How to solve for theta?












0












$begingroup$


$$-12sinθ-5cosθ=0$$



Is it possible? or do I need to generate a graph? I heard the Newton Raphson Method can be used...not sure how to proceed please advise.



I'm trying to calculate the minimum and maximum values of: $y=12cosθ-5sinθ$



The equation above is a derivitive of $y$.






calculus







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    You can solve $asin theta+bcostheta=c$ in general (see this question).
    $endgroup$
    – user26486
    Jun 16 '15 at 8:06










  • $begingroup$
    @user26486. I am sorry ! I did not go through the link since I was supposing a different approach. I deleted my answer !
    $endgroup$
    – Claude Leibovici
    Jun 16 '15 at 8:50
















0












$begingroup$


$$-12sinθ-5cosθ=0$$



Is it possible? or do I need to generate a graph? I heard the Newton Raphson Method can be used...not sure how to proceed please advise.



I'm trying to calculate the minimum and maximum values of: $y=12cosθ-5sinθ$



The equation above is a derivitive of $y$.






calculus







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    You can solve $asin theta+bcostheta=c$ in general (see this question).
    $endgroup$
    – user26486
    Jun 16 '15 at 8:06










  • $begingroup$
    @user26486. I am sorry ! I did not go through the link since I was supposing a different approach. I deleted my answer !
    $endgroup$
    – Claude Leibovici
    Jun 16 '15 at 8:50














0












0








0





$begingroup$


$$-12sinθ-5cosθ=0$$



Is it possible? or do I need to generate a graph? I heard the Newton Raphson Method can be used...not sure how to proceed please advise.



I'm trying to calculate the minimum and maximum values of: $y=12cosθ-5sinθ$



The equation above is a derivitive of $y$.






calculus







share|cite|improve this question









$endgroup$




$$-12sinθ-5cosθ=0$$



Is it possible? or do I need to generate a graph? I heard the Newton Raphson Method can be used...not sure how to proceed please advise.



I'm trying to calculate the minimum and maximum values of: $y=12cosθ-5sinθ$



The equation above is a derivitive of $y$.






calculus




calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jun 16 '15 at 7:59









ModriscoModrisco

182110




182110








  • 4




    $begingroup$
    You can solve $asin theta+bcostheta=c$ in general (see this question).
    $endgroup$
    – user26486
    Jun 16 '15 at 8:06










  • $begingroup$
    @user26486. I am sorry ! I did not go through the link since I was supposing a different approach. I deleted my answer !
    $endgroup$
    – Claude Leibovici
    Jun 16 '15 at 8:50














  • 4




    $begingroup$
    You can solve $asin theta+bcostheta=c$ in general (see this question).
    $endgroup$
    – user26486
    Jun 16 '15 at 8:06










  • $begingroup$
    @user26486. I am sorry ! I did not go through the link since I was supposing a different approach. I deleted my answer !
    $endgroup$
    – Claude Leibovici
    Jun 16 '15 at 8:50








4




4




$begingroup$
You can solve $asin theta+bcostheta=c$ in general (see this question).
$endgroup$
– user26486
Jun 16 '15 at 8:06




$begingroup$
You can solve $asin theta+bcostheta=c$ in general (see this question).
$endgroup$
– user26486
Jun 16 '15 at 8:06












$begingroup$
@user26486. I am sorry ! I did not go through the link since I was supposing a different approach. I deleted my answer !
$endgroup$
– Claude Leibovici
Jun 16 '15 at 8:50




$begingroup$
@user26486. I am sorry ! I did not go through the link since I was supposing a different approach. I deleted my answer !
$endgroup$
– Claude Leibovici
Jun 16 '15 at 8:50










5 Answers
5






active

oldest

votes


















0












$begingroup$

I don't think you should differentiate.



Your $y$ can be written as
$$
y(theta)=13 cos(theta+arctan(5/12)).
$$
Can you see from this what the maximum and minimum values are?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I didn't do it that way, but something like it, I calculated the arctan of -5/12...then I minused that value from 180 and 360 degrees to get the min and max points...then subed those values back into y, and received the answers -13 and 13....I'm mind boggled here but I got the answer at the back of the book anyway!
    $endgroup$
    – Modrisco
    Jun 16 '15 at 12:56





















2












$begingroup$

Hint: Check if $theta=pi/2,3pi/2$ (I'm considering the interval $[0,2pi]$ ) are solutions, and then divide by $cos$ and solve for $tan$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    $$12sintheta+5costheta=0\impliesfrac{12}{13}sintheta+frac5{13}costheta=0$$
    Now let $cos a=frac{12}{13},sin a=frac{5}{13}$. This works because $sin^2a+cos^2a=1$. We now have
    $$cos asintheta+sin acostheta=0\impliessin(theta+a)=0
    \impliestheta=pi n-a$$ for integer $n$.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Yes, it is possible, though the answer might not look nice. First divide by $-12$: $$begin{align}&sintheta + frac 5{12}costheta = 0implies \ implies &sintheta = -frac 5{12}costheta \[2ex] implies &sin^2theta = frac{25}{144}cos^2theta \[2ex] implies &1 - cos^2theta = frac{25}{144}cos^2thetaend{align}$$
      Can you take it from here?






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        y = 12cosx -5sinx



        y'= -12sinx -5cosx



        put y'=0



        -12sinx -5cosx = 0



        12sinx + 5cosx =0 (multiplied by -1 both sides )



        12sinx = -5cosx



        sinx/cosx=-5/12



        tanx=-5/12



        X = tan inverse (-5/12)



        Now can you proceed?






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Please see math.meta.stackexchange.com/questions/5020
          $endgroup$
          – Lord Shark the Unknown
          Dec 30 '18 at 19:44











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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        0












        $begingroup$

        I don't think you should differentiate.



        Your $y$ can be written as
        $$
        y(theta)=13 cos(theta+arctan(5/12)).
        $$
        Can you see from this what the maximum and minimum values are?






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          I didn't do it that way, but something like it, I calculated the arctan of -5/12...then I minused that value from 180 and 360 degrees to get the min and max points...then subed those values back into y, and received the answers -13 and 13....I'm mind boggled here but I got the answer at the back of the book anyway!
          $endgroup$
          – Modrisco
          Jun 16 '15 at 12:56


















        0












        $begingroup$

        I don't think you should differentiate.



        Your $y$ can be written as
        $$
        y(theta)=13 cos(theta+arctan(5/12)).
        $$
        Can you see from this what the maximum and minimum values are?






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          I didn't do it that way, but something like it, I calculated the arctan of -5/12...then I minused that value from 180 and 360 degrees to get the min and max points...then subed those values back into y, and received the answers -13 and 13....I'm mind boggled here but I got the answer at the back of the book anyway!
          $endgroup$
          – Modrisco
          Jun 16 '15 at 12:56
















        0












        0








        0





        $begingroup$

        I don't think you should differentiate.



        Your $y$ can be written as
        $$
        y(theta)=13 cos(theta+arctan(5/12)).
        $$
        Can you see from this what the maximum and minimum values are?






        share|cite|improve this answer









        $endgroup$



        I don't think you should differentiate.



        Your $y$ can be written as
        $$
        y(theta)=13 cos(theta+arctan(5/12)).
        $$
        Can you see from this what the maximum and minimum values are?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jun 16 '15 at 8:04









        mickepmickep

        18.5k12250




        18.5k12250












        • $begingroup$
          I didn't do it that way, but something like it, I calculated the arctan of -5/12...then I minused that value from 180 and 360 degrees to get the min and max points...then subed those values back into y, and received the answers -13 and 13....I'm mind boggled here but I got the answer at the back of the book anyway!
          $endgroup$
          – Modrisco
          Jun 16 '15 at 12:56




















        • $begingroup$
          I didn't do it that way, but something like it, I calculated the arctan of -5/12...then I minused that value from 180 and 360 degrees to get the min and max points...then subed those values back into y, and received the answers -13 and 13....I'm mind boggled here but I got the answer at the back of the book anyway!
          $endgroup$
          – Modrisco
          Jun 16 '15 at 12:56


















        $begingroup$
        I didn't do it that way, but something like it, I calculated the arctan of -5/12...then I minused that value from 180 and 360 degrees to get the min and max points...then subed those values back into y, and received the answers -13 and 13....I'm mind boggled here but I got the answer at the back of the book anyway!
        $endgroup$
        – Modrisco
        Jun 16 '15 at 12:56






        $begingroup$
        I didn't do it that way, but something like it, I calculated the arctan of -5/12...then I minused that value from 180 and 360 degrees to get the min and max points...then subed those values back into y, and received the answers -13 and 13....I'm mind boggled here but I got the answer at the back of the book anyway!
        $endgroup$
        – Modrisco
        Jun 16 '15 at 12:56













        2












        $begingroup$

        Hint: Check if $theta=pi/2,3pi/2$ (I'm considering the interval $[0,2pi]$ ) are solutions, and then divide by $cos$ and solve for $tan$.






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          Hint: Check if $theta=pi/2,3pi/2$ (I'm considering the interval $[0,2pi]$ ) are solutions, and then divide by $cos$ and solve for $tan$.






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            Hint: Check if $theta=pi/2,3pi/2$ (I'm considering the interval $[0,2pi]$ ) are solutions, and then divide by $cos$ and solve for $tan$.






            share|cite|improve this answer









            $endgroup$



            Hint: Check if $theta=pi/2,3pi/2$ (I'm considering the interval $[0,2pi]$ ) are solutions, and then divide by $cos$ and solve for $tan$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jun 16 '15 at 8:01









            Vincenzo OlivaVincenzo Oliva

            5,27411234




            5,27411234























                2












                $begingroup$

                $$12sintheta+5costheta=0\impliesfrac{12}{13}sintheta+frac5{13}costheta=0$$
                Now let $cos a=frac{12}{13},sin a=frac{5}{13}$. This works because $sin^2a+cos^2a=1$. We now have
                $$cos asintheta+sin acostheta=0\impliessin(theta+a)=0
                \impliestheta=pi n-a$$ for integer $n$.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  $$12sintheta+5costheta=0\impliesfrac{12}{13}sintheta+frac5{13}costheta=0$$
                  Now let $cos a=frac{12}{13},sin a=frac{5}{13}$. This works because $sin^2a+cos^2a=1$. We now have
                  $$cos asintheta+sin acostheta=0\impliessin(theta+a)=0
                  \impliestheta=pi n-a$$ for integer $n$.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    $$12sintheta+5costheta=0\impliesfrac{12}{13}sintheta+frac5{13}costheta=0$$
                    Now let $cos a=frac{12}{13},sin a=frac{5}{13}$. This works because $sin^2a+cos^2a=1$. We now have
                    $$cos asintheta+sin acostheta=0\impliessin(theta+a)=0
                    \impliestheta=pi n-a$$ for integer $n$.






                    share|cite|improve this answer









                    $endgroup$



                    $$12sintheta+5costheta=0\impliesfrac{12}{13}sintheta+frac5{13}costheta=0$$
                    Now let $cos a=frac{12}{13},sin a=frac{5}{13}$. This works because $sin^2a+cos^2a=1$. We now have
                    $$cos asintheta+sin acostheta=0\impliessin(theta+a)=0
                    \impliestheta=pi n-a$$ for integer $n$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jun 16 '15 at 8:19









                    Pauly BPauly B

                    3,9121818




                    3,9121818























                        1












                        $begingroup$

                        Yes, it is possible, though the answer might not look nice. First divide by $-12$: $$begin{align}&sintheta + frac 5{12}costheta = 0implies \ implies &sintheta = -frac 5{12}costheta \[2ex] implies &sin^2theta = frac{25}{144}cos^2theta \[2ex] implies &1 - cos^2theta = frac{25}{144}cos^2thetaend{align}$$
                        Can you take it from here?






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Yes, it is possible, though the answer might not look nice. First divide by $-12$: $$begin{align}&sintheta + frac 5{12}costheta = 0implies \ implies &sintheta = -frac 5{12}costheta \[2ex] implies &sin^2theta = frac{25}{144}cos^2theta \[2ex] implies &1 - cos^2theta = frac{25}{144}cos^2thetaend{align}$$
                          Can you take it from here?






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Yes, it is possible, though the answer might not look nice. First divide by $-12$: $$begin{align}&sintheta + frac 5{12}costheta = 0implies \ implies &sintheta = -frac 5{12}costheta \[2ex] implies &sin^2theta = frac{25}{144}cos^2theta \[2ex] implies &1 - cos^2theta = frac{25}{144}cos^2thetaend{align}$$
                            Can you take it from here?






                            share|cite|improve this answer









                            $endgroup$



                            Yes, it is possible, though the answer might not look nice. First divide by $-12$: $$begin{align}&sintheta + frac 5{12}costheta = 0implies \ implies &sintheta = -frac 5{12}costheta \[2ex] implies &sin^2theta = frac{25}{144}cos^2theta \[2ex] implies &1 - cos^2theta = frac{25}{144}cos^2thetaend{align}$$
                            Can you take it from here?







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jun 16 '15 at 8:06









                            GPerezGPerez

                            4,24111440




                            4,24111440























                                0












                                $begingroup$

                                y = 12cosx -5sinx



                                y'= -12sinx -5cosx



                                put y'=0



                                -12sinx -5cosx = 0



                                12sinx + 5cosx =0 (multiplied by -1 both sides )



                                12sinx = -5cosx



                                sinx/cosx=-5/12



                                tanx=-5/12



                                X = tan inverse (-5/12)



                                Now can you proceed?






                                share|cite|improve this answer









                                $endgroup$













                                • $begingroup$
                                  Please see math.meta.stackexchange.com/questions/5020
                                  $endgroup$
                                  – Lord Shark the Unknown
                                  Dec 30 '18 at 19:44
















                                0












                                $begingroup$

                                y = 12cosx -5sinx



                                y'= -12sinx -5cosx



                                put y'=0



                                -12sinx -5cosx = 0



                                12sinx + 5cosx =0 (multiplied by -1 both sides )



                                12sinx = -5cosx



                                sinx/cosx=-5/12



                                tanx=-5/12



                                X = tan inverse (-5/12)



                                Now can you proceed?






                                share|cite|improve this answer









                                $endgroup$













                                • $begingroup$
                                  Please see math.meta.stackexchange.com/questions/5020
                                  $endgroup$
                                  – Lord Shark the Unknown
                                  Dec 30 '18 at 19:44














                                0












                                0








                                0





                                $begingroup$

                                y = 12cosx -5sinx



                                y'= -12sinx -5cosx



                                put y'=0



                                -12sinx -5cosx = 0



                                12sinx + 5cosx =0 (multiplied by -1 both sides )



                                12sinx = -5cosx



                                sinx/cosx=-5/12



                                tanx=-5/12



                                X = tan inverse (-5/12)



                                Now can you proceed?






                                share|cite|improve this answer









                                $endgroup$



                                y = 12cosx -5sinx



                                y'= -12sinx -5cosx



                                put y'=0



                                -12sinx -5cosx = 0



                                12sinx + 5cosx =0 (multiplied by -1 both sides )



                                12sinx = -5cosx



                                sinx/cosx=-5/12



                                tanx=-5/12



                                X = tan inverse (-5/12)



                                Now can you proceed?







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Dec 30 '18 at 19:19









                                Shankar SubramanianShankar Subramanian

                                15




                                15












                                • $begingroup$
                                  Please see math.meta.stackexchange.com/questions/5020
                                  $endgroup$
                                  – Lord Shark the Unknown
                                  Dec 30 '18 at 19:44


















                                • $begingroup$
                                  Please see math.meta.stackexchange.com/questions/5020
                                  $endgroup$
                                  – Lord Shark the Unknown
                                  Dec 30 '18 at 19:44
















                                $begingroup$
                                Please see math.meta.stackexchange.com/questions/5020
                                $endgroup$
                                – Lord Shark the Unknown
                                Dec 30 '18 at 19:44




                                $begingroup$
                                Please see math.meta.stackexchange.com/questions/5020
                                $endgroup$
                                – Lord Shark the Unknown
                                Dec 30 '18 at 19:44


















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