Solve for common points in a circle and curve
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
[problem]
I got following equation of a circle..
$x^2+y^2=9 ..(1)$
and a equation of curve as follows..
$tan^4x + cot^4 x + 2 =4sin^2y....(2)$
I have to find points which satisfy both the equation..i.e points which are on the curve and inscribed inside or on the circle in common
[ My approach]
For solving following equation I first equate for trignometric function,So that i can obtain equation(2) in terms of x and y...Mathmetically I targeted for this.
$y=frac arcsin sqrt(tan^4x + cot^4 x + 2 ) 2$
(from equation (2))
I am trying to equate $sqrt(tan^4x + cot^4 x + 2 )$ in terms of $sin a$ where "a" is any polynomial expression ..
[HELP]
How should i do so..I cant break that in terms or sin function..It's going lengthy and complex . Or if any other less costly approach could be made
trigonometry quadratics
add a comment |Â
up vote
2
down vote
favorite
[problem]
I got following equation of a circle..
$x^2+y^2=9 ..(1)$
and a equation of curve as follows..
$tan^4x + cot^4 x + 2 =4sin^2y....(2)$
I have to find points which satisfy both the equation..i.e points which are on the curve and inscribed inside or on the circle in common
[ My approach]
For solving following equation I first equate for trignometric function,So that i can obtain equation(2) in terms of x and y...Mathmetically I targeted for this.
$y=frac arcsin sqrt(tan^4x + cot^4 x + 2 ) 2$
(from equation (2))
I am trying to equate $sqrt(tan^4x + cot^4 x + 2 )$ in terms of $sin a$ where "a" is any polynomial expression ..
[HELP]
How should i do so..I cant break that in terms or sin function..It's going lengthy and complex . Or if any other less costly approach could be made
trigonometry quadratics
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
[problem]
I got following equation of a circle..
$x^2+y^2=9 ..(1)$
and a equation of curve as follows..
$tan^4x + cot^4 x + 2 =4sin^2y....(2)$
I have to find points which satisfy both the equation..i.e points which are on the curve and inscribed inside or on the circle in common
[ My approach]
For solving following equation I first equate for trignometric function,So that i can obtain equation(2) in terms of x and y...Mathmetically I targeted for this.
$y=frac arcsin sqrt(tan^4x + cot^4 x + 2 ) 2$
(from equation (2))
I am trying to equate $sqrt(tan^4x + cot^4 x + 2 )$ in terms of $sin a$ where "a" is any polynomial expression ..
[HELP]
How should i do so..I cant break that in terms or sin function..It's going lengthy and complex . Or if any other less costly approach could be made
trigonometry quadratics
[problem]
I got following equation of a circle..
$x^2+y^2=9 ..(1)$
and a equation of curve as follows..
$tan^4x + cot^4 x + 2 =4sin^2y....(2)$
I have to find points which satisfy both the equation..i.e points which are on the curve and inscribed inside or on the circle in common
[ My approach]
For solving following equation I first equate for trignometric function,So that i can obtain equation(2) in terms of x and y...Mathmetically I targeted for this.
$y=frac arcsin sqrt(tan^4x + cot^4 x + 2 ) 2$
(from equation (2))
I am trying to equate $sqrt(tan^4x + cot^4 x + 2 )$ in terms of $sin a$ where "a" is any polynomial expression ..
[HELP]
How should i do so..I cant break that in terms or sin function..It's going lengthy and complex . Or if any other less costly approach could be made
trigonometry quadratics
trigonometry quadratics
edited Sep 5 at 10:54
Kevin
5,138722
5,138722
asked Sep 5 at 10:52
Shivanshu Raj
184
184
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
Hint:
For real $x,$
$$dfractan^4x+cot^4x2gesqrttan^4xcot^4x=?$$
$$impliessin^2yge1impliessin^2y=1$$ for real$y$
did you left + sign in equation of inequilaty or multiplying tan cot??
â Shivanshu Raj
Sep 5 at 15:19
@Raj, I have used Inequality_of_arithmetic_and_geometric_means
â lab bhattacharjee
Sep 5 at 15:47
I still can't get till here....I am still confusing lot
â Shivanshu Raj
Sep 6 at 10:38
@ShivanshuRaj, $$(tan^2x-cot^2x)^2ge0$$ for real positive $tan^2x,cot^2x$. Following this logic, we need $$sin^2y=1impliescos y=0$$ right? So, $y=?$
â lab bhattacharjee
Sep 6 at 10:45
y became a linear equation
â Shivanshu Raj
Sep 7 at 2:42
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint:
For real $x,$
$$dfractan^4x+cot^4x2gesqrttan^4xcot^4x=?$$
$$impliessin^2yge1impliessin^2y=1$$ for real$y$
did you left + sign in equation of inequilaty or multiplying tan cot??
â Shivanshu Raj
Sep 5 at 15:19
@Raj, I have used Inequality_of_arithmetic_and_geometric_means
â lab bhattacharjee
Sep 5 at 15:47
I still can't get till here....I am still confusing lot
â Shivanshu Raj
Sep 6 at 10:38
@ShivanshuRaj, $$(tan^2x-cot^2x)^2ge0$$ for real positive $tan^2x,cot^2x$. Following this logic, we need $$sin^2y=1impliescos y=0$$ right? So, $y=?$
â lab bhattacharjee
Sep 6 at 10:45
y became a linear equation
â Shivanshu Raj
Sep 7 at 2:42
add a comment |Â
up vote
1
down vote
Hint:
For real $x,$
$$dfractan^4x+cot^4x2gesqrttan^4xcot^4x=?$$
$$impliessin^2yge1impliessin^2y=1$$ for real$y$
did you left + sign in equation of inequilaty or multiplying tan cot??
â Shivanshu Raj
Sep 5 at 15:19
@Raj, I have used Inequality_of_arithmetic_and_geometric_means
â lab bhattacharjee
Sep 5 at 15:47
I still can't get till here....I am still confusing lot
â Shivanshu Raj
Sep 6 at 10:38
@ShivanshuRaj, $$(tan^2x-cot^2x)^2ge0$$ for real positive $tan^2x,cot^2x$. Following this logic, we need $$sin^2y=1impliescos y=0$$ right? So, $y=?$
â lab bhattacharjee
Sep 6 at 10:45
y became a linear equation
â Shivanshu Raj
Sep 7 at 2:42
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint:
For real $x,$
$$dfractan^4x+cot^4x2gesqrttan^4xcot^4x=?$$
$$impliessin^2yge1impliessin^2y=1$$ for real$y$
Hint:
For real $x,$
$$dfractan^4x+cot^4x2gesqrttan^4xcot^4x=?$$
$$impliessin^2yge1impliessin^2y=1$$ for real$y$
answered Sep 5 at 14:10
lab bhattacharjee
216k14153266
216k14153266
did you left + sign in equation of inequilaty or multiplying tan cot??
â Shivanshu Raj
Sep 5 at 15:19
@Raj, I have used Inequality_of_arithmetic_and_geometric_means
â lab bhattacharjee
Sep 5 at 15:47
I still can't get till here....I am still confusing lot
â Shivanshu Raj
Sep 6 at 10:38
@ShivanshuRaj, $$(tan^2x-cot^2x)^2ge0$$ for real positive $tan^2x,cot^2x$. Following this logic, we need $$sin^2y=1impliescos y=0$$ right? So, $y=?$
â lab bhattacharjee
Sep 6 at 10:45
y became a linear equation
â Shivanshu Raj
Sep 7 at 2:42
add a comment |Â
did you left + sign in equation of inequilaty or multiplying tan cot??
â Shivanshu Raj
Sep 5 at 15:19
@Raj, I have used Inequality_of_arithmetic_and_geometric_means
â lab bhattacharjee
Sep 5 at 15:47
I still can't get till here....I am still confusing lot
â Shivanshu Raj
Sep 6 at 10:38
@ShivanshuRaj, $$(tan^2x-cot^2x)^2ge0$$ for real positive $tan^2x,cot^2x$. Following this logic, we need $$sin^2y=1impliescos y=0$$ right? So, $y=?$
â lab bhattacharjee
Sep 6 at 10:45
y became a linear equation
â Shivanshu Raj
Sep 7 at 2:42
did you left + sign in equation of inequilaty or multiplying tan cot??
â Shivanshu Raj
Sep 5 at 15:19
did you left + sign in equation of inequilaty or multiplying tan cot??
â Shivanshu Raj
Sep 5 at 15:19
@Raj, I have used Inequality_of_arithmetic_and_geometric_means
â lab bhattacharjee
Sep 5 at 15:47
@Raj, I have used Inequality_of_arithmetic_and_geometric_means
â lab bhattacharjee
Sep 5 at 15:47
I still can't get till here....I am still confusing lot
â Shivanshu Raj
Sep 6 at 10:38
I still can't get till here....I am still confusing lot
â Shivanshu Raj
Sep 6 at 10:38
@ShivanshuRaj, $$(tan^2x-cot^2x)^2ge0$$ for real positive $tan^2x,cot^2x$. Following this logic, we need $$sin^2y=1impliescos y=0$$ right? So, $y=?$
â lab bhattacharjee
Sep 6 at 10:45
@ShivanshuRaj, $$(tan^2x-cot^2x)^2ge0$$ for real positive $tan^2x,cot^2x$. Following this logic, we need $$sin^2y=1impliescos y=0$$ right? So, $y=?$
â lab bhattacharjee
Sep 6 at 10:45
y became a linear equation
â Shivanshu Raj
Sep 7 at 2:42
y became a linear equation
â Shivanshu Raj
Sep 7 at 2:42
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2906129%2fsolve-for-common-points-in-a-circle-and-curve%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password