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Finding Value of Inverse Trig Funct

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data

    find the exact value of cos(arctan(2)+arctan(3))

    3. The attempt at a solution

    I know that arctan(2) is equivalent to tanx=2 and arctan(3) is equal to tanx=3

    I was thinking perhaps sinx/cosx=2 and 3 but I am not sure what numbers to use. I'm kind of weak in my trig rules =( can someone explain to me how I would go about this problem? I know the answer is -sqrt2/2 but I want to know HOW to do it.
     
  2. jcsd
  3. Feb 22, 2009 #2
    I would say tan(y) = 3 since they are different angles.
    It may be difficult to evaluate the cosine of the two angles if there's a quotient involved. Try separating the two angles by using the identities for adding two angles. Then you will just need to evaluate terms like cos(arctan(2)), for which you can draw a right triangle with a height of 2 and a base of 1. The angle of magnitude arctan(2) would be the angle adjacent to the base. You can then read the cosine right off the triangle.
     
  4. Feb 22, 2009 #3
    what do you mean?
     
  5. Feb 22, 2009 #4
    Try using the identity cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
     
  6. Feb 22, 2009 #5
    How would i solve for x and y then? because like thinking it in terms of sin/cos=2 or 3 seems kinda hard
     
  7. Feb 22, 2009 #6

    robphy

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    As slider142 says, use the identity for cos(A+B), where tan A=2 and tan B=3.
    The idea now is to express the identity for cos(A+B) in terms of tan A and tan B.
    For example, can you write cos A in terms of tan A?
     
  8. Feb 22, 2009 #7
    OHhhh yes think of it like a triangle, the angle would be arctan(2) the sine would be equal to 2 and the cos would be equal to 1 and the hypotenuse is equal to sqrt 5. then when you use CAH it would be 1/sqrt(5) for cos(arctan(2) right?Yes i got the answer =D thanks
     
    Last edited: Feb 22, 2009
  9. Feb 22, 2009 #8
    I got a new problem now ummm sine(2arcsec(13/5))

    I went about using the similar method above but without the identity. but because of the 2 there I don't know how to get the answer...

    do i use sine(2a)=2sinacosa?
     
  10. Feb 22, 2009 #9
    ArcTan(3) + ArcTan(2) = 3Pi/4
     
  11. Feb 22, 2009 #10
    That would be a straightforward method. :smile:
     
  12. Feb 22, 2009 #11
    Thanks so much for all your help ^^.
     
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