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What is the sin(Acos(x))?

  1. Aug 23, 2011 #1
    What is the sin(acos(X))?
  2. jcsd
  3. Aug 23, 2011 #2


    Staff: Mentor

    Is this a homework problem?

    Also, by "acos" do you mean arccos, or inverse cosine, or is a a constant?
  4. Aug 23, 2011 #3
    Is acos(x) the inverse cosine (arccos x), or is it [itex]a\cos x[/itex]? In the latter case, I don't think there's a particular simplification for the expression -- sine takes angles as arguments, and [itex]a\cos(x)[/itex] is not interpreted as an angle.

    If you mean [itex]\sin(\arccos(x))[/itex], then x is an angle. Let [itex]\theta = \arccos(x)[/itex] so that [itex]x = \cos \theta[/itex]. We can draw a right-triangle containing the angle [itex]\theta[/itex] in the following manner: the side adjacent to [itex]\theta[/itex] has length x, and the hypotenuse has length 1. Then the third side (the side opposite [itex]\theta[/itex]) has length [itex]\sqrt{1 - x^2}[/itex]. Note that this is possible, since [itex]-1 \leq x \leq 1[/itex]. Also, [itex]{-\pi \over 2} \leq \theta \leq {\pi \over 2}[/itex] because of the restricted range of arccos, which enabled us to use the right-triangle as a diagram. Now, observing the triangle, [itex]\sin(\arccos x) = \sin \theta = \sqrt{1 - x^2}[/itex].
  5. Aug 23, 2011 #4


    Staff: Mentor

    This problem is confusing. In the title, you have sin(Acos(x)), which implies to me that A is a constant. In the text of post #1 you have sin(acos(x)), which made me think you were asking about the arccos function. Which is it?
  6. Aug 24, 2011 #5


    User Avatar
    Science Advisor

    Take x to be the length of one leg in a right triangle having hypotenuse of length 1. "acos(x)" is the angle adjacent to x. Use the Pythagorean theorem to find the length of the other leg and then use the definition of sine to find sin(acos(x)).
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