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Trig prob

  1. Jun 8, 2006 #1

    [tex] \arcsin( \frac{\sqrt{36-x^2}}{6})= \arccos(?) [/tex]

    I was wondering how one would solve this equation; it was given as an extra credit problem on a test I took today. Being the impulsive person I am, I can't wait until monday to receive the solution.
    Last edited: Jun 8, 2006
  2. jcsd
  3. Jun 8, 2006 #2
    Think about the simplest possible solution. [tex]\arcsin(x)=\arccos(?)[/tex] has what solution?
  4. Jun 8, 2006 #3
    for some reason, I feel it would be cos^-1(x/6)...
  5. Jun 8, 2006 #4
    what reason?

  6. Jun 8, 2006 #5
    That would be correct. For the reason think about what the arcsince function is, it is the angle whose since you are inputting, then you can draw a right triangle from the information given in the problem and find what the cosine of that angle would be.
  7. Jun 8, 2006 #6
    well, since I know the sin of that triangle is sqrt(36-x^2)/6


    using pathagorean theorem x^2 + y^2=6^2
    y^2= 36-x^2

    y=opposite leg
    x=adjacent leg, so
    I concluded that arccos(x/6).
  8. Jun 8, 2006 #7
    Yep, that's pretty much it.
  9. Jun 8, 2006 #8
    the funny thing about this problem is I answered it completely different, which was also wrong; I guess it was spur of the moment, and I was enduring a brain fart of epic proportion.
  10. Jun 9, 2006 #9


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    Homework Helper

    The more elegant answer has already been given, but suppose you're not thinking of that triangle and still want to solve it. You could take the sine of both sides, the LHS then gives: sqrt(36-x²)/6.

    For the RHS, you start with sin(arccos(y)) but you can use the fundamental identity cos²a+sin²a = 1 to rewrite (I'll take the positive root): sin(a) = sqrt(1-cos²(a)). So:

    sin(arccos(y)) = sqrt(1-cos²(arccos(y))) = sqrt(1-y²)

    Now you can solve sqrt(1-y²) = sqrt(36-x²)/6, but by inspection it's allready easy to see that a solution for y = x/6.
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