Finding the roots of a polynomial involving trigonometry












1












$begingroup$


(a) Show that a root of $sin (x) − x/5 = 0$ lies between $x = 2$
and $x = 3$.



(b) Use the method of halving the interval twice to find an estimate of the root.



I tried to do $f(x)=sin(2)-2/54$, but I got $-0.3651005033$



The answers has $ 0.51 $?



Once I know how to do part (a), I can do part (b).



Thank you in advance!!






trigonometry polynomials







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$endgroup$












  • $begingroup$
    I think you were using degrees?
    $endgroup$
    – Zachary Hunter
    Dec 30 '18 at 0:13










  • $begingroup$
    Can’t (b) be done independently of (a) ?
    $endgroup$
    – T. Fo
    Dec 30 '18 at 0:15










  • $begingroup$
    oh sorry yes i was using degrees. thank you!!
    $endgroup$
    – Emily
    Dec 30 '18 at 0:19










  • $begingroup$
    The function has a change of sign between 2 and 3. This indicates that at least 1 root exists, since the function is continuous.
    $endgroup$
    – NoChance
    Dec 30 '18 at 0:29
















1












$begingroup$


(a) Show that a root of $sin (x) − x/5 = 0$ lies between $x = 2$
and $x = 3$.



(b) Use the method of halving the interval twice to find an estimate of the root.



I tried to do $f(x)=sin(2)-2/54$, but I got $-0.3651005033$



The answers has $ 0.51 $?



Once I know how to do part (a), I can do part (b).



Thank you in advance!!






trigonometry polynomials







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think you were using degrees?
    $endgroup$
    – Zachary Hunter
    Dec 30 '18 at 0:13










  • $begingroup$
    Can’t (b) be done independently of (a) ?
    $endgroup$
    – T. Fo
    Dec 30 '18 at 0:15










  • $begingroup$
    oh sorry yes i was using degrees. thank you!!
    $endgroup$
    – Emily
    Dec 30 '18 at 0:19










  • $begingroup$
    The function has a change of sign between 2 and 3. This indicates that at least 1 root exists, since the function is continuous.
    $endgroup$
    – NoChance
    Dec 30 '18 at 0:29














1












1








1





$begingroup$


(a) Show that a root of $sin (x) − x/5 = 0$ lies between $x = 2$
and $x = 3$.



(b) Use the method of halving the interval twice to find an estimate of the root.



I tried to do $f(x)=sin(2)-2/54$, but I got $-0.3651005033$



The answers has $ 0.51 $?



Once I know how to do part (a), I can do part (b).



Thank you in advance!!






trigonometry polynomials







share|cite|improve this question











$endgroup$




(a) Show that a root of $sin (x) − x/5 = 0$ lies between $x = 2$
and $x = 3$.



(b) Use the method of halving the interval twice to find an estimate of the root.



I tried to do $f(x)=sin(2)-2/54$, but I got $-0.3651005033$



The answers has $ 0.51 $?



Once I know how to do part (a), I can do part (b).



Thank you in advance!!






trigonometry polynomials




trigonometry polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 0:41









Larry

2,53531131




2,53531131










asked Dec 30 '18 at 0:10









EmilyEmily

84




84












  • $begingroup$
    I think you were using degrees?
    $endgroup$
    – Zachary Hunter
    Dec 30 '18 at 0:13










  • $begingroup$
    Can’t (b) be done independently of (a) ?
    $endgroup$
    – T. Fo
    Dec 30 '18 at 0:15










  • $begingroup$
    oh sorry yes i was using degrees. thank you!!
    $endgroup$
    – Emily
    Dec 30 '18 at 0:19










  • $begingroup$
    The function has a change of sign between 2 and 3. This indicates that at least 1 root exists, since the function is continuous.
    $endgroup$
    – NoChance
    Dec 30 '18 at 0:29


















  • $begingroup$
    I think you were using degrees?
    $endgroup$
    – Zachary Hunter
    Dec 30 '18 at 0:13










  • $begingroup$
    Can’t (b) be done independently of (a) ?
    $endgroup$
    – T. Fo
    Dec 30 '18 at 0:15










  • $begingroup$
    oh sorry yes i was using degrees. thank you!!
    $endgroup$
    – Emily
    Dec 30 '18 at 0:19










  • $begingroup$
    The function has a change of sign between 2 and 3. This indicates that at least 1 root exists, since the function is continuous.
    $endgroup$
    – NoChance
    Dec 30 '18 at 0:29
















$begingroup$
I think you were using degrees?
$endgroup$
– Zachary Hunter
Dec 30 '18 at 0:13




$begingroup$
I think you were using degrees?
$endgroup$
– Zachary Hunter
Dec 30 '18 at 0:13












$begingroup$
Can’t (b) be done independently of (a) ?
$endgroup$
– T. Fo
Dec 30 '18 at 0:15




$begingroup$
Can’t (b) be done independently of (a) ?
$endgroup$
– T. Fo
Dec 30 '18 at 0:15












$begingroup$
oh sorry yes i was using degrees. thank you!!
$endgroup$
– Emily
Dec 30 '18 at 0:19




$begingroup$
oh sorry yes i was using degrees. thank you!!
$endgroup$
– Emily
Dec 30 '18 at 0:19












$begingroup$
The function has a change of sign between 2 and 3. This indicates that at least 1 root exists, since the function is continuous.
$endgroup$
– NoChance
Dec 30 '18 at 0:29




$begingroup$
The function has a change of sign between 2 and 3. This indicates that at least 1 root exists, since the function is continuous.
$endgroup$
– NoChance
Dec 30 '18 at 0:29










1 Answer
1






active

oldest

votes


















2












$begingroup$

If you got -0.36 but the answer is supposed to have 0.51, then you probably have the wrong mode in your calculator. Check if the answer is supposed to be in degrees or radians and if your calculator is in the same mode as your answer is supposed to be in. To help you with a, to show that a root exists between x = 2 and x = 3, plug in those values into the function. Since the function is continuous between those x-values, you can use the Intermediate Value Theorem to show that a root exists between those two x-values.



Edit: From NoChance's comment, here is a link to the theorem I referenced if you want more information: https://en.wikipedia.org/wiki/Intermediate_value_theorem






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you!! Yes, my calculator was in degrees instead of radians.
    $endgroup$
    – Emily
    Dec 30 '18 at 0:26










  • $begingroup$
    To enhance your answer, it may be better to include a reference to the theorems you have mentioned as hyperlinks. Thanks.
    $endgroup$
    – NoChance
    Dec 30 '18 at 0:28










  • $begingroup$
    @Emily, if you think this answer helped you please mark it as such, Thanks.
    $endgroup$
    – NoChance
    Dec 30 '18 at 0:28











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

If you got -0.36 but the answer is supposed to have 0.51, then you probably have the wrong mode in your calculator. Check if the answer is supposed to be in degrees or radians and if your calculator is in the same mode as your answer is supposed to be in. To help you with a, to show that a root exists between x = 2 and x = 3, plug in those values into the function. Since the function is continuous between those x-values, you can use the Intermediate Value Theorem to show that a root exists between those two x-values.



Edit: From NoChance's comment, here is a link to the theorem I referenced if you want more information: https://en.wikipedia.org/wiki/Intermediate_value_theorem






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you!! Yes, my calculator was in degrees instead of radians.
    $endgroup$
    – Emily
    Dec 30 '18 at 0:26










  • $begingroup$
    To enhance your answer, it may be better to include a reference to the theorems you have mentioned as hyperlinks. Thanks.
    $endgroup$
    – NoChance
    Dec 30 '18 at 0:28










  • $begingroup$
    @Emily, if you think this answer helped you please mark it as such, Thanks.
    $endgroup$
    – NoChance
    Dec 30 '18 at 0:28
















2












$begingroup$

If you got -0.36 but the answer is supposed to have 0.51, then you probably have the wrong mode in your calculator. Check if the answer is supposed to be in degrees or radians and if your calculator is in the same mode as your answer is supposed to be in. To help you with a, to show that a root exists between x = 2 and x = 3, plug in those values into the function. Since the function is continuous between those x-values, you can use the Intermediate Value Theorem to show that a root exists between those two x-values.



Edit: From NoChance's comment, here is a link to the theorem I referenced if you want more information: https://en.wikipedia.org/wiki/Intermediate_value_theorem






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you!! Yes, my calculator was in degrees instead of radians.
    $endgroup$
    – Emily
    Dec 30 '18 at 0:26










  • $begingroup$
    To enhance your answer, it may be better to include a reference to the theorems you have mentioned as hyperlinks. Thanks.
    $endgroup$
    – NoChance
    Dec 30 '18 at 0:28










  • $begingroup$
    @Emily, if you think this answer helped you please mark it as such, Thanks.
    $endgroup$
    – NoChance
    Dec 30 '18 at 0:28














2












2








2





$begingroup$

If you got -0.36 but the answer is supposed to have 0.51, then you probably have the wrong mode in your calculator. Check if the answer is supposed to be in degrees or radians and if your calculator is in the same mode as your answer is supposed to be in. To help you with a, to show that a root exists between x = 2 and x = 3, plug in those values into the function. Since the function is continuous between those x-values, you can use the Intermediate Value Theorem to show that a root exists between those two x-values.



Edit: From NoChance's comment, here is a link to the theorem I referenced if you want more information: https://en.wikipedia.org/wiki/Intermediate_value_theorem






share|cite|improve this answer











$endgroup$



If you got -0.36 but the answer is supposed to have 0.51, then you probably have the wrong mode in your calculator. Check if the answer is supposed to be in degrees or radians and if your calculator is in the same mode as your answer is supposed to be in. To help you with a, to show that a root exists between x = 2 and x = 3, plug in those values into the function. Since the function is continuous between those x-values, you can use the Intermediate Value Theorem to show that a root exists between those two x-values.



Edit: From NoChance's comment, here is a link to the theorem I referenced if you want more information: https://en.wikipedia.org/wiki/Intermediate_value_theorem







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 30 '18 at 0:33

























answered Dec 30 '18 at 0:21









Eliot Eliot

676




676












  • $begingroup$
    Thank you!! Yes, my calculator was in degrees instead of radians.
    $endgroup$
    – Emily
    Dec 30 '18 at 0:26










  • $begingroup$
    To enhance your answer, it may be better to include a reference to the theorems you have mentioned as hyperlinks. Thanks.
    $endgroup$
    – NoChance
    Dec 30 '18 at 0:28










  • $begingroup$
    @Emily, if you think this answer helped you please mark it as such, Thanks.
    $endgroup$
    – NoChance
    Dec 30 '18 at 0:28


















  • $begingroup$
    Thank you!! Yes, my calculator was in degrees instead of radians.
    $endgroup$
    – Emily
    Dec 30 '18 at 0:26










  • $begingroup$
    To enhance your answer, it may be better to include a reference to the theorems you have mentioned as hyperlinks. Thanks.
    $endgroup$
    – NoChance
    Dec 30 '18 at 0:28










  • $begingroup$
    @Emily, if you think this answer helped you please mark it as such, Thanks.
    $endgroup$
    – NoChance
    Dec 30 '18 at 0:28
















$begingroup$
Thank you!! Yes, my calculator was in degrees instead of radians.
$endgroup$
– Emily
Dec 30 '18 at 0:26




$begingroup$
Thank you!! Yes, my calculator was in degrees instead of radians.
$endgroup$
– Emily
Dec 30 '18 at 0:26












$begingroup$
To enhance your answer, it may be better to include a reference to the theorems you have mentioned as hyperlinks. Thanks.
$endgroup$
– NoChance
Dec 30 '18 at 0:28




$begingroup$
To enhance your answer, it may be better to include a reference to the theorems you have mentioned as hyperlinks. Thanks.
$endgroup$
– NoChance
Dec 30 '18 at 0:28












$begingroup$
@Emily, if you think this answer helped you please mark it as such, Thanks.
$endgroup$
– NoChance
Dec 30 '18 at 0:28




$begingroup$
@Emily, if you think this answer helped you please mark it as such, Thanks.
$endgroup$
– NoChance
Dec 30 '18 at 0:28


















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