How to show that $cos(sin^-1(x))$ is $sqrt1-x^2$? [duplicate]

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  • $cos(arcsin(x)) = sqrt1 - x^2$. How?

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How to show that $cos(sin^-1(x))$ is $sqrt1-x^2$?




I remember having to draw a triangle, but I'm not sure anymore.




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    The function $sin()$ takes a ratio, $x/1$ in this case, and returns an angle $theta = sin(x/1)$. Draw a triangle that models this, and then say what $cos(theta)$ is.
    – Mike Pierce
    Aug 4 at 1:33















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  • $cos(arcsin(x)) = sqrt1 - x^2$. How?

    7 answers




How to show that $cos(sin^-1(x))$ is $sqrt1-x^2$?




I remember having to draw a triangle, but I'm not sure anymore.




trigonometry




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marked as duplicate by Simply Beautiful Art, Lord Shark the Unknown, lab bhattacharjee trigonometry
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    The function $sin()$ takes a ratio, $x/1$ in this case, and returns an angle $theta = sin(x/1)$. Draw a triangle that models this, and then say what $cos(theta)$ is.
    – Mike Pierce
    Aug 4 at 1:33













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up vote
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This question already has an answer here:



  • $cos(arcsin(x)) = sqrt1 - x^2$. How?

    7 answers




How to show that $cos(sin^-1(x))$ is $sqrt1-x^2$?




I remember having to draw a triangle, but I'm not sure anymore.




trigonometry




share|cite|improve this question














This question already has an answer here:



  • $cos(arcsin(x)) = sqrt1 - x^2$. How?

    7 answers




How to show that $cos(sin^-1(x))$ is $sqrt1-x^2$?




I remember having to draw a triangle, but I'm not sure anymore.





This question already has an answer here:



  • $cos(arcsin(x)) = sqrt1 - x^2$. How?

    7 answers





trigonometry





share|cite|improve this question












share|cite|improve this question




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edited Aug 4 at 2:02









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asked Aug 4 at 1:30









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marked as duplicate by Simply Beautiful Art, Lord Shark the Unknown, lab bhattacharjee trigonometry
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  • 2




    The function $sin()$ takes a ratio, $x/1$ in this case, and returns an angle $theta = sin(x/1)$. Draw a triangle that models this, and then say what $cos(theta)$ is.
    – Mike Pierce
    Aug 4 at 1:33













  • 2




    The function $sin()$ takes a ratio, $x/1$ in this case, and returns an angle $theta = sin(x/1)$. Draw a triangle that models this, and then say what $cos(theta)$ is.
    – Mike Pierce
    Aug 4 at 1:33








2




2




The function $sin()$ takes a ratio, $x/1$ in this case, and returns an angle $theta = sin(x/1)$. Draw a triangle that models this, and then say what $cos(theta)$ is.
– Mike Pierce
Aug 4 at 1:33





The function $sin()$ takes a ratio, $x/1$ in this case, and returns an angle $theta = sin(x/1)$. Draw a triangle that models this, and then say what $cos(theta)$ is.
– Mike Pierce
Aug 4 at 1:33











2 Answers
2






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2
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accepted










Given $cos(sin^-1x)$



Let $sin^-1x=theta$
$$impliessintheta=x\sin^2theta=x^2\1-cos^2theta=x^2\cos^2theta=1-x^2\costheta=pmsqrt1-x^2\theta=cos^-1pmsqrt1-x^2$$
Now, plug in $theta=cos^-1pmsqrt1-x^2$ in $cos(sin^-1x)$ and we get$$pmsqrt1-x^2$$But note that $sin^-1x$ is in $left[dfrac-pi2,dfracpi2right]$ so $cos(sin^-1x)ge0$



Therefore, $cos(sin^-1x)=sqrt1-x^2$






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    enter image description here



    $cos(sin^-1(x)) = cos(sin^-1(x/1))$



    Since sin is opposite hypotenuse we know $x$ is the opposite side and the hypotenuse is 1. To get cos we would need the adjacent side of the right triangle, so it equals $sqrt1^2 - x^2$






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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      2
      down vote



      accepted










      Given $cos(sin^-1x)$



      Let $sin^-1x=theta$
      $$impliessintheta=x\sin^2theta=x^2\1-cos^2theta=x^2\cos^2theta=1-x^2\costheta=pmsqrt1-x^2\theta=cos^-1pmsqrt1-x^2$$
      Now, plug in $theta=cos^-1pmsqrt1-x^2$ in $cos(sin^-1x)$ and we get$$pmsqrt1-x^2$$But note that $sin^-1x$ is in $left[dfrac-pi2,dfracpi2right]$ so $cos(sin^-1x)ge0$



      Therefore, $cos(sin^-1x)=sqrt1-x^2$






      share|cite|improve this answer

























        up vote
        2
        down vote



        accepted










        Given $cos(sin^-1x)$



        Let $sin^-1x=theta$
        $$impliessintheta=x\sin^2theta=x^2\1-cos^2theta=x^2\cos^2theta=1-x^2\costheta=pmsqrt1-x^2\theta=cos^-1pmsqrt1-x^2$$
        Now, plug in $theta=cos^-1pmsqrt1-x^2$ in $cos(sin^-1x)$ and we get$$pmsqrt1-x^2$$But note that $sin^-1x$ is in $left[dfrac-pi2,dfracpi2right]$ so $cos(sin^-1x)ge0$



        Therefore, $cos(sin^-1x)=sqrt1-x^2$






        share|cite|improve this answer























          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          Given $cos(sin^-1x)$



          Let $sin^-1x=theta$
          $$impliessintheta=x\sin^2theta=x^2\1-cos^2theta=x^2\cos^2theta=1-x^2\costheta=pmsqrt1-x^2\theta=cos^-1pmsqrt1-x^2$$
          Now, plug in $theta=cos^-1pmsqrt1-x^2$ in $cos(sin^-1x)$ and we get$$pmsqrt1-x^2$$But note that $sin^-1x$ is in $left[dfrac-pi2,dfracpi2right]$ so $cos(sin^-1x)ge0$



          Therefore, $cos(sin^-1x)=sqrt1-x^2$






          share|cite|improve this answer













          Given $cos(sin^-1x)$



          Let $sin^-1x=theta$
          $$impliessintheta=x\sin^2theta=x^2\1-cos^2theta=x^2\cos^2theta=1-x^2\costheta=pmsqrt1-x^2\theta=cos^-1pmsqrt1-x^2$$
          Now, plug in $theta=cos^-1pmsqrt1-x^2$ in $cos(sin^-1x)$ and we get$$pmsqrt1-x^2$$But note that $sin^-1x$ is in $left[dfrac-pi2,dfracpi2right]$ so $cos(sin^-1x)ge0$



          Therefore, $cos(sin^-1x)=sqrt1-x^2$







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Aug 4 at 1:48









          Key Flex

          3,663422




          3,663422




















              up vote
              4
              down vote













              enter image description here



              $cos(sin^-1(x)) = cos(sin^-1(x/1))$



              Since sin is opposite hypotenuse we know $x$ is the opposite side and the hypotenuse is 1. To get cos we would need the adjacent side of the right triangle, so it equals $sqrt1^2 - x^2$






              share|cite|improve this answer



























                up vote
                4
                down vote













                enter image description here



                $cos(sin^-1(x)) = cos(sin^-1(x/1))$



                Since sin is opposite hypotenuse we know $x$ is the opposite side and the hypotenuse is 1. To get cos we would need the adjacent side of the right triangle, so it equals $sqrt1^2 - x^2$






                share|cite|improve this answer

























                  up vote
                  4
                  down vote










                  up vote
                  4
                  down vote









                  enter image description here



                  $cos(sin^-1(x)) = cos(sin^-1(x/1))$



                  Since sin is opposite hypotenuse we know $x$ is the opposite side and the hypotenuse is 1. To get cos we would need the adjacent side of the right triangle, so it equals $sqrt1^2 - x^2$






                  share|cite|improve this answer















                  enter image description here



                  $cos(sin^-1(x)) = cos(sin^-1(x/1))$



                  Since sin is opposite hypotenuse we know $x$ is the opposite side and the hypotenuse is 1. To get cos we would need the adjacent side of the right triangle, so it equals $sqrt1^2 - x^2$







                  share|cite|improve this answer















                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Aug 4 at 2:22









                  Simply Beautiful Art

                  49k571169




                  49k571169











                  answered Aug 4 at 1:44









                  Bas bas

                  18911




                  18911












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