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 $y$.










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 $y$.










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 $y$.










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






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











        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1327211%2fhow-to-solve-for-theta%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        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


















                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1327211%2fhow-to-solve-for-theta%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                Popular posts from this blog

                                Human spaceflight

                                Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                                張江高科駅