Choosing signs in inverse trigonometric composing

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I understood why he chose the positive square root in the sin but why the tan is also positive ? Isn't the tan positive and negative in this interval ?
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    I think you misunderstood the comment. It said choose the + sign for the sin. For the tan, the sign depends on the sign of x. For negative x where the angle would be in the interval $[fracpi2,pi]$ the tan would be negative.
    – herb steinberg
    Sep 10 at 21:27














up vote
0
down vote

favorite












I understood why he chose the positive square root in the sin but why the tan is also positive ? Isn't the tan positive and negative in this interval ?
enter image description here






trigonometry







share|cite|improve this question

















  • 1




    I think you misunderstood the comment. It said choose the + sign for the sin. For the tan, the sign depends on the sign of x. For negative x where the angle would be in the interval $[fracpi2,pi]$ the tan would be negative.
    – herb steinberg
    Sep 10 at 21:27












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I understood why he chose the positive square root in the sin but why the tan is also positive ? Isn't the tan positive and negative in this interval ?
enter image description here






trigonometry







share|cite|improve this question













I understood why he chose the positive square root in the sin but why the tan is also positive ? Isn't the tan positive and negative in this interval ?
enter image description here






trigonometry




trigonometry






share|cite|improve this question













share|cite|improve this question











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asked Sep 10 at 21:13









Ahmed M. Elsonbaty

474




474







  • 1




    I think you misunderstood the comment. It said choose the + sign for the sin. For the tan, the sign depends on the sign of x. For negative x where the angle would be in the interval $[fracpi2,pi]$ the tan would be negative.
    – herb steinberg
    Sep 10 at 21:27












  • 1




    I think you misunderstood the comment. It said choose the + sign for the sin. For the tan, the sign depends on the sign of x. For negative x where the angle would be in the interval $[fracpi2,pi]$ the tan would be negative.
    – herb steinberg
    Sep 10 at 21:27







1




1




I think you misunderstood the comment. It said choose the + sign for the sin. For the tan, the sign depends on the sign of x. For negative x where the angle would be in the interval $[fracpi2,pi]$ the tan would be negative.
– herb steinberg
Sep 10 at 21:27




I think you misunderstood the comment. It said choose the + sign for the sin. For the tan, the sign depends on the sign of x. For negative x where the angle would be in the interval $[fracpi2,pi]$ the tan would be negative.
– herb steinberg
Sep 10 at 21:27










2 Answers
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Note that when $-1 <x < 0$, $$tan(cos^-1(x)) = fracsqrt1-x^2x < 0.$$






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    Note that, for $thetain[0,pi]$ and $thetanepi/2$,
    $$
    tantheta=fracsinthetacostheta
    $$
    Since you have already established that $sinarccos x=sqrt1-x^2$, you can directly conclude that
    $$
    tanarccos x=fracsqrt1-x^2x
    $$
    because $cosarccos x=x$ by definition.



    Using $pm$ is misleading, in my opinion.






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      2 Answers
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      2 Answers
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      up vote
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      Note that when $-1 <x < 0$, $$tan(cos^-1(x)) = fracsqrt1-x^2x < 0.$$






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        up vote
        3
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        Note that when $-1 <x < 0$, $$tan(cos^-1(x)) = fracsqrt1-x^2x < 0.$$






        share|cite|improve this answer






















          up vote
          3
          down vote










          up vote
          3
          down vote









          Note that when $-1 <x < 0$, $$tan(cos^-1(x)) = fracsqrt1-x^2x < 0.$$






          share|cite|improve this answer












          Note that when $-1 <x < 0$, $$tan(cos^-1(x)) = fracsqrt1-x^2x < 0.$$







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          answered Sep 10 at 21:23









          Math Lover

          12.9k31333




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              up vote
              0
              down vote













              Note that, for $thetain[0,pi]$ and $thetanepi/2$,
              $$
              tantheta=fracsinthetacostheta
              $$
              Since you have already established that $sinarccos x=sqrt1-x^2$, you can directly conclude that
              $$
              tanarccos x=fracsqrt1-x^2x
              $$
              because $cosarccos x=x$ by definition.



              Using $pm$ is misleading, in my opinion.






              share|cite|improve this answer
























                up vote
                0
                down vote













                Note that, for $thetain[0,pi]$ and $thetanepi/2$,
                $$
                tantheta=fracsinthetacostheta
                $$
                Since you have already established that $sinarccos x=sqrt1-x^2$, you can directly conclude that
                $$
                tanarccos x=fracsqrt1-x^2x
                $$
                because $cosarccos x=x$ by definition.



                Using $pm$ is misleading, in my opinion.






                share|cite|improve this answer






















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Note that, for $thetain[0,pi]$ and $thetanepi/2$,
                  $$
                  tantheta=fracsinthetacostheta
                  $$
                  Since you have already established that $sinarccos x=sqrt1-x^2$, you can directly conclude that
                  $$
                  tanarccos x=fracsqrt1-x^2x
                  $$
                  because $cosarccos x=x$ by definition.



                  Using $pm$ is misleading, in my opinion.






                  share|cite|improve this answer












                  Note that, for $thetain[0,pi]$ and $thetanepi/2$,
                  $$
                  tantheta=fracsinthetacostheta
                  $$
                  Since you have already established that $sinarccos x=sqrt1-x^2$, you can directly conclude that
                  $$
                  tanarccos x=fracsqrt1-x^2x
                  $$
                  because $cosarccos x=x$ by definition.



                  Using $pm$ is misleading, in my opinion.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 10 at 21:50









                  egreg

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