Trigonometric equation over unit circle [closed]
$begingroup$
Let us consider the equations
$$
{p_0}^2 + {q_0}^2 =1 qquadtext{and}qquad {-a} p_0sin s + b q_0cos s = 0
$$
How do i solve for $p_0, q_0$?
calculus trigonometry
edited Dec 10 '18 at 14:16
Harry49
6,17331132
asked Dec 10 '18 at 14:06
pablo_mathscobarpablo_mathscobar
996
$endgroup$
closed as off-topic by Saad, Davide Giraudo, Jyrki Lahtonen, metamorphy, Andrei Dec 18 '18 at 16:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Davide Giraudo, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let us consider the equations
$$
{p_0}^2 + {q_0}^2 =1 qquadtext{and}qquad {-a} p_0sin s + b q_0cos s = 0
$$
How do i solve for $p_0, q_0$?
calculus trigonometry
edited Dec 10 '18 at 14:16
Harry49
6,17331132
asked Dec 10 '18 at 14:06
pablo_mathscobarpablo_mathscobar
996
$endgroup$
closed as off-topic by Saad, Davide Giraudo, Jyrki Lahtonen, metamorphy, Andrei Dec 18 '18 at 16:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Davide Giraudo, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
I've edited your equation assuming you meant $asin(cdot)$ and not $operatorname{asin}(cdot)$. Please revert it if that is incorrect,
$endgroup$
– Jam
Dec 10 '18 at 14:11
$begingroup$
Presumably $a$ and $b$ are fixed constants. But does the second equation hold for all real $s$?
$endgroup$
– Ben W
Dec 10 '18 at 14:14
$begingroup$
@BenW yeah the second equation holds for all real s
$endgroup$
– pablo_mathscobar
Dec 10 '18 at 14:15
$begingroup$
The only way the second equation holds for all real s is: $a p_0 = 0$ and $b q_0 = 0$. If $a ne 0$ and $ b=0$, then $p_0 = 0$ and $q_0 = pm 1$, and if $b ne 0$ and $a=0$, then $q_0 = 0$ and $p_0 = pm 1$. If neither a nor b is zero, then both $p_0$ and $q_0$ must be $0$ which is a contradiction. If both a and b are zero, then the second equation is an identity and all points on the unit circle satisfy both equations.
$endgroup$
– Alex Smart
Dec 10 '18 at 14:34
add a comment |
$begingroup$
Let us consider the equations
$$
{p_0}^2 + {q_0}^2 =1 qquadtext{and}qquad {-a} p_0sin s + b q_0cos s = 0
$$
How do i solve for $p_0, q_0$?
calculus trigonometry
edited Dec 10 '18 at 14:16
Harry49
6,17331132
asked Dec 10 '18 at 14:06
pablo_mathscobarpablo_mathscobar
996
$endgroup$
Let us consider the equations
$$
{p_0}^2 + {q_0}^2 =1 qquadtext{and}qquad {-a} p_0sin s + b q_0cos s = 0
$$
How do i solve for $p_0, q_0$?
calculus trigonometry
calculus trigonometry
edited Dec 10 '18 at 14:16
Harry49
6,17331132
asked Dec 10 '18 at 14:06
pablo_mathscobarpablo_mathscobar
996
edited Dec 10 '18 at 14:16
Harry49
6,17331132
asked Dec 10 '18 at 14:06
pablo_mathscobarpablo_mathscobar
996
edited Dec 10 '18 at 14:16
Harry49
6,17331132
edited Dec 10 '18 at 14:16
Harry49
6,17331132
edited Dec 10 '18 at 14:16
Harry49
6,17331132
6,17331132
asked Dec 10 '18 at 14:06
pablo_mathscobarpablo_mathscobar
996
asked Dec 10 '18 at 14:06
pablo_mathscobarpablo_mathscobar
996
asked Dec 10 '18 at 14:06
pablo_mathscobarpablo_mathscobar
996
996
closed as off-topic by Saad, Davide Giraudo, Jyrki Lahtonen, metamorphy, Andrei Dec 18 '18 at 16:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Davide Giraudo, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Davide Giraudo, Jyrki Lahtonen, metamorphy, Andrei Dec 18 '18 at 16:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Davide Giraudo, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
I've edited your equation assuming you meant $asin(cdot)$ and not $operatorname{asin}(cdot)$. Please revert it if that is incorrect,
$endgroup$
– Jam
Dec 10 '18 at 14:11
$begingroup$
Presumably $a$ and $b$ are fixed constants. But does the second equation hold for all real $s$?
$endgroup$
– Ben W
Dec 10 '18 at 14:14
$begingroup$
@BenW yeah the second equation holds for all real s
$endgroup$
– pablo_mathscobar
Dec 10 '18 at 14:15
$begingroup$
The only way the second equation holds for all real s is: $a p_0 = 0$ and $b q_0 = 0$. If $a ne 0$ and $ b=0$, then $p_0 = 0$ and $q_0 = pm 1$, and if $b ne 0$ and $a=0$, then $q_0 = 0$ and $p_0 = pm 1$. If neither a nor b is zero, then both $p_0$ and $q_0$ must be $0$ which is a contradiction. If both a and b are zero, then the second equation is an identity and all points on the unit circle satisfy both equations.
$endgroup$
– Alex Smart
Dec 10 '18 at 14:34
add a comment |
$begingroup$
I've edited your equation assuming you meant $asin(cdot)$ and not $operatorname{asin}(cdot)$. Please revert it if that is incorrect,
$endgroup$
– Jam
Dec 10 '18 at 14:11
$begingroup$
Presumably $a$ and $b$ are fixed constants. But does the second equation hold for all real $s$?
$endgroup$
– Ben W
Dec 10 '18 at 14:14
$begingroup$
@BenW yeah the second equation holds for all real s
$endgroup$
– pablo_mathscobar
Dec 10 '18 at 14:15
$begingroup$
The only way the second equation holds for all real s is: $a p_0 = 0$ and $b q_0 = 0$. If $a ne 0$ and $ b=0$, then $p_0 = 0$ and $q_0 = pm 1$, and if $b ne 0$ and $a=0$, then $q_0 = 0$ and $p_0 = pm 1$. If neither a nor b is zero, then both $p_0$ and $q_0$ must be $0$ which is a contradiction. If both a and b are zero, then the second equation is an identity and all points on the unit circle satisfy both equations.
$endgroup$
– Alex Smart
Dec 10 '18 at 14:34
$begingroup$
I've edited your equation assuming you meant $asin(cdot)$ and not $operatorname{asin}(cdot)$. Please revert it if that is incorrect,
$endgroup$
– Jam
Dec 10 '18 at 14:11
$begingroup$
I've edited your equation assuming you meant $asin(cdot)$ and not $operatorname{asin}(cdot)$. Please revert it if that is incorrect,
$endgroup$
– Jam
Dec 10 '18 at 14:11
$begingroup$
Presumably $a$ and $b$ are fixed constants. But does the second equation hold for all real $s$?
$endgroup$
– Ben W
Dec 10 '18 at 14:14
$begingroup$
Presumably $a$ and $b$ are fixed constants. But does the second equation hold for all real $s$?
$endgroup$
– Ben W
Dec 10 '18 at 14:14
$begingroup$
@BenW yeah the second equation holds for all real s
$endgroup$
– pablo_mathscobar
Dec 10 '18 at 14:15
$begingroup$
@BenW yeah the second equation holds for all real s
$endgroup$
– pablo_mathscobar
Dec 10 '18 at 14:15
$begingroup$
The only way the second equation holds for all real s is: $a p_0 = 0$ and $b q_0 = 0$. If $a ne 0$ and $ b=0$, then $p_0 = 0$ and $q_0 = pm 1$, and if $b ne 0$ and $a=0$, then $q_0 = 0$ and $p_0 = pm 1$. If neither a nor b is zero, then both $p_0$ and $q_0$ must be $0$ which is a contradiction. If both a and b are zero, then the second equation is an identity and all points on the unit circle satisfy both equations.
$endgroup$
– Alex Smart
Dec 10 '18 at 14:34
$begingroup$
The only way the second equation holds for all real s is: $a p_0 = 0$ and $b q_0 = 0$. If $a ne 0$ and $ b=0$, then $p_0 = 0$ and $q_0 = pm 1$, and if $b ne 0$ and $a=0$, then $q_0 = 0$ and $p_0 = pm 1$. If neither a nor b is zero, then both $p_0$ and $q_0$ must be $0$ which is a contradiction. If both a and b are zero, then the second equation is an identity and all points on the unit circle satisfy both equations.
$endgroup$
– Alex Smart
Dec 10 '18 at 14:34
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $s=0$ so that we have $bq_0=0$, and by letting $s=pi/2$ we get $-ap_0=0$. Then either $b=0$ or $q_0=0$. If the latter, then $p_0in{-1,1}$ by the first equation. Similarly, if $p_0=0$ then $q_0in{-1,1}$. So the solutions $(p_0,q_0)$ must be among $(-1,0)$, $(1,0)$ if $a=0$ and $(0,-1)$, $(0,1)$ if $b=0$. It is routine to verify that these are indeed solutions in the given cases.
answered Dec 10 '18 at 14:21
Ben WBen W
2,234615
$endgroup$
add a comment |
$begingroup$
The first equation is the unit circle and the second is a line passing through the origin. You can solve the two equations to find the points of intersections.
You can parametrize the unit circle $ x^2 + y^2 = 1 $ as $(x, y) = (cos theta, sin theta) $ for $theta in [0, 2 pi)$.
Substituting in the second equation, you have $ tan theta = frac{a}{b} tan s$, or $theta = arctan(frac{a}{b} tan s)$ or $pi +arctan(frac{a}{b} tan s)$.
answered Dec 10 '18 at 14:24
Alex SmartAlex Smart
614
$endgroup$
add a comment |
$begingroup$
Let us rewrite the system in the simple form
$$begin{cases}p^2+q^2=1,\up=vq.end{cases}$$
Then
$$v^2p^2+v^2q^2=v^2$$ is
$$(u^2+v^2)p^2=v^2.$$
The rest is yours.
answered Dec 10 '18 at 15:08
Yves DaoustYves Daoust
125k671223
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $s=0$ so that we have $bq_0=0$, and by letting $s=pi/2$ we get $-ap_0=0$. Then either $b=0$ or $q_0=0$. If the latter, then $p_0in{-1,1}$ by the first equation. Similarly, if $p_0=0$ then $q_0in{-1,1}$. So the solutions $(p_0,q_0)$ must be among $(-1,0)$, $(1,0)$ if $a=0$ and $(0,-1)$, $(0,1)$ if $b=0$. It is routine to verify that these are indeed solutions in the given cases.
answered Dec 10 '18 at 14:21
Ben WBen W
2,234615
$endgroup$
add a comment |
$begingroup$
Let $s=0$ so that we have $bq_0=0$, and by letting $s=pi/2$ we get $-ap_0=0$. Then either $b=0$ or $q_0=0$. If the latter, then $p_0in{-1,1}$ by the first equation. Similarly, if $p_0=0$ then $q_0in{-1,1}$. So the solutions $(p_0,q_0)$ must be among $(-1,0)$, $(1,0)$ if $a=0$ and $(0,-1)$, $(0,1)$ if $b=0$. It is routine to verify that these are indeed solutions in the given cases.
answered Dec 10 '18 at 14:21
Ben WBen W
2,234615
$endgroup$
add a comment |
$begingroup$
Let $s=0$ so that we have $bq_0=0$, and by letting $s=pi/2$ we get $-ap_0=0$. Then either $b=0$ or $q_0=0$. If the latter, then $p_0in{-1,1}$ by the first equation. Similarly, if $p_0=0$ then $q_0in{-1,1}$. So the solutions $(p_0,q_0)$ must be among $(-1,0)$, $(1,0)$ if $a=0$ and $(0,-1)$, $(0,1)$ if $b=0$. It is routine to verify that these are indeed solutions in the given cases.
answered Dec 10 '18 at 14:21
Ben WBen W
2,234615
$endgroup$
Let $s=0$ so that we have $bq_0=0$, and by letting $s=pi/2$ we get $-ap_0=0$. Then either $b=0$ or $q_0=0$. If the latter, then $p_0in{-1,1}$ by the first equation. Similarly, if $p_0=0$ then $q_0in{-1,1}$. So the solutions $(p_0,q_0)$ must be among $(-1,0)$, $(1,0)$ if $a=0$ and $(0,-1)$, $(0,1)$ if $b=0$. It is routine to verify that these are indeed solutions in the given cases.
answered Dec 10 '18 at 14:21
Ben WBen W
2,234615
answered Dec 10 '18 at 14:21
Ben WBen W
2,234615
answered Dec 10 '18 at 14:21
Ben WBen W
2,234615
answered Dec 10 '18 at 14:21
Ben WBen W
2,234615
2,234615
add a comment |
add a comment |
$begingroup$
The first equation is the unit circle and the second is a line passing through the origin. You can solve the two equations to find the points of intersections.
You can parametrize the unit circle $ x^2 + y^2 = 1 $ as $(x, y) = (cos theta, sin theta) $ for $theta in [0, 2 pi)$.
Substituting in the second equation, you have $ tan theta = frac{a}{b} tan s$, or $theta = arctan(frac{a}{b} tan s)$ or $pi +arctan(frac{a}{b} tan s)$.
answered Dec 10 '18 at 14:24
Alex SmartAlex Smart
614
$endgroup$
add a comment |
$begingroup$
The first equation is the unit circle and the second is a line passing through the origin. You can solve the two equations to find the points of intersections.
You can parametrize the unit circle $ x^2 + y^2 = 1 $ as $(x, y) = (cos theta, sin theta) $ for $theta in [0, 2 pi)$.
Substituting in the second equation, you have $ tan theta = frac{a}{b} tan s$, or $theta = arctan(frac{a}{b} tan s)$ or $pi +arctan(frac{a}{b} tan s)$.
answered Dec 10 '18 at 14:24
Alex SmartAlex Smart
614
$endgroup$
add a comment |
$begingroup$
The first equation is the unit circle and the second is a line passing through the origin. You can solve the two equations to find the points of intersections.
You can parametrize the unit circle $ x^2 + y^2 = 1 $ as $(x, y) = (cos theta, sin theta) $ for $theta in [0, 2 pi)$.
Substituting in the second equation, you have $ tan theta = frac{a}{b} tan s$, or $theta = arctan(frac{a}{b} tan s)$ or $pi +arctan(frac{a}{b} tan s)$.
answered Dec 10 '18 at 14:24
Alex SmartAlex Smart
614
$endgroup$
The first equation is the unit circle and the second is a line passing through the origin. You can solve the two equations to find the points of intersections.
You can parametrize the unit circle $ x^2 + y^2 = 1 $ as $(x, y) = (cos theta, sin theta) $ for $theta in [0, 2 pi)$.
Substituting in the second equation, you have $ tan theta = frac{a}{b} tan s$, or $theta = arctan(frac{a}{b} tan s)$ or $pi +arctan(frac{a}{b} tan s)$.
answered Dec 10 '18 at 14:24
Alex SmartAlex Smart
614
answered Dec 10 '18 at 14:24
Alex SmartAlex Smart
614
answered Dec 10 '18 at 14:24
Alex SmartAlex Smart
614
answered Dec 10 '18 at 14:24
Alex SmartAlex Smart
614
614
add a comment |
add a comment |
$begingroup$
Let us rewrite the system in the simple form
$$begin{cases}p^2+q^2=1,\up=vq.end{cases}$$
Then
$$v^2p^2+v^2q^2=v^2$$ is
$$(u^2+v^2)p^2=v^2.$$
The rest is yours.
answered Dec 10 '18 at 15:08
Yves DaoustYves Daoust
125k671223
$endgroup$
add a comment |
$begingroup$
Let us rewrite the system in the simple form
$$begin{cases}p^2+q^2=1,\up=vq.end{cases}$$
Then
$$v^2p^2+v^2q^2=v^2$$ is
$$(u^2+v^2)p^2=v^2.$$
The rest is yours.
answered Dec 10 '18 at 15:08
Yves DaoustYves Daoust
125k671223
$endgroup$
add a comment |
$begingroup$
Let us rewrite the system in the simple form
$$begin{cases}p^2+q^2=1,\up=vq.end{cases}$$
Then
$$v^2p^2+v^2q^2=v^2$$ is
$$(u^2+v^2)p^2=v^2.$$
The rest is yours.
answered Dec 10 '18 at 15:08
Yves DaoustYves Daoust
125k671223
$endgroup$
Let us rewrite the system in the simple form
$$begin{cases}p^2+q^2=1,\up=vq.end{cases}$$
Then
$$v^2p^2+v^2q^2=v^2$$ is
$$(u^2+v^2)p^2=v^2.$$
The rest is yours.
answered Dec 10 '18 at 15:08
Yves DaoustYves Daoust
125k671223
answered Dec 10 '18 at 15:08
Yves DaoustYves Daoust
125k671223
answered Dec 10 '18 at 15:08
Yves DaoustYves Daoust
125k671223
answered Dec 10 '18 at 15:08
Yves DaoustYves Daoust
125k671223
125k671223
add a comment |
add a comment |
$begingroup$
I've edited your equation assuming you meant $asin(cdot)$ and not $operatorname{asin}(cdot)$. Please revert it if that is incorrect,
$endgroup$
– Jam
Dec 10 '18 at 14:11
$begingroup$
Presumably $a$ and $b$ are fixed constants. But does the second equation hold for all real $s$?
$endgroup$
– Ben W
Dec 10 '18 at 14:14
$begingroup$
@BenW yeah the second equation holds for all real s
$endgroup$
– pablo_mathscobar
Dec 10 '18 at 14:15
$begingroup$
The only way the second equation holds for all real s is: $a p_0 = 0$ and $b q_0 = 0$. If $a ne 0$ and $ b=0$, then $p_0 = 0$ and $q_0 = pm 1$, and if $b ne 0$ and $a=0$, then $q_0 = 0$ and $p_0 = pm 1$. If neither a nor b is zero, then both $p_0$ and $q_0$ must be $0$ which is a contradiction. If both a and b are zero, then the second equation is an identity and all points on the unit circle satisfy both equations.
$endgroup$
– Alex Smart
Dec 10 '18 at 14:34