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







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    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
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          active

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          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























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            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






























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