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Inverse function of sin , cos and tan

  1. Oct 24, 2005 #1
    Solve the following question. Give your answer in cert form .Thank you .
    sin-1 x + sin-1 (2x) = pie/3
     
  2. jcsd
  3. Oct 24, 2005 #2

    Tide

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    Tell us what you have tried so far so we can help you! :)
     
  4. Oct 24, 2005 #3
    You could use numerical methods. Example:
    f(x) = arcsinx+arcsin2x = pi/3 =>let x(n+1) = sin( pi/3 - arcsinxn), that should work...
     
  5. Oct 24, 2005 #4

    Tide

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    I don't think you have to resort to numerical methods. I'd like to see what the original poster has tried.
     
  6. Oct 26, 2005 #5
    Can you show me the step by step calculations . Thank You .
     
  7. Oct 26, 2005 #6

    Tide

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    If you take the sine of both sides and use the basic properties of inverse trig functions you will find:

    [tex]x \left( \sqrt{1-4x^2} +2 \sqrt {1 - x^2} \right) = \frac {\sqrt 3}{2}[/tex]

    I believe you can solve this equation without resorting to numerical approximation.
     
  8. Oct 27, 2005 #7
    I still not understand .Can you explain more detail ? Thank You .
     
  9. Oct 27, 2005 #8
    Think about a right triangle. If sin-1(x)=θ, then the hypotenuse must be 1 and the adjacent side [itex]\sqrt{1-x^2}[/itex]. If the sine is 2x, then the hypotenuse is still 1, but the cosine (the adjacent side) is [itex]\sqrt{1-4x^2}[/itex]. Now try using that to rewrite your equation:

    [tex]\sin{\theta}=x[/tex]

    [tex]\sin{\phi}=2x=2\sin{\theta}[/tex]
     
  10. Oct 27, 2005 #9

    Hurkyl

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    How much do you understand? Have you tried working through the suggested step yourself? Where did you get stuck?
     
  11. Oct 29, 2005 #10
    I do not know what I need to do with the sin-1(x) +sin-1 (2x) .Is it these two have relationship ? If yes please tell me . The suggested step is to short , I cannot understand it . So counld you give me more details on that .
     
  12. Oct 29, 2005 #11

    Tide

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    ng,

    When I suggested taking the sine of both sides I assumed you would recognize you need to use a basic trig identity for the sum of angles:

    [tex]\sin a + b = \sin a \cos b + \cos a \sin b[/tex]

    Also, you will need basic relations for the inverse trig functions:

    [tex]\sin \sin^{-1} x = x[/tex]

    [tex]\sin \cos^{-1} x = \sqrt {1-x^2}[/tex]

    and similarly for the cosines. Does that help?
     
  13. Oct 29, 2005 #12

    Hurkyl

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    I was hoping when one poster suggested that you take the sine of both sides, you would have gotten at least as far as

    sin( sin-1 x + sin-1 (2x) ) = sin( pie/3 )
     
  14. Oct 30, 2005 #13
    Can you prove for me why [tex]\sin \cos^{-1} x = \sqrt {1-x^2}[/tex] . Thank You .
     
  15. Oct 30, 2005 #14

    Hurkyl

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    Have you tried to do it yourself? How far did you get? Have you consulted your book at all?
     
  16. Oct 30, 2005 #15

    Tide

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    I certainly can but I like Hurkyl's questions! Ask again if you get stuck. :smile:
     
  17. Oct 30, 2005 #16
    You might try starting with the identity:

    [tex]\sin{x}=\sqrt{1-\cos^2{x}}[/tex]
     
  18. Oct 30, 2005 #17
    My book does not teach me that . It only have sin-1 A + cos-1 B = pie/2 . So can you please prove for me because I have no idea about that ?
     
  19. Oct 30, 2005 #18
    [tex]\sin{\left(\cos^{-1}{x}\right)}=\sqrt{1-\cos^2{\left(\cos^{-1}{x}}\right)}[/tex]

    You know that [itex]\cos{\left(\cos^{-1}{x}\right)}=x[/itex], so that above equation becomes:

    [tex]\sqrt{1-x^2}[/tex]

    You could also do this geometrically by drawing a right triangle and using the Pythagorean Theorem.

    In your last post, did you mean?:

    [tex]\sin^{-1}{x}+\cos^{-1}{x}=\frac{\pi}{2}[/tex]
     
    Last edited: Oct 30, 2005
  20. Oct 30, 2005 #19
    Oh yes . I am sorry about the typing error . I have seen your answer but how can we know [tex]\sin{\left(\cos^{-1}{x}\right)}=\sqrt{1-\cos^2{\left(\cos^{-1}{x}}\right)}[/tex]
     
  21. Oct 30, 2005 #20

    Tide

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    That should be obvious: [itex]\cos \cos^{-1} x = x[/itex] so just square it! :)
     
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