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How is ## \tan ( \arcsin x) = \frac{x}{\sqrt{1-x^2}} ## ?

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  1. Aug 17, 2016 #1

    Rectifier

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    • Member warned about posting without the template and with no effort shown
    How can I write ## \tan ( \arcsin x) ## as ## \frac{x}{\sqrt{1-x^2}} ##? This is not a problem in itself but a part of a solution to a problem.
     
  2. jcsd
  3. Aug 17, 2016 #2

    jedishrfu

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    Start by using the sin / cos definition of tan and you should begin to see the derivation.
     
  4. Aug 17, 2016 #3

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    ## \tan ( \arcsin x) = \frac{\sin ( \arcsin x)}{\cos ( \arcsin x)} = \frac{x}{\cos ( \arcsin x)} ##

    I draw a triangle with sides x and 1. Therefore the hypotenuse is ## \sqrt{1+x^2}##.

    ## \cos \left( v \right) = \frac{1}{\sqrt{1+x^2}} ##

    This is where I get stuck. Please help :,(
     
  5. Aug 17, 2016 #4
    If ##\theta=arcsin x##, what is ##\sin \theta?##
     
  6. Aug 17, 2016 #5
    You drew the wrong triangle. You need a triangle with an angle whose sine is x. Then which of the sides needs to be x and 1?
     
  7. Aug 17, 2016 #6

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    If ## \theta = \arcsin (\sin x)##

    ##x## is the opposite side of the angle ## \theta ## and 1 is the hypotenuse?.
     
  8. Aug 17, 2016 #7
    Does that get you the answer?
     
  9. Aug 17, 2016 #8
    No. What is the sine of the angle whose sine is x?
     
  10. Aug 17, 2016 #9
    I think he got it correct this time.
     
  11. Aug 17, 2016 #10

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    Hmm....
    ## \theta = \arcsin (\sin x)##

    Then ##\sin \theta = \sin x##

    But this looks wrong :eek:

    Sorry, this is hard for me :S
     
  12. Aug 17, 2016 #11
    Well you drew the triangle correct, so what's ##cos(\theta)## ?
     
  13. Aug 17, 2016 #12

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    Wait, so one side of the triangle is equal to the angle?

    ##\cos x = \sqrt{1-x^2}##
     
  14. Aug 17, 2016 #13
    So ##tan(\theta)## is?
     
  15. Aug 17, 2016 #14

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    ##\frac{x}{\sqrt{1-x^2}}## :D
     
  16. Aug 17, 2016 #15
    No. ##sin \theta = x##
     
  17. Aug 17, 2016 #16
    In post #8, I asked you "What is the sine of the angle whose sine is x?" You answered sin x. What would your answer have been if I asked "What is the name of the person whose name is John?"
     
  18. Aug 17, 2016 #17

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    It is x :)
     
  19. Aug 17, 2016 #18
    Good. So, in terms of x, what is cos theta? Then , in terms of x, what is tan theta?
     
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