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Homework Help: Difference Formula Involving Inverse Trgi Functions

  1. Aug 11, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the exact value of : sin[sin^-1 (3/5) - cos^-1 (-4/5)]

    2. Relevant equations
    sin(A-B) = sinA*cosB-cosA*sinB

    3. The attempt at a solution
    sin A = (3/5) sin B = (-3/5)
    cos A = (4/5) cos B = (-4/5)

    (3/5)*(-4/5) - (-3/5)(4/5)
    (-12/25) - (-12/25) = 0

    I did that and the book says that the answer is (-24/25). What did I do wrong?
  2. jcsd
  3. Aug 11, 2011 #2
    I get something slightly different for sin B.
  4. Aug 11, 2011 #3
    How did you get sin B? from what I understand, -pi/2<B<0. I'm so confused though
  5. Aug 11, 2011 #4


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    Well... the inverse sin and inverse cosine are multivalued functions. In other words, there are many different angles that have, say, a cosine of -4/5. So when you are asked to find the inverse cosine of -4/5 (which is B in your case), which one should you choose? It makes a difference because not all the angles have the same sine. Depending on which angle you choose for B, you might get [itex]\sin B = -\frac{3}{5}[/itex] or you might get something different - that's what Bohrok was alluding to.

    The way mathematicians resolve this is the following convention: unless there is some special reason to do otherwise, [itex]\cos^{-1} X[/itex] means that you pick the angle between 0 and 180 degrees whose cosine is [itex]X[/itex]. This is called the "principal value" (as opposed to all the other, "non-principal" angles which have cosine X but which are not between 0 and 180 degrees). So in this case, to get the angle B, think about the angle between 0 and 180 degrees that has a cosine of -4/5. It might help to actually draw out a unit circle and look at it to figure out what quadrant the angle is in. Now, what is the sine of that angle?

    By the way, for the inverse sine, the principal value is the one between -90 and +90 degrees. So to find the inverse sine of 3/5 (i.e. to find the angle A), you'd pick the angle between -90 and +90 degrees that has a sine of 3/5.
    Last edited: Aug 12, 2011
  6. Aug 11, 2011 #5


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    Staff: Mentor

    Were I marking this question, I would want to see more than one answer. Don't forget that in one sweep around the origin, there will always be more than one angle whose sine is, for example, 3/5.

    Similarly, between 0 and 2.Pi there will be more than one answer to cos-1(-4/5).
    That's one answer. 0 is also correct. Any more?

    It is always a good idea to work things out on your calculator, for comparison. Or for reassurance :smile:

    Always bear in mind that your calculator will display only one answer, even when there are multiple solutions.
    Last edited: Aug 11, 2011
  7. Aug 11, 2011 #6
    Draw a right triangle and let sinθ = x where θ is one of the acute angles and label all the sides. How could you express cosθ? Or,
    sin2θ + cos2θ = 1 → sinθ = ±√(1 - cos2θ)
    Let x = cosθ → θ = cos-1x, then substitute that into the above equation. Then you'll have to choose + or - for the square root based on the fact that arcsin is taken to be between -90° and 90° as diazona said.
  8. Aug 11, 2011 #7


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    What is sin(cos-1(-4/5) ?
    If -1 < x < 0 π/2 < cos-1(x) < π, i.e. cos-1(x) terminates in the second quadrant on the unit circle. Therefore, sin(cos-1(x)) is non-negative.​

    Similarly cos(sin-1(x)) is also non-negative.

    BTW: arcsin & arccos are well defined functions. As such, arcsin(x) has only one value for any x in the domain of arcsin. Similarly, arccos(x) has only one value for any x in its domain.
  9. Aug 12, 2011 #8


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    I guess I meant "inverse sine" and "inverse cosine" then. In my circles people tend to use the terms synonymously ;-) I'll go back and edit.
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