs of $sin$ and $cos$ look like for all $x$?
– spaceisdarkgreen
Aug 3 at 4:17
8
Note: "it is reflected and shifted by $À/2$" holds for every trigonometric function when you change function to cofunction or the other way around. Quite useful. The answer uses this formula with $cos$ and $sin$ to obtain it for $cot$.
– Kamil Maciorowski
Aug 3 at 6:50
3
@EthanChan: In case you were taught the definition of $sin,cos$ in terms of the right-angle triangle, here is the correct way to extend it to other angles. Consider the unit circle $C$ in the $x,y$-plane (centred at origin with radius $1$). For angle $t$, draw a ray $R$ through the origin that is the result of rotating the $x$-axis by $t$ anti-clockwise. Let $P$ be the intersection of $R$ and $C$. Then define $cos(x),sin(x)$ as the $x,y$ coordinates of $P$. Equivalently, $P$ is the result of starting from $(1,0)$ and moving anti-clockwise along $C$ by a distance (arc-length) of $t$.
– user21820
Aug 3 at 11:59
2
@EthanChan: Then observe that if you change the angle from $t$ to $À/2-t$ it is equivalent to reflecting $P$ across the line $x=y$. This is the geometric reason why $cos(t) = sin(À/2-t)$ and $sin(t) = cos(À/2-t)$, because the $x,y$-coordinates are swapped in such a reflection.
– user21820
Aug 3 at 12:02
But how do we know this till holds when theta, or x in this case, is greater than pi/2? For instance, what if x was 3pi/4? How would this still hold? Can you prove it me please?
– Ethan Chan
Aug 3 at 4:07
But how do we know this till holds when theta, or x in this case, is greater than pi/2? For instance, what if x was 3pi/4? How would this still hold? Can you prove it me please?
– Ethan Chan
Aug 3 at 4:07
6
6
@EthanChan Do you know what the graphs of $sin$ and $cos$ look like for all $x$?
– spaceisdarkgreen
Aug 3 at 4:17
@EthanChan Do you know what the graphs of $sin$ and $cos$ look like for all $x$?
– spaceisdarkgreen
Aug 3 at 4:17
8
8
Note: "it is reflected and shifted by $À/2$" holds for every trigonometric function when you change function to cofunction or the other way around. Quite useful. The answer uses this formula with $cos$ and $sin$ to obtain it for $cot$.
– Kamil Maciorowski
Aug 3 at 6:50
Note: "it is reflected and shifted by $À/2$" holds for every trigonometric function when you change function to cofunction or the other way around. Quite useful. The answer uses this formula with $cos$ and $sin$ to obtain it for $cot$.
– Kamil Maciorowski
Aug 3 at 6:50
3
3
@EthanChan: In case you were taught the definition of $sin,cos$ in terms of the right-angle triangle, here is the correct way to extend it to other angles. Consider the unit circle $C$ in the $x,y$-plane (centred at origin with radius $1$). For angle $t$, draw a ray $R$ through the origin that is the result of rotating the $x$-axis by $t$ anti-clockwise. Let $P$ be the intersection of $R$ and $C$. Then define $cos(x),sin(x)$ as the $x,y$ coordinates of $P$. Equivalently, $P$ is the result of starting from $(1,0)$ and moving anti-clockwise along $C$ by a distance (arc-length) of $t$.
– user21820
Aug 3 at 11:59
@EthanChan: In case you were taught the definition of $sin,cos$ in terms of the right-angle triangle, here is the correct way to extend it to other angles. Consider the unit circle $C$ in the $x,y$-plane (centred at origin with radius $1$). For angle $t$, draw a ray $R$ through the origin that is the result of rotating the $x$-axis by $t$ anti-clockwise. Let $P$ be the intersection of $R$ and $C$. Then define $cos(x),sin(x)$ as the $x,y$ coordinates of $P$. Equivalently, $P$ is the result of starting from $(1,0)$ and moving anti-clockwise along $C$ by a distance (arc-length) of $t$.
– user21820
Aug 3 at 11:59
2
2
@EthanChan: Then observe that if you change the angle from $t$ to $À/2-t$ it is equivalent to reflecting $P$ across the line $x=y$. This is the geometric reason why $cos(t) = sin(À/2-t)$ and $sin(t) = cos(À/2-t)$, because the $x,y$-coordinates are swapped in such a reflection.
– user21820
Aug 3 at 12:02
@EthanChan: Then observe that if you change the angle from $t$ to $À/2-t$ it is equivalent to reflecting $P$ across the line $x=y$. This is the geometric reason why $cos(t) = sin(À/2-t)$ and $sin(t) = cos(À/2-t)$, because the $x,y$-coordinates are swapped in such a reflection.
– user21820
Aug 3 at 12:02
add a comment |Â
up vote
4
down vote
$$cottheta = fraccosthetasintheta = fracsin(fracpi2-theta)cos(fracpi2-theta) = tanBig(fracpi2-thetaBig)$$
answered Aug 3 at 3:57
mr_e_man
843119
Okay. Last thing, how does this work for values of theta greater than 90 degrees?
– Ethan Chan
Aug 3 at 3:58
These identities are always true, as long as $sinthetaneq0$.
– mr_e_man
Aug 3 at 3:59
But can you prove why they will always be true?
– Ethan Chan
Aug 3 at 4:08
1
That depends on your definitions of the trig functions. Cotangent is always cosine over sine, and tangent is always sine over cosine. All that remains is whether $costheta = sin(pi/2-theta)$ (which is equivalent to the same with $cos$ and $sin$ swapped; try $theta=pi/2-phi$).
– mr_e_man
Aug 3 at 4:10
If you're using the unit circle, defining $costheta=x$ and $sintheta=y$, then swapping $sintheta$ and $costheta$ is the same as reflecting across the line $y=x$. This reflects the angle relative to $pi/4$. $$theta'-fracpi4=-Big(theta-fracpi4Big)$$ $$theta'=fracpi2-theta$$ $$costheta=sintheta'=sin(pi/2-theta)$$
– mr_e_man
Aug 3 at 4:23
add a comment |Â
up vote
4
down vote
$$cottheta = fraccosthetasintheta = fracsin(fracpi2-theta)cos(fracpi2-theta) = tanBig(fracpi2-thetaBig)$$
answered Aug 3 at 3:57
mr_e_man
843119
Okay. Last thing, how does this work for values of theta greater than 90 degrees?
– Ethan Chan
Aug 3 at 3:58
These identities are always true, as long as $sinthetaneq0$.
– mr_e_man
Aug 3 at 3:59
But can you prove why they will always be true?
– Ethan Chan
Aug 3 at 4:08
1
That depends on your definitions of the trig functions. Cotangent is always cosine over sine, and tangent is always sine over cosine. All that remains is whether $costheta = sin(pi/2-theta)$ (which is equivalent to the same with $cos$ and $sin$ swapped; try $theta=pi/2-phi$).
– mr_e_man
Aug 3 at 4:10
If you're using the unit circle, defining $costheta=x$ and $sintheta=y$, then swapping $sintheta$ and $costheta$ is the same as reflecting across the line $y=x$. This reflects the angle relative to $pi/4$. $$theta'-fracpi4=-Big(theta-fracpi4Big)$$ $$theta'=fracpi2-theta$$ $$costheta=sintheta'=sin(pi/2-theta)$$
– mr_e_man
Aug 3 at 4:23
add a comment |Â
up vote
4
down vote
up vote
4
down vote
$$cottheta = fraccosthetasintheta = fracsin(fracpi2-theta)cos(fracpi2-theta) = tanBig(fracpi2-thetaBig)$$
answered Aug 3 at 3:57
mr_e_man
843119
$$cottheta = fraccosthetasintheta = fracsin(fracpi2-theta)cos(fracpi2-theta) = tanBig(fracpi2-thetaBig)$$
answered Aug 3 at 3:57
mr_e_man
843119
answered Aug 3 at 3:57
mr_e_man
843119
answered Aug 3 at 3:57
mr_e_man
843119
answered Aug 3 at 3:57
mr_e_man
843119
843119
Okay. Last thing, how does this work for values of theta greater than 90 degrees?
– Ethan Chan
Aug 3 at 3:58
These identities are always true, as long as $sinthetaneq0$.
– mr_e_man
Aug 3 at 3:59
But can you prove why they will always be true?
– Ethan Chan
Aug 3 at 4:08
1
That depends on your definitions of the trig functions. Cotangent is always cosine over sine, and tangent is always sine over cosine. All that remains is whether $costheta = sin(pi/2-theta)$ (which is equivalent to the same with $cos$ and $sin$ swapped; try $theta=pi/2-phi$).
– mr_e_man
Aug 3 at 4:10
If you're using the unit circle, defining $costheta=x$ and $sintheta=y$, then swapping $sintheta$ and $costheta$ is the same as reflecting across the line $y=x$. This reflects the angle relative to $pi/4$. $$theta'-fracpi4=-Big(theta-fracpi4Big)$$ $$theta'=fracpi2-theta$$ $$costheta=sintheta'=sin(pi/2-theta)$$
– mr_e_man
Aug 3 at 4:23
add a comment |Â
Okay. Last thing, how does this work for values of theta greater than 90 degrees?
– Ethan Chan
Aug 3 at 3:58
These identities are always true, as long as $sinthetaneq0$.
– mr_e_man
Aug 3 at 3:59
But can you prove why they will always be true?
– Ethan Chan
Aug 3 at 4:08
1
That depends on your definitions of the trig functions. Cotangent is always cosine over sine, and tangent is always sine over cosine. All that remains is whether $costheta = sin(pi/2-theta)$ (which is equivalent to the same with $cos$ and $sin$ swapped; try $theta=pi/2-phi$).
– mr_e_man
Aug 3 at 4:10
If you're using the unit circle, defining $costheta=x$ and $sintheta=y$, then swapping $sintheta$ and $costheta$ is the same as reflecting across the line $y=x$. This reflects the angle relative to $pi/4$. $$theta'-fracpi4=-Big(theta-fracpi4Big)$$ $$theta'=fracpi2-theta$$ $$costheta=sintheta'=sin(pi/2-theta)$$
– mr_e_man
Aug 3 at 4:23
Okay. Last thing, how does this work for values of theta greater than 90 degrees?
– Ethan Chan
Aug 3 at 3:58
Okay. Last thing, how does this work for values of theta greater than 90 degrees?
– Ethan Chan
Aug 3 at 3:58
These identities are always true, as long as $sinthetaneq0$.
– mr_e_man
Aug 3 at 3:59
These identities are always true, as long as $sinthetaneq0$.
– mr_e_man
Aug 3 at 3:59
But can you prove why they will always be true?
– Ethan Chan
Aug 3 at 4:08
But can you prove why they will always be true?
– Ethan Chan
Aug 3 at 4:08
1
1
That depends on your definitions of the trig functions. Cotangent is always cosine over sine, and tangent is always sine over cosine. All that remains is whether $costheta = sin(pi/2-theta)$ (which is equivalent to the same with $cos$ and $sin$ swapped; try $theta=pi/2-phi$).
– mr_e_man
Aug 3 at 4:10
That depends on your definitions of the trig functions. Cotangent is always cosine over sine, and tangent is always sine over cosine. All that remains is whether $costheta = sin(pi/2-theta)$ (which is equivalent to the same with $cos$ and $sin$ swapped; try $theta=pi/2-phi$).
– mr_e_man
Aug 3 at 4:10
If you're using the unit circle, defining $costheta=x$ and $sintheta=y$, then swapping $sintheta$ and $costheta$ is the same as reflecting across the line $y=x$. This reflects the angle relative to $pi/4$. $$theta'-fracpi4=-Big(theta-fracpi4Big)$$ $$theta'=fracpi2-theta$$ $$costheta=sintheta'=sin(pi/2-theta)$$
– mr_e_man
Aug 3 at 4:23
If you're using the unit circle, defining $costheta=x$ and $sintheta=y$, then swapping $sintheta$ and $costheta$ is the same as reflecting across the line $y=x$. This reflects the angle relative to $pi/4$. $$theta'-fracpi4=-Big(theta-fracpi4Big)$$ $$theta'=fracpi2-theta$$ $$costheta=sintheta'=sin(pi/2-theta)$$
– mr_e_man
Aug 3 at 4:23
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up vote
4
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Just as a visual complement for the answers you already have, notice the following animation.
Green angle: $theta$, yellow angle: $pi/2 - theta$ and $cot(x)$ is the red curve.
For more details regarding these types of constructions, see this question.
answered Aug 3 at 14:59
A. Arredondo
656
add a comment |Â
up vote
4
down vote
Just as a visual complement for the answers you already have, notice the following animation.
Green angle: $theta$, yellow angle: $pi/2 - theta$ and $cot(x)$ is the red curve.
For more details regarding these types of constructions, see this question.
answered Aug 3 at 14:59
A. Arredondo
656
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Just as a visual complement for the answers you already have, notice the following animation.
Green angle: $theta$, yellow angle: $pi/2 - theta$ and $cot(x)$ is the red curve.
For more details regarding these types of constructions, see this question.
answered Aug 3 at 14:59
A. Arredondo
656
Just as a visual complement for the answers you already have, notice the following animation.
Green angle: $theta$, yellow angle: $pi/2 - theta$ and $cot(x)$ is the red curve.
For more details regarding these types of constructions, see this question.
answered Aug 3 at 14:59
A. Arredondo
656
answered Aug 3 at 14:59
A. Arredondo
656
answered Aug 3 at 14:59
A. Arredondo
656
answered Aug 3 at 14:59
A. Arredondo
656
656
add a comment |Â
add a comment |Â
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1
Not sure how helpful this is (especially because of how it's inverted) but the cotangent function can be thought of as recording the x value of the intersection of the line $y=1$ and the line that travels through origin to a point on the unit circle. Here is a visual of this: desmos.com/calculator/qbdj61u2t9
– Mason
Aug 3 at 4:16