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Drawing a right triangle to simpliy the given expressions

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Sin(sec^-1(sqrt(x^2+16)/4))

    2. Relevant equations



    3. The attempt at a solution
    I did the math and ended up getting x^2-1 as the opposite, but the answers on the back of the book say other wise.
     
  2. jcsd
  3. Feb 19, 2012 #2
    For future reference, trigonometry is usually precalculus material. Posting in the appropriate forum may be a good idea.

    Assuming a is the adjacent side, b the opposite, and c the hypotenuse, you need to use

    sec x = c / a (so arcsec c / a = x)
    b2 = c2 - a2
    sin x = b / c

    Carefully resolve.
     
  4. Feb 19, 2012 #3
    My solution: b^2=(x^2+16/16) + 1

    I dont know what to do from there!
     
  5. Feb 19, 2012 #4

    Mark44

    Staff: Mentor

    This doesn't make any sense to me.

    Draw a right triangle, with one acute angle labeled θ. Label the two sides and hypotenuse so that sec θ = √(x2 + 16)/4. With appropriate restrictions, this equation is equivalent to θ = sec-1[√(x2 + 16)/4].

    After labeling the sides, find sinθ, and you're done.
     
  6. Feb 19, 2012 #5
    How do i get the opposite value?
     
  7. Feb 19, 2012 #6

    Mark44

    Staff: Mentor

    It's a right triangle. If you know any two sides, you should be able to get the third.
     
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