Solve a system of trigonometric equations
How can I solve this system of trigonometric equations analytically? It is from physics class.
$$
begin{cases}
30tcos{alpha}=50\
-30tsin{alpha}-4.9t^2=0
end{cases}
$$
trigonometry systems-of-equations nonlinear-system
asked Nov 28 '18 at 17:03
StepiiStepii
106
add a comment |
How can I solve this system of trigonometric equations analytically? It is from physics class.
$$
begin{cases}
30tcos{alpha}=50\
-30tsin{alpha}-4.9t^2=0
end{cases}
$$
trigonometry systems-of-equations nonlinear-system
asked Nov 28 '18 at 17:03
StepiiStepii
106
Hint: try moving the $t^2$ term to the other side, squaring the two equations, and adding them together. What happens to the trigonometric terms?
– Josh B.
Nov 28 '18 at 17:16
add a comment |
How can I solve this system of trigonometric equations analytically? It is from physics class.
$$
begin{cases}
30tcos{alpha}=50\
-30tsin{alpha}-4.9t^2=0
end{cases}
$$
trigonometry systems-of-equations nonlinear-system
asked Nov 28 '18 at 17:03
StepiiStepii
106
How can I solve this system of trigonometric equations analytically? It is from physics class.
$$
begin{cases}
30tcos{alpha}=50\
-30tsin{alpha}-4.9t^2=0
end{cases}
$$
trigonometry systems-of-equations nonlinear-system
trigonometry systems-of-equations nonlinear-system
asked Nov 28 '18 at 17:03
StepiiStepii
106
asked Nov 28 '18 at 17:03
StepiiStepii
106
edited Nov 28 '18 at 17:12
Stepii
asked Nov 28 '18 at 17:03
StepiiStepii
106
asked Nov 28 '18 at 17:03
StepiiStepii
106
asked Nov 28 '18 at 17:03
StepiiStepii
106
106
Hint: try moving the $t^2$ term to the other side, squaring the two equations, and adding them together. What happens to the trigonometric terms?
– Josh B.
Nov 28 '18 at 17:16
add a comment |
Hint: try moving the $t^2$ term to the other side, squaring the two equations, and adding them together. What happens to the trigonometric terms?
– Josh B.
Nov 28 '18 at 17:16
Hint: try moving the $t^2$ term to the other side, squaring the two equations, and adding them together. What happens to the trigonometric terms?
– Josh B.
Nov 28 '18 at 17:16
Hint: try moving the $t^2$ term to the other side, squaring the two equations, and adding them together. What happens to the trigonometric terms?
– Josh B.
Nov 28 '18 at 17:16
add a comment |
3 Answers
3
active
oldest
votes
Hint: Squaring both the equations, you will get $900t^2cos^2{alpha}=2500\
900t^2sin^2{alpha}={4.9}^2t^4$.
Note that $sin^2{alpha}+cos^2{alpha}=1$.
So add both the equations and solve for $t$ using the substitution $t^2=u$.
answered Nov 28 '18 at 17:18
Thomas ShelbyThomas Shelby
1,891217
1
Turn it into a quadratic is the gist of my answer but this is much nicer+faster.
– Mason
Nov 28 '18 at 17:22
@Mason Thank you.
– Thomas Shelby
Nov 28 '18 at 17:24
add a comment |
$30tcos{alpha}=50 implies t=frac{5}{3} secalpha$
You can plug this information into the other equation and solve:
$$-30tsin{alpha}-4.9t^2=0implies -30(frac{5}{3} secalpha)sin{alpha}-4.9(frac{5}{3} secalpha)^2=0$$
$$-50(tanalpha)-4.9(frac{5}{3} secalpha)^2=0$$
$$-50(tanalpha)-4.9frac{25}{9} sec^2alpha=0$$
$$-50(tanalpha)-4.9frac{25}{9} (tan^2alpha+1)=0$$
Taking $y=tanalpha$ you can solve a quadratic equation.
$$-50(y)-4.9frac{25}{9} (y^2+1)=0$$
I think you're probably in good shape from here?
answered Nov 28 '18 at 17:20
MasonMason
1,9621530
This results in negative alpha but its a real physics problem, so it must be positive.
– Stepii
Nov 28 '18 at 17:27
The angle cannot be negative? Add $2pi$?
– Mason
Nov 28 '18 at 17:29
Adding 2π gives angles greater than 90°
– Stepii
Nov 28 '18 at 17:38
@Stepii Which also doesn't make sense in the context of the problem?
– Mason
Nov 28 '18 at 17:43
What did you get for $y$. Maybe I made an error... Here's y
– Mason
Nov 28 '18 at 17:44
|
show 10 more comments
Sorry. This is not an answer. Just part of a chat conversation that we are not really suppose to be having via comments.
This is the image from the link you shared. Unless I am reading this graph wrong it says: That when $a<0$ then $t>0$
answered Nov 28 '18 at 18:22
MasonMason
1,9621530
The extraneous solutions are when $a>0$ because it implies that $t<0$ which as you have commented makes little sense often in physics problems.
– Mason
Nov 28 '18 at 18:23
But that solutions (with the angle and time both positive) satisfy the given system
– Stepii
Nov 28 '18 at 18:29
add a comment |
Your Answer
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3 Answers
3
active
oldest
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3 Answers
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active
oldest
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Hint: Squaring both the equations, you will get $900t^2cos^2{alpha}=2500\
900t^2sin^2{alpha}={4.9}^2t^4$.
Note that $sin^2{alpha}+cos^2{alpha}=1$.
So add both the equations and solve for $t$ using the substitution $t^2=u$.
answered Nov 28 '18 at 17:18
Thomas ShelbyThomas Shelby
1,891217
1
Turn it into a quadratic is the gist of my answer but this is much nicer+faster.
– Mason
Nov 28 '18 at 17:22
@Mason Thank you.
– Thomas Shelby
Nov 28 '18 at 17:24
add a comment |
Hint: Squaring both the equations, you will get $900t^2cos^2{alpha}=2500\
900t^2sin^2{alpha}={4.9}^2t^4$.
Note that $sin^2{alpha}+cos^2{alpha}=1$.
So add both the equations and solve for $t$ using the substitution $t^2=u$.
answered Nov 28 '18 at 17:18
Thomas ShelbyThomas Shelby
1,891217
1
Turn it into a quadratic is the gist of my answer but this is much nicer+faster.
– Mason
Nov 28 '18 at 17:22
@Mason Thank you.
– Thomas Shelby
Nov 28 '18 at 17:24
add a comment |
Hint: Squaring both the equations, you will get $900t^2cos^2{alpha}=2500\
900t^2sin^2{alpha}={4.9}^2t^4$.
Note that $sin^2{alpha}+cos^2{alpha}=1$.
So add both the equations and solve for $t$ using the substitution $t^2=u$.
answered Nov 28 '18 at 17:18
Thomas ShelbyThomas Shelby
1,891217
Hint: Squaring both the equations, you will get $900t^2cos^2{alpha}=2500\
900t^2sin^2{alpha}={4.9}^2t^4$.
Note that $sin^2{alpha}+cos^2{alpha}=1$.
So add both the equations and solve for $t$ using the substitution $t^2=u$.
answered Nov 28 '18 at 17:18
Thomas ShelbyThomas Shelby
1,891217
answered Nov 28 '18 at 17:18
Thomas ShelbyThomas Shelby
1,891217
answered Nov 28 '18 at 17:18
Thomas ShelbyThomas Shelby
1,891217
answered Nov 28 '18 at 17:18
Thomas ShelbyThomas Shelby
1,891217
1,891217
1
Turn it into a quadratic is the gist of my answer but this is much nicer+faster.
– Mason
Nov 28 '18 at 17:22
@Mason Thank you.
– Thomas Shelby
Nov 28 '18 at 17:24
add a comment |
1
Turn it into a quadratic is the gist of my answer but this is much nicer+faster.
– Mason
Nov 28 '18 at 17:22
@Mason Thank you.
– Thomas Shelby
Nov 28 '18 at 17:24
1
1
Turn it into a quadratic is the gist of my answer but this is much nicer+faster.
– Mason
Nov 28 '18 at 17:22
Turn it into a quadratic is the gist of my answer but this is much nicer+faster.
– Mason
Nov 28 '18 at 17:22
@Mason Thank you.
– Thomas Shelby
Nov 28 '18 at 17:24
@Mason Thank you.
– Thomas Shelby
Nov 28 '18 at 17:24
add a comment |
$30tcos{alpha}=50 implies t=frac{5}{3} secalpha$
You can plug this information into the other equation and solve:
$$-30tsin{alpha}-4.9t^2=0implies -30(frac{5}{3} secalpha)sin{alpha}-4.9(frac{5}{3} secalpha)^2=0$$
$$-50(tanalpha)-4.9(frac{5}{3} secalpha)^2=0$$
$$-50(tanalpha)-4.9frac{25}{9} sec^2alpha=0$$
$$-50(tanalpha)-4.9frac{25}{9} (tan^2alpha+1)=0$$
Taking $y=tanalpha$ you can solve a quadratic equation.
$$-50(y)-4.9frac{25}{9} (y^2+1)=0$$
I think you're probably in good shape from here?
answered Nov 28 '18 at 17:20
MasonMason
1,9621530
This results in negative alpha but its a real physics problem, so it must be positive.
– Stepii
Nov 28 '18 at 17:27
The angle cannot be negative? Add $2pi$?
– Mason
Nov 28 '18 at 17:29
Adding 2π gives angles greater than 90°
– Stepii
Nov 28 '18 at 17:38
@Stepii Which also doesn't make sense in the context of the problem?
– Mason
Nov 28 '18 at 17:43
What did you get for $y$. Maybe I made an error... Here's y
– Mason
Nov 28 '18 at 17:44
|
show 10 more comments
$30tcos{alpha}=50 implies t=frac{5}{3} secalpha$
You can plug this information into the other equation and solve:
$$-30tsin{alpha}-4.9t^2=0implies -30(frac{5}{3} secalpha)sin{alpha}-4.9(frac{5}{3} secalpha)^2=0$$
$$-50(tanalpha)-4.9(frac{5}{3} secalpha)^2=0$$
$$-50(tanalpha)-4.9frac{25}{9} sec^2alpha=0$$
$$-50(tanalpha)-4.9frac{25}{9} (tan^2alpha+1)=0$$
Taking $y=tanalpha$ you can solve a quadratic equation.
$$-50(y)-4.9frac{25}{9} (y^2+1)=0$$
I think you're probably in good shape from here?
answered Nov 28 '18 at 17:20
MasonMason
1,9621530
This results in negative alpha but its a real physics problem, so it must be positive.
– Stepii
Nov 28 '18 at 17:27
The angle cannot be negative? Add $2pi$?
– Mason
Nov 28 '18 at 17:29
Adding 2π gives angles greater than 90°
– Stepii
Nov 28 '18 at 17:38
@Stepii Which also doesn't make sense in the context of the problem?
– Mason
Nov 28 '18 at 17:43
What did you get for $y$. Maybe I made an error... Here's y
– Mason
Nov 28 '18 at 17:44
|
show 10 more comments
$30tcos{alpha}=50 implies t=frac{5}{3} secalpha$
You can plug this information into the other equation and solve:
$$-30tsin{alpha}-4.9t^2=0implies -30(frac{5}{3} secalpha)sin{alpha}-4.9(frac{5}{3} secalpha)^2=0$$
$$-50(tanalpha)-4.9(frac{5}{3} secalpha)^2=0$$
$$-50(tanalpha)-4.9frac{25}{9} sec^2alpha=0$$
$$-50(tanalpha)-4.9frac{25}{9} (tan^2alpha+1)=0$$
Taking $y=tanalpha$ you can solve a quadratic equation.
$$-50(y)-4.9frac{25}{9} (y^2+1)=0$$
I think you're probably in good shape from here?
answered Nov 28 '18 at 17:20
MasonMason
1,9621530
$30tcos{alpha}=50 implies t=frac{5}{3} secalpha$
You can plug this information into the other equation and solve:
$$-30tsin{alpha}-4.9t^2=0implies -30(frac{5}{3} secalpha)sin{alpha}-4.9(frac{5}{3} secalpha)^2=0$$
$$-50(tanalpha)-4.9(frac{5}{3} secalpha)^2=0$$
$$-50(tanalpha)-4.9frac{25}{9} sec^2alpha=0$$
$$-50(tanalpha)-4.9frac{25}{9} (tan^2alpha+1)=0$$
Taking $y=tanalpha$ you can solve a quadratic equation.
$$-50(y)-4.9frac{25}{9} (y^2+1)=0$$
I think you're probably in good shape from here?
answered Nov 28 '18 at 17:20
MasonMason
1,9621530
answered Nov 28 '18 at 17:20
MasonMason
1,9621530
answered Nov 28 '18 at 17:20
MasonMason
1,9621530
answered Nov 28 '18 at 17:20
MasonMason
1,9621530
1,9621530
This results in negative alpha but its a real physics problem, so it must be positive.
– Stepii
Nov 28 '18 at 17:27
The angle cannot be negative? Add $2pi$?
– Mason
Nov 28 '18 at 17:29
Adding 2π gives angles greater than 90°
– Stepii
Nov 28 '18 at 17:38
@Stepii Which also doesn't make sense in the context of the problem?
– Mason
Nov 28 '18 at 17:43
What did you get for $y$. Maybe I made an error... Here's y
– Mason
Nov 28 '18 at 17:44
|
show 10 more comments
This results in negative alpha but its a real physics problem, so it must be positive.
– Stepii
Nov 28 '18 at 17:27
The angle cannot be negative? Add $2pi$?
– Mason
Nov 28 '18 at 17:29
Adding 2π gives angles greater than 90°
– Stepii
Nov 28 '18 at 17:38
@Stepii Which also doesn't make sense in the context of the problem?
– Mason
Nov 28 '18 at 17:43
What did you get for $y$. Maybe I made an error... Here's y
– Mason
Nov 28 '18 at 17:44
This results in negative alpha but its a real physics problem, so it must be positive.
– Stepii
Nov 28 '18 at 17:27
This results in negative alpha but its a real physics problem, so it must be positive.
– Stepii
Nov 28 '18 at 17:27
The angle cannot be negative? Add $2pi$?
– Mason
Nov 28 '18 at 17:29
The angle cannot be negative? Add $2pi$?
– Mason
Nov 28 '18 at 17:29
Adding 2π gives angles greater than 90°
– Stepii
Nov 28 '18 at 17:38
Adding 2π gives angles greater than 90°
– Stepii
Nov 28 '18 at 17:38
@Stepii Which also doesn't make sense in the context of the problem?
– Mason
Nov 28 '18 at 17:43
@Stepii Which also doesn't make sense in the context of the problem?
– Mason
Nov 28 '18 at 17:43
What did you get for $y$. Maybe I made an error... Here's y
– Mason
Nov 28 '18 at 17:44
What did you get for $y$. Maybe I made an error... Here's y
– Mason
Nov 28 '18 at 17:44
|
show 10 more comments
Sorry. This is not an answer. Just part of a chat conversation that we are not really suppose to be having via comments.
This is the image from the link you shared. Unless I am reading this graph wrong it says: That when $a<0$ then $t>0$
answered Nov 28 '18 at 18:22
MasonMason
1,9621530
The extraneous solutions are when $a>0$ because it implies that $t<0$ which as you have commented makes little sense often in physics problems.
– Mason
Nov 28 '18 at 18:23
But that solutions (with the angle and time both positive) satisfy the given system
– Stepii
Nov 28 '18 at 18:29
add a comment |
Sorry. This is not an answer. Just part of a chat conversation that we are not really suppose to be having via comments.
This is the image from the link you shared. Unless I am reading this graph wrong it says: That when $a<0$ then $t>0$
answered Nov 28 '18 at 18:22
MasonMason
1,9621530
The extraneous solutions are when $a>0$ because it implies that $t<0$ which as you have commented makes little sense often in physics problems.
– Mason
Nov 28 '18 at 18:23
But that solutions (with the angle and time both positive) satisfy the given system
– Stepii
Nov 28 '18 at 18:29
add a comment |
Sorry. This is not an answer. Just part of a chat conversation that we are not really suppose to be having via comments.
This is the image from the link you shared. Unless I am reading this graph wrong it says: That when $a<0$ then $t>0$
answered Nov 28 '18 at 18:22
MasonMason
1,9621530
Sorry. This is not an answer. Just part of a chat conversation that we are not really suppose to be having via comments.
This is the image from the link you shared. Unless I am reading this graph wrong it says: That when $a<0$ then $t>0$
answered Nov 28 '18 at 18:22
MasonMason
1,9621530
answered Nov 28 '18 at 18:22
MasonMason
1,9621530
answered Nov 28 '18 at 18:22
MasonMason
1,9621530
answered Nov 28 '18 at 18:22
MasonMason
1,9621530
1,9621530
The extraneous solutions are when $a>0$ because it implies that $t<0$ which as you have commented makes little sense often in physics problems.
– Mason
Nov 28 '18 at 18:23
But that solutions (with the angle and time both positive) satisfy the given system
– Stepii
Nov 28 '18 at 18:29
add a comment |
The extraneous solutions are when $a>0$ because it implies that $t<0$ which as you have commented makes little sense often in physics problems.
– Mason
Nov 28 '18 at 18:23
But that solutions (with the angle and time both positive) satisfy the given system
– Stepii
Nov 28 '18 at 18:29
The extraneous solutions are when $a>0$ because it implies that $t<0$ which as you have commented makes little sense often in physics problems.
– Mason
Nov 28 '18 at 18:23
The extraneous solutions are when $a>0$ because it implies that $t<0$ which as you have commented makes little sense often in physics problems.
– Mason
Nov 28 '18 at 18:23
But that solutions (with the angle and time both positive) satisfy the given system
– Stepii
Nov 28 '18 at 18:29
But that solutions (with the angle and time both positive) satisfy the given system
– Stepii
Nov 28 '18 at 18:29
add a comment |
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Hint: try moving the $t^2$ term to the other side, squaring the two equations, and adding them together. What happens to the trigonometric terms?
– Josh B.
Nov 28 '18 at 17:16