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Elementary Trigonometry, Help with

  1. Sep 22, 2007 #1
    Hello, I have been trying to solve this type of question but the results of my efforts are frequently disappointing. Since this is considered elementary and simple I have no doubt that someone can help me with this:

    Solve arcsin (3/5) + arctan (1/7).

    What if it had said 641/19 instead of 3/5, and 3/711 instead of 1/7? My teachers guided me towards:
    tan x = tan (arcsin 3/5 + arctan 1/7) =, if arcsin 3/5 = A and arctan 1/7 = B, =
    sin(A + B)/cos(A + B) which I develop to [(sinAcosB + cosAsinB)/(cosAcosB + sinAsinB)].
    I do not know what to do next, and I certainly do not see how this is elementary - or am I trying to solve it the wrong way? I know the answer but the method eludes me.

    I apologize if above looks messy; some posts I have read contains very structured mathematical language and it seems to be in a different font - larger and bolder, but I don't know how to write that.

    Thanks in advance.
     
  2. jcsd
  3. Sep 22, 2007 #2
    cos(A + B) = cosAcosB + sinAsinB ... Is it true? Check it.

    If this message is beneficial for you, send a message. If it is not, explain what you do not understand.

    Also, you can use a triangle to find cosA, sinB,... since you know that sinA=3/5 because of arcsin 3/5=A
     
    Last edited: Sep 22, 2007
  4. Sep 22, 2007 #3
    cos(A + B) = cos(A)cos(B) + sin(A)sin(B). At least I believe it is so.
    Your message was unfortunately not beneficial for me as I find your triangle hint rather vague.
     
  5. Sep 22, 2007 #4

    mathman

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    You got the sign wrong! It should be:cos(A + B) = cos(A)cos(B) - sin(A)sin(B).
     
  6. Sep 23, 2007 #5
    Oh sorry. I tried it again using that but I still fail: I get arctan(25/31). Help?
     
  7. Sep 23, 2007 #6
    i dont know if you call this elementary// that
    arctanx +arctany=arctan(x+y)/1-xy

    you can express arcsin(3/5) in arctan
    let arcsin(3/5)=A
    SinA=3/5
    CosA=squareroot(1-sqr(3/5))=4/5
    or, TanA=(3/5)/(4/5)=3/4
    A=arctan(3/4)
    or arcsin(3/5)=arctan(3/4)

    now

    arctan(3/4)+arctan(1/7)=arctan[ {(3/4)+(1/7)} / {1-(3/4)x(1/7)} ]
    =arctan[ (25/28)/(25/28) ]
    =arctan(1)
    =Pie/4
    ------------------------

    or why not to use Tan(A+B)=TanA+TanB/1-TanATanB

    you got arcsin(3/5)=A
    n TanA=3/4
    Similarly TanB=1/7

    Tan(A+B)=(3/4+1/7){1-(3/4)x(1/7)}
    ..... and the same answer..
    i hope this is correct
    i think you have done some calculation or tranformation mistake while solving Sin(A+B)/cos(A+B)

    if you post your steps ..... we can see the mistake thereafter
     
  8. Sep 23, 2007 #7
    Thank you for your help! I solved it now but with the method I tried at first: I had made sloppy mistakes somewhere. Since (thanks to flora) cos(A + B) = cosAcosB - sinAsinB I got (sinA + cosAtanB)/(cosA - sinAtanB) and, calculating cosA through (1 - sinAsinA)^½ I get that sinA + cosAtanB = cosA - sinAtanB, why x = arctan 1 which is pi/4. Thanks again!
     
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