Mixing
Trigonometry Related Problem involving Quadratic Equations [closed]
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If $alpha$ and $beta$ are the solutions of the equations $$sin^2x+asin x+b=0$$
and $$cos^2x+c cos x+d=0$$ then $sin(alpha+beta)=$? I cannot figure out how to start the sum and proceed with it.
trigonometry
edited Jul 27 at 11:42
Bernard
110k635102
asked Jul 27 at 11:37
Wonder
276
closed as off-topic by Yves Daoust, José Carlos Santos, John Ma, amWhy, Adrian Keister Jul 28 at 0:08
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, amWhy, Adrian Keister
add a comment |Â
up vote
0
down vote
favorite
If $alpha$ and $beta$ are the solutions of the equations $$sin^2x+asin x+b=0$$
and $$cos^2x+c cos x+d=0$$ then $sin(alpha+beta)=$? I cannot figure out how to start the sum and proceed with it.
trigonometry
edited Jul 27 at 11:42
Bernard
110k635102
asked Jul 27 at 11:37
Wonder
276
closed as off-topic by Yves Daoust, José Carlos Santos, John Ma, amWhy, Adrian Keister Jul 28 at 0:08
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, amWhy, Adrian Keister
3
Do you mean that $alpha$ and $beta$ satisfy both equations or is $alpha$ a solution of the former and $beta$ of the latter?
– egreg
Jul 27 at 11:56
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $alpha$ and $beta$ are the solutions of the equations $$sin^2x+asin x+b=0$$
and $$cos^2x+c cos x+d=0$$ then $sin(alpha+beta)=$? I cannot figure out how to start the sum and proceed with it.
trigonometry
edited Jul 27 at 11:42
Bernard
110k635102
asked Jul 27 at 11:37
Wonder
276
If $alpha$ and $beta$ are the solutions of the equations $$sin^2x+asin x+b=0$$
and $$cos^2x+c cos x+d=0$$ then $sin(alpha+beta)=$? I cannot figure out how to start the sum and proceed with it.
trigonometry
edited Jul 27 at 11:42
Bernard
110k635102
asked Jul 27 at 11:37
Wonder
276
edited Jul 27 at 11:42
Bernard
110k635102
edited Jul 27 at 11:42
Bernard
110k635102
edited Jul 27 at 11:42
Bernard
110k635102
110k635102
asked Jul 27 at 11:37
Wonder
276
asked Jul 27 at 11:37
Wonder
276
asked Jul 27 at 11:37
Wonder
276
276
closed as off-topic by Yves Daoust, José Carlos Santos, John Ma, amWhy, Adrian Keister Jul 28 at 0:08
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, amWhy, Adrian Keister
closed as off-topic by Yves Daoust, José Carlos Santos, John Ma, amWhy, Adrian Keister Jul 28 at 0:08
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, amWhy, Adrian Keister
3
Do you mean that $alpha$ and $beta$ satisfy both equations or is $alpha$ a solution of the former and $beta$ of the latter?
– egreg
Jul 27 at 11:56
add a comment |Â
3
Do you mean that $alpha$ and $beta$ satisfy both equations or is $alpha$ a solution of the former and $beta$ of the latter?
– egreg
Jul 27 at 11:56
3
3
Do you mean that $alpha$ and $beta$ satisfy both equations or is $alpha$ a solution of the former and $beta$ of the latter?
– egreg
Jul 27 at 11:56
Do you mean that $alpha$ and $beta$ satisfy both equations or is $alpha$ a solution of the former and $beta$ of the latter?
– egreg
Jul 27 at 11:56
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Hint :
$$beginalign &sin^2 alpha + asin alpha + b = sin^2 beta + asin beta + b = 0 \&to sin^2 alpha - sin^2 beta +asin alpha - asin beta = 0 \&to (sin alpha - sin beta)(sin alpha + sin beta + a) = 0endalign$$
In the same manner :
$(cos alpha - cos beta)(cos alpha + cos beta + c) = 0$
edited Jul 27 at 12:00
Davide Morgante
1,751220
answered Jul 27 at 11:48
S.H.W
1,0431720
add a comment |Â
up vote
0
down vote
$$impliessinalpha+sinbeta=-dfrac a1$$
and $$cosalpha+cosbeta=-dfrac c1$$
On division and using Prosthaphaeresis Formulas, $$dfrac ac=tandfracalpha+beta2$$ assuming $cosdfracalpha-beta2ne0$
Now $sin2A=dfrac2tan A1+tan^2A$
answered Jul 27 at 12:35
lab bhattacharjee
214k14152263
See also : math.stackexchange.com/questions/2021356/… and math.stackexchange.com/questions/1490652/…
– lab bhattacharjee
Jul 27 at 12:37
No, the roots are $alpha$ and $beta$, not their sines or cosines.
– Yves Daoust
Jul 27 at 12:44
@YvesDaoust, The accepted answer doesn't seem to mean so.
– lab bhattacharjee
Jul 27 at 12:50
@labbhattacharjee the accepted answer clearly shows that alpha and beta are the roots as they have been replaced with x in the solution.
– Wonder
Jul 27 at 13:03
Right, my bad...
– Yves Daoust
Jul 27 at 13:54
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint :
$$beginalign &sin^2 alpha + asin alpha + b = sin^2 beta + asin beta + b = 0 \&to sin^2 alpha - sin^2 beta +asin alpha - asin beta = 0 \&to (sin alpha - sin beta)(sin alpha + sin beta + a) = 0endalign$$
In the same manner :
$(cos alpha - cos beta)(cos alpha + cos beta + c) = 0$
edited Jul 27 at 12:00
Davide Morgante
1,751220
answered Jul 27 at 11:48
S.H.W
1,0431720
add a comment |Â
up vote
1
down vote
accepted
Hint :
$$beginalign &sin^2 alpha + asin alpha + b = sin^2 beta + asin beta + b = 0 \&to sin^2 alpha - sin^2 beta +asin alpha - asin beta = 0 \&to (sin alpha - sin beta)(sin alpha + sin beta + a) = 0endalign$$
In the same manner :
$(cos alpha - cos beta)(cos alpha + cos beta + c) = 0$
edited Jul 27 at 12:00
Davide Morgante
1,751220
answered Jul 27 at 11:48
S.H.W
1,0431720
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint :
$$beginalign &sin^2 alpha + asin alpha + b = sin^2 beta + asin beta + b = 0 \&to sin^2 alpha - sin^2 beta +asin alpha - asin beta = 0 \&to (sin alpha - sin beta)(sin alpha + sin beta + a) = 0endalign$$
In the same manner :
$(cos alpha - cos beta)(cos alpha + cos beta + c) = 0$
edited Jul 27 at 12:00
Davide Morgante
1,751220
answered Jul 27 at 11:48
S.H.W
1,0431720
Hint :
$$beginalign &sin^2 alpha + asin alpha + b = sin^2 beta + asin beta + b = 0 \&to sin^2 alpha - sin^2 beta +asin alpha - asin beta = 0 \&to (sin alpha - sin beta)(sin alpha + sin beta + a) = 0endalign$$
In the same manner :
$(cos alpha - cos beta)(cos alpha + cos beta + c) = 0$
edited Jul 27 at 12:00
Davide Morgante
1,751220
answered Jul 27 at 11:48
S.H.W
1,0431720
edited Jul 27 at 12:00
Davide Morgante
1,751220
edited Jul 27 at 12:00
Davide Morgante
1,751220
edited Jul 27 at 12:00
Davide Morgante
1,751220
1,751220
answered Jul 27 at 11:48
S.H.W
1,0431720
answered Jul 27 at 11:48
S.H.W
1,0431720
answered Jul 27 at 11:48
S.H.W
1,0431720
1,0431720
add a comment |Â
add a comment |Â
up vote
0
down vote
$$impliessinalpha+sinbeta=-dfrac a1$$
and $$cosalpha+cosbeta=-dfrac c1$$
On division and using Prosthaphaeresis Formulas, $$dfrac ac=tandfracalpha+beta2$$ assuming $cosdfracalpha-beta2ne0$
Now $sin2A=dfrac2tan A1+tan^2A$
answered Jul 27 at 12:35
lab bhattacharjee
214k14152263
See also : math.stackexchange.com/questions/2021356/… and math.stackexchange.com/questions/1490652/…
– lab bhattacharjee
Jul 27 at 12:37
No, the roots are $alpha$ and $beta$, not their sines or cosines.
– Yves Daoust
Jul 27 at 12:44
@YvesDaoust, The accepted answer doesn't seem to mean so.
– lab bhattacharjee
Jul 27 at 12:50
@labbhattacharjee the accepted answer clearly shows that alpha and beta are the roots as they have been replaced with x in the solution.
– Wonder
Jul 27 at 13:03
Right, my bad...
– Yves Daoust
Jul 27 at 13:54
add a comment |Â
up vote
0
down vote
$$impliessinalpha+sinbeta=-dfrac a1$$
and $$cosalpha+cosbeta=-dfrac c1$$
On division and using Prosthaphaeresis Formulas, $$dfrac ac=tandfracalpha+beta2$$ assuming $cosdfracalpha-beta2ne0$
Now $sin2A=dfrac2tan A1+tan^2A$
answered Jul 27 at 12:35
lab bhattacharjee
214k14152263
See also : math.stackexchange.com/questions/2021356/… and math.stackexchange.com/questions/1490652/…
– lab bhattacharjee
Jul 27 at 12:37
No, the roots are $alpha$ and $beta$, not their sines or cosines.
– Yves Daoust
Jul 27 at 12:44
@YvesDaoust, The accepted answer doesn't seem to mean so.
– lab bhattacharjee
Jul 27 at 12:50
@labbhattacharjee the accepted answer clearly shows that alpha and beta are the roots as they have been replaced with x in the solution.
– Wonder
Jul 27 at 13:03
Right, my bad...
– Yves Daoust
Jul 27 at 13:54
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$impliessinalpha+sinbeta=-dfrac a1$$
and $$cosalpha+cosbeta=-dfrac c1$$
On division and using Prosthaphaeresis Formulas, $$dfrac ac=tandfracalpha+beta2$$ assuming $cosdfracalpha-beta2ne0$
Now $sin2A=dfrac2tan A1+tan^2A$
answered Jul 27 at 12:35
lab bhattacharjee
214k14152263
$$impliessinalpha+sinbeta=-dfrac a1$$
and $$cosalpha+cosbeta=-dfrac c1$$
On division and using Prosthaphaeresis Formulas, $$dfrac ac=tandfracalpha+beta2$$ assuming $cosdfracalpha-beta2ne0$
Now $sin2A=dfrac2tan A1+tan^2A$
answered Jul 27 at 12:35
lab bhattacharjee
214k14152263
answered Jul 27 at 12:35
lab bhattacharjee
214k14152263
answered Jul 27 at 12:35
lab bhattacharjee
214k14152263
answered Jul 27 at 12:35
lab bhattacharjee
214k14152263
214k14152263
See also : math.stackexchange.com/questions/2021356/… and math.stackexchange.com/questions/1490652/…
– lab bhattacharjee
Jul 27 at 12:37
No, the roots are $alpha$ and $beta$, not their sines or cosines.
– Yves Daoust
Jul 27 at 12:44
@YvesDaoust, The accepted answer doesn't seem to mean so.
– lab bhattacharjee
Jul 27 at 12:50
@labbhattacharjee the accepted answer clearly shows that alpha and beta are the roots as they have been replaced with x in the solution.
– Wonder
Jul 27 at 13:03
Right, my bad...
– Yves Daoust
Jul 27 at 13:54
add a comment |Â
See also : math.stackexchange.com/questions/2021356/… and math.stackexchange.com/questions/1490652/…
– lab bhattacharjee
Jul 27 at 12:37
No, the roots are $alpha$ and $beta$, not their sines or cosines.
– Yves Daoust
Jul 27 at 12:44
@YvesDaoust, The accepted answer doesn't seem to mean so.
– lab bhattacharjee
Jul 27 at 12:50
@labbhattacharjee the accepted answer clearly shows that alpha and beta are the roots as they have been replaced with x in the solution.
– Wonder
Jul 27 at 13:03
Right, my bad...
– Yves Daoust
Jul 27 at 13:54
See also : math.stackexchange.com/questions/2021356/… and math.stackexchange.com/questions/1490652/…
– lab bhattacharjee
Jul 27 at 12:37
See also : math.stackexchange.com/questions/2021356/… and math.stackexchange.com/questions/1490652/…
– lab bhattacharjee
Jul 27 at 12:37
No, the roots are $alpha$ and $beta$, not their sines or cosines.
– Yves Daoust
Jul 27 at 12:44
No, the roots are $alpha$ and $beta$, not their sines or cosines.
– Yves Daoust
Jul 27 at 12:44
@YvesDaoust, The accepted answer doesn't seem to mean so.
– lab bhattacharjee
Jul 27 at 12:50
@YvesDaoust, The accepted answer doesn't seem to mean so.
– lab bhattacharjee
Jul 27 at 12:50
@labbhattacharjee the accepted answer clearly shows that alpha and beta are the roots as they have been replaced with x in the solution.
– Wonder
Jul 27 at 13:03
@labbhattacharjee the accepted answer clearly shows that alpha and beta are the roots as they have been replaced with x in the solution.
– Wonder
Jul 27 at 13:03
Right, my bad...
– Yves Daoust
Jul 27 at 13:54
Right, my bad...
– Yves Daoust
Jul 27 at 13:54
add a comment |Â
3
Do you mean that $alpha$ and $beta$ satisfy both equations or is $alpha$ a solution of the former and $beta$ of the latter?
– egreg
Jul 27 at 11:56