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Homework Help: Cos(arcsin(-1/3)) how to do it by hand?

  1. Apr 5, 2007 #1
    cos(arcsin(-1/3)) how to do it by hand??

    1. The problem statement, all variables and given/known data
    Solve without a calculator

    3. The attempt at a solution
    now i have fount the solution to be (-2/3)(2^(1/2)) now not only did my calculator come up with that but between the SSS formula and pythagorean formula( i think is what we cal it, a^2+b^2....) i was able to get the same answer but the question says not to use a calculator and using those formulas of course im going to have to use a calculator to figure out some of the roots and what not.

    I havent had to do this stuff for quite sometime now im a ways past precalc but i was trying to help a friend and im not sure how to do this without a calculator or a root table and any other aids other than my mind. So being that this is precalc i would assume there is a easy way do do this beyond what im trying if the teacher expects it to be done without a calc so if you know how please enlighten me.
  2. jcsd
  3. Apr 5, 2007 #2
    It's actually very simple once you know how to do it. Forget about cosine for a moment. Some angle (x) equals the arcsine of -1/3 -- so x=arcsin(-1/3). So you know that the sin(x)=-1/3. Now you can make a right triangle in the IV quadrant and solve for the cosine value, since you are simply solving for the cos(arcsin(-1/3)) or cos(x).
    Last edited: Apr 5, 2007
  4. Apr 5, 2007 #3


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    Homework Helper

    You can also try to work out the general formula for cos(arcsin(x)). Here we go.
    Let [tex]\alpha = \arcsin(x) \Rightarrow \sin \alpha = x[/tex]
    cos(arcsin(x)) will become: [tex]\cos \alpha[/tex]

    Also, you should note that arcsin(x) will return the values in the interval [tex]\left[ - \frac{\pi}{2} ; \ \frac{\pi}{2}\right][/tex], i.e [tex]\alpha[/tex] is in I, and IV quadrant, and its cosine value should be positive. So we have:
    [tex]\cos \alpha = \sqrt{1 - \sin ^ 2 \alpha} = \sqrt{1 - x ^ 2}[/tex]

    Applying the formula to your question yielding:
    [tex]\cos \left( \arcsin \left( - \frac{1}{3} \right) \right)) = \sqrt{1 - \left( -\frac{1}{3} \right) ^ 2} = \sqrt{1 - \frac{1}{9}} = \frac{2\sqrt{2}}{3}[/tex]. :)

    You can do the same and find the general formula to:
    Can you get it? :)
  5. Apr 5, 2007 #4
    thanks for both of you for the replies. I especially apreciate yours VietDao29, that way seems to be the best route in finding the aswer via no calculator because of its simplicity, i know with the triangle method although easy not so much without the calculator.
  6. Apr 6, 2007 #5
    Yeah I think his method is better too. :biggrin:
    Last edited: Apr 6, 2007
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