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Y = Sin(x) + Cos(x) Transpose

  1. Jun 15, 2009 #1
    Hi Guys, Simple question; I'm trying to work out the transpose of Y = Sin(x) + Cos(x) to make x the subject. I thought it would be x = arccos(arcsin(y)) / 2 however I don't think thats right. Is there another theorem I'm missing?
     
  2. jcsd
  3. Jun 15, 2009 #2
    what's [tex]y^2[/tex]?
     
  4. Jun 15, 2009 #3
    that is meant to be divided by 2
     
  5. Jun 15, 2009 #4
    [tex]y = sin(x) + cos(x)[/tex]

    [tex]\Rightarrow y^2 = (sin(x) + cos(x))^2[/tex]

    [tex]\Rightarrow y^2 = sin^2(x) + 2sin(x)cos(x) + cos^2(x)[/tex]

    [tex]\Rightarrow y^2 = \left(sin^2(x) + cos^2(x)\right) + 2sin(x)cos(x)[/tex]

    [tex]\Rightarrow y^2 = 1 + 2sin(x)cos(x)[/tex]

    [tex]\Rightarrow y^2 - 1 = 2sin(x)cos(x)[/tex]

    [tex]\Rightarrow y^2 - 1 = sin(2x)[/tex]

    [tex]\Rightarrow 2x = \sin^{-1 }(y^2 - 1)[/tex]

    [tex]\Rightarrow x = \frac { \sin^{-1 } (y^2 - 1) } {2} [/tex]
     
    Last edited: Jun 15, 2009
  6. Jun 15, 2009 #5
    Thanks very much for that. Anyone looking for the rules for this, I've found them on http://math2.org/math/trig/identities.htm

    sin(2x) = 2 sin x cos x

    sin^2(x) = 1/2 - 1/2 cos(2x)

    cos^2(x) = 1/2 + 1/2 cos(2x)
     
  7. Jun 15, 2009 #6
    [tex]sin^2(x) + cos^2(x) = 1[/tex] should be one of the first identities you learn.
     
  8. Jun 15, 2009 #7

    Mute

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    You can put it into another form by noting:

    [tex]\sin x + \cos x = \sqrt{2}\left[\sin x \left(\frac{1}{\sqrt{2}}\right) + \cos x \left(\frac{1}{\sqrt{2}} \right)\right] = \sqrt{2}\left[\sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} \right] = \sqrt{2}\sin \left(x+\frac{\pi}{4}\right)[/tex].

    So, y is also equal to this expression and you can solve for x. Since you don't seem to be familiar with many trig identities, I used the sine "angle sum formula", [itex]\sin(a \pm b) = \sin a \cos b \pm \cos a \sin b[/itex].
     
    Last edited: Jun 15, 2009
  9. Aug 8, 2009 #8
    Suppose I was to change the equation to Y = a sin(x) + b cos(x). How would I transpose it to make x the subject? I've searched through identities on that site but can't find one that relates to having 2 different coefficients a and b.
     
  10. Aug 8, 2009 #9

    Hurkyl

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    Think hard about how Mute's example works....
     
  11. Aug 8, 2009 #10
    A bit more searching I've come across this:

    a sin x + b cos x = R sin (x + alpha) which is an R Formulae with R = Sqrt(a^2 + b^2) and alpha = atan(b / a). If I use that I should be able to solve for x.

    Having a look at Mute's post, what should I be looking for? It's been a few years since I left school so most of this is going back to that. I'd rather know how this thing works than just the answer so I'll keep looking.
     
  12. Aug 8, 2009 #11

    Hurkyl

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    To derive the formula you just found by searching -- his calculation is the derivation of that formula, just in a special case.
     
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