How to Determine Cosine, and Sine Given $tan^{-1}x$












0














The problem was, "for $0 < x < 1$, express and simplify in therms of $x$: $sin[2tan^{-1}(x)]$".



My professor figured out that, $sin[2tan^{-1}(x)] = frac{2x}{x^2+1}$, using the Double Angle Sine identity.



The Double Sine identity is, $sin2theta = 2sinthetacostheta$.



How did he know what values to use for Cosine and Sine in this equation?






algebra-precalculus trigonometry







share|cite|improve this question





























    0














    The problem was, "for $0 < x < 1$, express and simplify in therms of $x$: $sin[2tan^{-1}(x)]$".



    My professor figured out that, $sin[2tan^{-1}(x)] = frac{2x}{x^2+1}$, using the Double Angle Sine identity.



    The Double Sine identity is, $sin2theta = 2sinthetacostheta$.



    How did he know what values to use for Cosine and Sine in this equation?






    algebra-precalculus trigonometry







    share|cite|improve this question



























      0












      0








      0







      The problem was, "for $0 < x < 1$, express and simplify in therms of $x$: $sin[2tan^{-1}(x)]$".



      My professor figured out that, $sin[2tan^{-1}(x)] = frac{2x}{x^2+1}$, using the Double Angle Sine identity.



      The Double Sine identity is, $sin2theta = 2sinthetacostheta$.



      How did he know what values to use for Cosine and Sine in this equation?






      algebra-precalculus trigonometry







      share|cite|improve this question















      The problem was, "for $0 < x < 1$, express and simplify in therms of $x$: $sin[2tan^{-1}(x)]$".



      My professor figured out that, $sin[2tan^{-1}(x)] = frac{2x}{x^2+1}$, using the Double Angle Sine identity.



      The Double Sine identity is, $sin2theta = 2sinthetacostheta$.



      How did he know what values to use for Cosine and Sine in this equation?






      algebra-precalculus trigonometry




      algebra-precalculus trigonometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 26 at 21:28

























      asked Nov 26 at 21:17









      LuminousNutria

      1709




      1709






















          2 Answers
          2






          active

          oldest

          votes


















          3














          Hint:
          $$sin^2(y)+cos^2(y) = 1 implies tan^{2}(y)+1=frac{1}{cos^2(y)}.$$
          Consequently, if
          $$tan^{-1}(x) = y implies x=tan(y) implies cos(y) = frac{1}{sqrt{x^2+1}}, sin(y)=?$$






          share|cite|improve this answer





















          • There is a missing step: as is, $cos y$ is only known to be $pmfrac{1}{sqrt{x^2+1}}$. However, $mathrm{arctan},xin]-pi/2,pi/2[$, so the cosine is nonnegative.
            – Jean-Claude Arbaut
            Nov 26 at 21:56












          • @Jean-ClaudeArbaut I left it purposefully to let the OP figure out the details.
            – Math Lover
            Nov 26 at 21:57










          • That's a good point :) +1
            – Jean-Claude Arbaut
            Nov 26 at 21:58












          • @Jean-ClaudeArbaut If you want to use the “perverse notation” (as D. E. Knuth calls it), use arctan xinmathopen]-pi/2,pi/2mathclose[, that renders $arctan xinmathopen]-pi/2,pi/2mathclose[$.
            – egreg
            Nov 26 at 21:59












          • @egreg Thanks, good to know. Why perverse? That's the notation I learned and used since high school and through all my university curriculum. I find it clear, and more readable than alternative notations, especially in handwriting.
            – Jean-Claude Arbaut
            Nov 26 at 22:04





















          2














          Consider a right triangle whose other two angles are $theta =tan^{-1}x$ and $frac{pi}2-theta$.



          If the legs have lengths $1$ and $x$ (say the side opposite $theta$ has length $x$) then the hypotenuse has length $sqrt{1+x^2}$.



          It follows that $sintheta=frac x{sqrt{1+x^2}}$ and $costheta =frac1{sqrt{1+x^2}}$.






          share|cite|improve this answer























            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%2f3014940%2fhow-to-determine-cosine-and-sine-given-tan-1x%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3














            Hint:
            $$sin^2(y)+cos^2(y) = 1 implies tan^{2}(y)+1=frac{1}{cos^2(y)}.$$
            Consequently, if
            $$tan^{-1}(x) = y implies x=tan(y) implies cos(y) = frac{1}{sqrt{x^2+1}}, sin(y)=?$$






            share|cite|improve this answer





















            • There is a missing step: as is, $cos y$ is only known to be $pmfrac{1}{sqrt{x^2+1}}$. However, $mathrm{arctan},xin]-pi/2,pi/2[$, so the cosine is nonnegative.
              – Jean-Claude Arbaut
              Nov 26 at 21:56












            • @Jean-ClaudeArbaut I left it purposefully to let the OP figure out the details.
              – Math Lover
              Nov 26 at 21:57










            • That's a good point :) +1
              – Jean-Claude Arbaut
              Nov 26 at 21:58












            • @Jean-ClaudeArbaut If you want to use the “perverse notation” (as D. E. Knuth calls it), use arctan xinmathopen]-pi/2,pi/2mathclose[, that renders $arctan xinmathopen]-pi/2,pi/2mathclose[$.
              – egreg
              Nov 26 at 21:59












            • @egreg Thanks, good to know. Why perverse? That's the notation I learned and used since high school and through all my university curriculum. I find it clear, and more readable than alternative notations, especially in handwriting.
              – Jean-Claude Arbaut
              Nov 26 at 22:04


















            3














            Hint:
            $$sin^2(y)+cos^2(y) = 1 implies tan^{2}(y)+1=frac{1}{cos^2(y)}.$$
            Consequently, if
            $$tan^{-1}(x) = y implies x=tan(y) implies cos(y) = frac{1}{sqrt{x^2+1}}, sin(y)=?$$






            share|cite|improve this answer





















            • There is a missing step: as is, $cos y$ is only known to be $pmfrac{1}{sqrt{x^2+1}}$. However, $mathrm{arctan},xin]-pi/2,pi/2[$, so the cosine is nonnegative.
              – Jean-Claude Arbaut
              Nov 26 at 21:56












            • @Jean-ClaudeArbaut I left it purposefully to let the OP figure out the details.
              – Math Lover
              Nov 26 at 21:57










            • That's a good point :) +1
              – Jean-Claude Arbaut
              Nov 26 at 21:58












            • @Jean-ClaudeArbaut If you want to use the “perverse notation” (as D. E. Knuth calls it), use arctan xinmathopen]-pi/2,pi/2mathclose[, that renders $arctan xinmathopen]-pi/2,pi/2mathclose[$.
              – egreg
              Nov 26 at 21:59












            • @egreg Thanks, good to know. Why perverse? That's the notation I learned and used since high school and through all my university curriculum. I find it clear, and more readable than alternative notations, especially in handwriting.
              – Jean-Claude Arbaut
              Nov 26 at 22:04
















            3












            3








            3






            Hint:
            $$sin^2(y)+cos^2(y) = 1 implies tan^{2}(y)+1=frac{1}{cos^2(y)}.$$
            Consequently, if
            $$tan^{-1}(x) = y implies x=tan(y) implies cos(y) = frac{1}{sqrt{x^2+1}}, sin(y)=?$$






            share|cite|improve this answer












            Hint:
            $$sin^2(y)+cos^2(y) = 1 implies tan^{2}(y)+1=frac{1}{cos^2(y)}.$$
            Consequently, if
            $$tan^{-1}(x) = y implies x=tan(y) implies cos(y) = frac{1}{sqrt{x^2+1}}, sin(y)=?$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 26 at 21:42









            Math Lover

            13.7k31435




            13.7k31435












            • There is a missing step: as is, $cos y$ is only known to be $pmfrac{1}{sqrt{x^2+1}}$. However, $mathrm{arctan},xin]-pi/2,pi/2[$, so the cosine is nonnegative.
              – Jean-Claude Arbaut
              Nov 26 at 21:56












            • @Jean-ClaudeArbaut I left it purposefully to let the OP figure out the details.
              – Math Lover
              Nov 26 at 21:57










            • That's a good point :) +1
              – Jean-Claude Arbaut
              Nov 26 at 21:58












            • @Jean-ClaudeArbaut If you want to use the “perverse notation” (as D. E. Knuth calls it), use arctan xinmathopen]-pi/2,pi/2mathclose[, that renders $arctan xinmathopen]-pi/2,pi/2mathclose[$.
              – egreg
              Nov 26 at 21:59












            • @egreg Thanks, good to know. Why perverse? That's the notation I learned and used since high school and through all my university curriculum. I find it clear, and more readable than alternative notations, especially in handwriting.
              – Jean-Claude Arbaut
              Nov 26 at 22:04




















            • There is a missing step: as is, $cos y$ is only known to be $pmfrac{1}{sqrt{x^2+1}}$. However, $mathrm{arctan},xin]-pi/2,pi/2[$, so the cosine is nonnegative.
              – Jean-Claude Arbaut
              Nov 26 at 21:56












            • @Jean-ClaudeArbaut I left it purposefully to let the OP figure out the details.
              – Math Lover
              Nov 26 at 21:57










            • That's a good point :) +1
              – Jean-Claude Arbaut
              Nov 26 at 21:58












            • @Jean-ClaudeArbaut If you want to use the “perverse notation” (as D. E. Knuth calls it), use arctan xinmathopen]-pi/2,pi/2mathclose[, that renders $arctan xinmathopen]-pi/2,pi/2mathclose[$.
              – egreg
              Nov 26 at 21:59












            • @egreg Thanks, good to know. Why perverse? That's the notation I learned and used since high school and through all my university curriculum. I find it clear, and more readable than alternative notations, especially in handwriting.
              – Jean-Claude Arbaut
              Nov 26 at 22:04


















            There is a missing step: as is, $cos y$ is only known to be $pmfrac{1}{sqrt{x^2+1}}$. However, $mathrm{arctan},xin]-pi/2,pi/2[$, so the cosine is nonnegative.
            – Jean-Claude Arbaut
            Nov 26 at 21:56






            There is a missing step: as is, $cos y$ is only known to be $pmfrac{1}{sqrt{x^2+1}}$. However, $mathrm{arctan},xin]-pi/2,pi/2[$, so the cosine is nonnegative.
            – Jean-Claude Arbaut
            Nov 26 at 21:56














            @Jean-ClaudeArbaut I left it purposefully to let the OP figure out the details.
            – Math Lover
            Nov 26 at 21:57




            @Jean-ClaudeArbaut I left it purposefully to let the OP figure out the details.
            – Math Lover
            Nov 26 at 21:57












            That's a good point :) +1
            – Jean-Claude Arbaut
            Nov 26 at 21:58






            That's a good point :) +1
            – Jean-Claude Arbaut
            Nov 26 at 21:58














            @Jean-ClaudeArbaut If you want to use the “perverse notation” (as D. E. Knuth calls it), use arctan xinmathopen]-pi/2,pi/2mathclose[, that renders $arctan xinmathopen]-pi/2,pi/2mathclose[$.
            – egreg
            Nov 26 at 21:59






            @Jean-ClaudeArbaut If you want to use the “perverse notation” (as D. E. Knuth calls it), use arctan xinmathopen]-pi/2,pi/2mathclose[, that renders $arctan xinmathopen]-pi/2,pi/2mathclose[$.
            – egreg
            Nov 26 at 21:59














            @egreg Thanks, good to know. Why perverse? That's the notation I learned and used since high school and through all my university curriculum. I find it clear, and more readable than alternative notations, especially in handwriting.
            – Jean-Claude Arbaut
            Nov 26 at 22:04






            @egreg Thanks, good to know. Why perverse? That's the notation I learned and used since high school and through all my university curriculum. I find it clear, and more readable than alternative notations, especially in handwriting.
            – Jean-Claude Arbaut
            Nov 26 at 22:04













            2














            Consider a right triangle whose other two angles are $theta =tan^{-1}x$ and $frac{pi}2-theta$.



            If the legs have lengths $1$ and $x$ (say the side opposite $theta$ has length $x$) then the hypotenuse has length $sqrt{1+x^2}$.



            It follows that $sintheta=frac x{sqrt{1+x^2}}$ and $costheta =frac1{sqrt{1+x^2}}$.






            share|cite|improve this answer




























              2














              Consider a right triangle whose other two angles are $theta =tan^{-1}x$ and $frac{pi}2-theta$.



              If the legs have lengths $1$ and $x$ (say the side opposite $theta$ has length $x$) then the hypotenuse has length $sqrt{1+x^2}$.



              It follows that $sintheta=frac x{sqrt{1+x^2}}$ and $costheta =frac1{sqrt{1+x^2}}$.






              share|cite|improve this answer


























                2












                2








                2






                Consider a right triangle whose other two angles are $theta =tan^{-1}x$ and $frac{pi}2-theta$.



                If the legs have lengths $1$ and $x$ (say the side opposite $theta$ has length $x$) then the hypotenuse has length $sqrt{1+x^2}$.



                It follows that $sintheta=frac x{sqrt{1+x^2}}$ and $costheta =frac1{sqrt{1+x^2}}$.






                share|cite|improve this answer














                Consider a right triangle whose other two angles are $theta =tan^{-1}x$ and $frac{pi}2-theta$.



                If the legs have lengths $1$ and $x$ (say the side opposite $theta$ has length $x$) then the hypotenuse has length $sqrt{1+x^2}$.



                It follows that $sintheta=frac x{sqrt{1+x^2}}$ and $costheta =frac1{sqrt{1+x^2}}$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 26 at 21:54

























                answered Nov 26 at 21:47









                Chris Custer

                10.8k3724




                10.8k3724






























                    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.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • 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.


                    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%2f3014940%2fhow-to-determine-cosine-and-sine-given-tan-1x%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

                    Probability when a professor distributes a quiz and homework assignment to a class of n students.

                    Image
                    0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                    Read more

                    Aardman Animations

                    Image
                    Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                    Read more

                    Are they similar matrix

                    Image
                    0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                    Read more