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Homework Help: Trig factor law help

  1. Nov 18, 2008 #1
    1. The problem statement, all variables and given/known data
    i encountered some problems while solving this question.
    Hope someone would show me the necessary working and solution, thanks in advance (:

    Question:

    Given that sin x + sin y = a and cos x + cos y = a. where a[tex]\neq[/tex]0 , find sin x + cos x in terms of a.

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 18, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi icystrike! Welcome to PF! :smile:

    Hint: instead of x and y, write both equations in terms of (x + y)/2 and (x - y)/2. :wink:

    (or you could just subtract the two equations, and use a formula for sinx - cosx)
     
  4. Nov 18, 2008 #3
    Re: Trigonometry

    i tried tat method alr..
    factor law fellowed by double angle law

    i derive with 2(sin x/2)(cos y/2)=0

    but nothing can move on..
     
  5. Nov 18, 2008 #4

    tiny-tim

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    No, you shouldn't have got that :confused:

    show us your full calculation. :smile:
     
  6. Nov 18, 2008 #5
    Re: Trigonometry

    http://img143.imageshack.us/img143/2867/lalalodw5.jpg [Broken]
     
    Last edited by a moderator: May 3, 2017
  7. Nov 18, 2008 #6

    tiny-tim

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    ah … you went a bit doo-lally :uhh: after sin(x+y)/2 = cos(x+y)/2 …

    just divide by the RHS, and you get tan(x+y)/2 = 1,

    and so x+y = … ? :smile:
     
  8. Nov 18, 2008 #7
    Re: Trigonometry

    yup.
    (x+y)/2=45 deg , 225 deg
     
  9. Nov 18, 2008 #8

    tiny-tim

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    Go on then … so x+y = … ?

    and so sinx + cosx = … ? :smile:
     
  10. Nov 18, 2008 #9
    Re: Trigonometry

    eh. x+y=90,450 deg

    sinx+cosx
    i cant find..
     
  11. Nov 18, 2008 #10

    tiny-tim

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    i] for a start you can forget 450 … it's more than 360

    ii] so x + y = 90º … so siny = … ? :smile:
     
  12. Nov 18, 2008 #11
    Re: Trigonometry

    i think cant drop 450 deg cos they did not mention tat it must be larger or equal to 0 and smaller or equal to 360. cos it produces different answers in the end..

    siny=cosx or 2(siny/2)(cosy/2)
    theefore sinx+cosx=sinx+siny=a
     
  13. Nov 18, 2008 #12

    tiny-tim

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    Yup! :biggrin:
    But if x + y = 450, then y = 450 - x, so siny = sin(450-x) = sin(90-x) = cosx …

    multiples of 360 make no difference. :wink:
    uhh? :confused:

    2(siny/2)(cosy/2) = siny whatever x is … how does that help?
     
  14. Nov 18, 2008 #13
    Re: Trigonometry

    Rather doing all those things, I think I could solve it on much easier way:
    If [tex]sinx+siny=a[/tex] then [tex]sinx+\sqrt{1-cos^2y}=a[/tex], from the other equation [tex]cosy=a-cosx[/tex]:

    [tex]sinx+\sqrt{1-(a-cosx)^2}=a[/tex]

    [tex]\sqrt{1-(a^2-2acosx+cos^2x)}=a-sinx[/tex]

    [tex]1-a^2+2acosx-cos^2x=a^2-2asinx+sin^2x[/tex]

    [tex]sin^2x-a^2+2acosx=a^2-2asinx+sin^2x[/tex]

    [tex]-a^2+2acosx=a^2-2asinx[/tex]

    [tex]2acosx+2asinx=2a^2[/tex]

    I think,now, out of here you know the answer. Regards.
     
  15. Nov 18, 2008 #14
    Re: Trigonometry

    thank you guys very much..:smile:
    is there a way to rep you guys or something?
     
  16. Nov 19, 2008 #15

    tiny-tim

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    There's no repping on this forum …

    just be extra kind to the next goldfish you meet! :biggrin:
     
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