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Math Help - Arc values not on the unit circle

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    Arc values not on the unit circle

    How do you find cos (arctan4/3 - arcsin3/5)?
    Last edited by Nervous; October 13th 2011 at 10:41 AM.
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    Re: Arc values not on the unit circle

    Quote Originally Posted by Nervous View Post
    How do you find acrcos (arctan4/3 - arcsin3/5)?
    I think you really mean the value of cos[arctan(4/3)-arcsin(3/5)] ... correct?
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    Re: Arc values not on the unit circle

    Yes, I'll fix that.
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    Re: Arc values not on the unit circle

    I'm a bit confused by the title: "Arcvalues not on the unit circle". arctan(4/3) and arcsin(3/5) definitely are on the unit circle. I think you mean simply that those are not any of the "easy" values you learn- arcsin(\sqrt{2}/2}, cos(1/2), etc.

    However, it should be very easy to see that if we have a right triangle with one leg of length 3 and and another of length 4, then the hypotenuse has length 5. So, since sine is "opposite side over hypotenuse" while tangent is "opposite side over near side", it should be just as easy to see that "arctan(4/3)" and "arctan(3/5)" are opposite angles in such a triangle. Cosine of their sum would be very easy to find! For their difference, us the fact that cos(a- b)= cos(a)cos(b)+ sin(a)sin(b) and its easy to find sine and cosine of those two angles.
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    Re: Arc values not on the unit circle

    Thanks.
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