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Exact Value of Trigonometric Equations

  1. Nov 27, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the exact value of the given expressions.
    Code (Text):

    sin (2 sin[SUP]-1[/SUP] (4/5))
    or
    sin (2 arcsin (4/5))
     
    Just two different ways of writing it

    2. Relevant equations
    maybe


    3. The attempt at a solution
    I've only gotten this far:
    Code (Text):

    sin (2 arcsin (4/5)) = 2 sin (arcsin (4/5))
     
    I think i might need to use: if -1 <= x <= 1, then sin(arcsin (x)) = x and cos(arccos (x)) = x. But I am not entirely sure. Any help explaining this to me?
     
  2. jcsd
  3. Nov 27, 2012 #2

    micromass

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    You seem to think that [itex]\sin(2\alpha)=2\sin(\alpha)[/itex]. This is not true. It is one of those mistakes that newbies like to make, but you need to remember that it is not true.

    That said, do you know the actual formula for [itex]\sin(2\alpha)[/itex]?
     
  4. Nov 27, 2012 #3

    jedishrfu

    Staff: Mentor

    You need to use the identity: sin(2x) = 2 (sin x) (cos x) where x = arcsin(4/5)
     
  5. Nov 27, 2012 #4
    Okay thank you. Yes i am familiar with sin 2x = 2 sin(x)cos(x). So would this be correct:

    sin (2 arcsin (4/5)) = 2 sin (arcsin (4/5)) * cos (arcsin (4/5))
     
  6. Nov 27, 2012 #5

    jedishrfu

    Staff: Mentor

    Yes, now you can continue to simply the factors.
     
  7. Nov 27, 2012 #6
    Would my next step be this:

    Code (Text):

    [U]sin (2 arcsin (4/5))[/U]    = cos (arcsin (4/5))
    2 sin (arcsin (4/5))

    1 = cos (arcsin (4/5)) ?
     
     
    Last edited: Nov 27, 2012
  8. Nov 27, 2012 #7
    If you want the value of sin(2 arcsin(4/5)), you'd be better off calculating 2 sin(arcsin(4/5)) and cos(arcsin(4/5)) and multiplying those results. The first should be easy, and for the second I'd try using some equation that connects sin(x) and cos(x) and use that to get an expression for cos(x)=...
     
  9. Nov 27, 2012 #8
    Could the way i did be used? If so, would it become:
    Code (Text):

    1 = cos (arcsin (4/5))
     
    And I'm a little confused about arcsin. Does arcsin mean "some side" over "some side" just like cos means adj / hyp ?
     
  10. Nov 27, 2012 #9
    You shouldn't lose sight of what you're aiming for, i.e. the value of sin(2 arcsin(4/5)), so putting that in a fraction will lead you away from your goal. Ans you still seem to think that 2 sin(...) is the same as sin(2 ...) which it isn't.

    No. arcsin is the inverse of sin, so when y=sin(x), you have arcsin(y)=x. It's like y=x² meaning the same as √y=x, and like that, it's true only if you keep to certain values for x and y. (IN your case, you only look at one value, 4/5, for the arcsin function, so you shouldn't have any problems.)
     
  11. Nov 27, 2012 #10

    Mark44

    Staff: Mentor

    No, not quite. √(x2) is not the same as x.
     
  12. Nov 27, 2012 #11

    LCKurtz

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    Remember that arcsin(x) is the [principle value] of the angle whose sine is x. So when you want to calculate sin (arcsin (4/5)) you are asking for the sine of the angle whose sine is x. Much like asking for the color of a horse whose color is brown.
     
  13. Nov 27, 2012 #12
    So you're this far: [tex]sin(2u)=2sin(u)cos(u)[/tex] and here we are working with u=arcsin of 4/5. So far we have:
    [tex]2sin(arcsin(4/5))cos(arcsin(4/5))=8cos(arcsin(4/5))/5[/tex]

    So now the problem is determining cos(arcsin(4/5)). The first thing that you need to do is think about what arcsin(4/5) really is. This value is the measure of the angle formed in a right triangle whose opposite side is 4 and the hypotenuse is 5, because [tex]sin(x)=\frac{opposite}{hypotenuse}[/tex]

    Think about this triangle. We know that it is a right triangle, and we know two of the sides. Thus, you can calculate the third side using the pythagorean theorem. After that, you can use the definition of cosine. [tex]cos(arcsin(4/5))=\frac{adjacent}{hypotenuse}[/tex]
    We know that the hypotenuse is 5 and the adjacent side is the one that we have not yet found. Can you finish the problem from here?
     
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