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How to make the equation sin(x-60degree)-cos(30degree-x)=1 into cosx=k?

  1. Mar 3, 2007 #1

    inv

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    [Solved]How to make the equation sin(x-60degree)-cos(30degree-x)=1 into cosx=k?

    1. The problem statement, all variables and given/known data
    Hi. How do this- show that the equation sin(x-60degree)-cos(30degree-x)=1 can be written in the form cosx=k,where k is a constant?


    2. Relevant equations
    Trigonometry rules or something


    3. The attempt at a solution
    1=sinxcos60 +cosxcos30-cos30cosx+cos60sinx
    =sinx (STuck*)
     
    Last edited: Mar 3, 2007
  2. jcsd
  3. Mar 3, 2007 #2

    AlephZero

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    Check your formulas for sin(a-b) and cos(a-b). You seem to have some of the signs wrong.
     
  4. Mar 3, 2007 #3

    arildno

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    Where is your sin*sin term, and where is the sin(60)cos(x) term??
    How do you think you'll ever manage such problems when you evidently don't bother to learn the addition formulae PROPERLY?

    This problem is TRIVIAL, it is only your own negligence that causes you to get stuck.
     
  5. Mar 3, 2007 #4

    inv

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    I start from here:
    1=sin(x-60)-cos(30-x)
    =sinxcos(-60)-cosx(sin(-60)-cos30cos(-x)-sin30sin(-x) (*I then apply the odd properties for cos and sin and their exact values & *Arildno my sin*sin term is at the last term here)
    obtaining-
    Telling me what's wrong here'll be helpful.What's wrong?
     
    Last edited: Mar 3, 2007
  6. Mar 3, 2007 #5

    arildno

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    You should learn basic arithmetical rules first!
    For example,
    "sin(x-60)=sinxcos(-60)-cosx(sin(-60)"
    This is totally incorrect!
    To do it PROPERLY, you should use EITHER:
    sin(a-b)=sin(a)cos(b)-sin(b)cos(a), yielding
    sin(x-60)=sin(x)cos(60)-sin(60)cos(x)
    OR:
    sin(a-b)=sin(a+(-b))=sin(a)cos(-b)+sin(-b)cos(a), yielding:
    sin(x-60)=sin(x+(-60))=sin(x)cos(-60)+sin(-60)cos(x), yielding the same answer as above, when invoking the even and odd properties of cosine and sine.
     
  7. Mar 3, 2007 #6

    HallsofIvy

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    The first thing that was wrong was

    !!
    Precisely what "trigonometry rules" are you using and what do they say?

    How do you expect us to tell you what is wrong when you don't show us what you have done! What are cos(60), cos(30), sin(60), sin(30)?
    What formulas did you use for cos(x- 60), sin(x-60), cos(30-x), and sin(30-x)?
     
  8. Mar 3, 2007 #7

    inv

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    I'm using sin(A+-B)=sinAcosB+-cosAsinb & cos(A+-B)=cosAcosB-+sinAsinB ,odd properties sin(-theta)=-sin(theta),cos(-theta)=cos(theta) & cos 60=.5,sin30=.5,cos30=square root of 3 over 2,sin60=cos 30 trigonometry rules HallfofIvy.They say the way of solving this question. I used the odd properties and the exact values I specified above,and it worked hence this problem is solved.Thanks indeed to whoever posted here some if not all with fast response.Been helpful arildno and aleph on this also considerate.Cya.
     
  9. Mar 3, 2007 #8

    arildno

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    What you wrote was totally incorrect, irrespective of how the end result accords with the "answer".
     
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