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Trig equations in the form

  1. Aug 24, 2006 #1

    danago

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    Gold Member

    Hey. The latest thing we've been studying in trig classes is equations of the form:

    [tex]
    a\cos \theta + b\sin \theta = c
    [/tex]

    The book explains that to solve an equation like this, i should write it in the form:

    [tex]
    a\cos \theta + b\sin \theta = \sqrt {a^2 + b^2 } \cos (\theta - \alpha )
    [/tex]

    Now. The teacher has tried all she can to explain how the second form can be derived, but nobody seems to understand. Could someone please go through it?

    Thanks,
    Dan.
     
  2. jcsd
  3. Aug 24, 2006 #2

    benorin

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    Homework Helper

    It seems easier to solve such equations this way:

    [tex]a\cos \theta + b\sin \theta = c[/tex]

    recall that the Pythagorean identity gives [tex]\sin\theta =\pm\sqrt{1-\cos^{2}\theta}[/tex] so substitute this in to get

    [tex]a\cos \theta \pm b\sqrt{1-\cos^{2}\theta} = c[/tex]

    rearrange to get

    [tex] \pm b\sqrt{1-\cos^{2}\theta} = c-a\cos \theta [/tex]

    square both sides

    [tex] b^2(1-\cos^{2}\theta) = c^2-2ac\cos \theta+a^2\cos^{2}\theta [/tex]

    cleaning it up gives

    [tex](a^2+b^2)\cos^{2}\theta -2ac\cos\theta+c^2-b^2=0[/tex]

    which is a quadratic equation in [tex]\cos\theta[/tex] so use the quadratic formula to solve for

    [tex]\cos\theta = \frac{2ac\pm\sqrt{4a^2c^2-4(a^2+b^2)(c^2-b^2)}}{2(a^2+b^2)}= \frac{ac\pm b\sqrt{a^2+b^2-c^2}}{a^2+b^2}[/tex]

    which is messy, but it works.
     
  4. Aug 24, 2006 #3

    Integral

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    Staff Emeritus
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    Gold Member

    Dango,
    I have 2 questions:
    What happened to the c in the first equation?

    Where did the alpha in the second come from?
     
  5. Aug 24, 2006 #4
    The idea behind this rewriting is that by you can always make [tex]\sqrt {a^2 + b^2 } \cos (\theta - \alpha )[/tex] equal to [tex]a\cos \theta + b\sin \theta[/tex] by choosing [tex]\alpha[/tex] correctly. We can see this by considering the angle addition formula for cosines:

    [tex]\cos (\theta + \phi) = \cos \theta \cos \phi - \sin \theta \sin \phi[/tex]

    If we replace [tex]\phi[/tex] with [tex]-\alpha[/tex], we can expand the right hand side of the equation and find the value of [tex]\alpha[/tex] that makes the equation true (i.e. that makes the coefficients of [tex]\sin \theta[/tex] and [tex]\cos \theta[/tex] correct).
     
  6. Sep 11, 2006 #5
    [tex]\alpha[/tex] is an obtuse angle which is equal to:

    [tex]
    {\tan ^{ - 1} \left( { - \frac{b}{a}} \right)}
    [/tex]

    as for the rule maybe this link will help you http://mathworld.wolfram.com/HarmonicAdditionTheorem.html
    (credit for this link is to be given to J77)

    this rule is reffered to as the harmonic addition formula

    hope this helps
     
  7. Sep 11, 2006 #6

    HallsofIvy

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    sin(a- b)= sin(a)cos(b)- cos(a)sin(b).

    If you are given Asin(x)- Bcos(x), you can take a= x and then try to find b such that sin(b)= A, cos(b)= B. Of course that will only work if A and B are between -1 and 1 and A2+ B2= 1. Okay, for general A and B, factor out something so that is true: write A sin(x)- B cos(x) as
    [tex]\sqrt{A^2+ B^2}(\frac{A}{\sqrt{A^2+ B^2}}cos(x)- \frac{B}{\sqrt{A^2+ B^2}}sin(x)[/tex]
    Then if [itex]sin(b)= \frac{A}{\sqrt{A^2+ B^2}}[/itex], it follows that [itex]cos(b)= \frac{B}{\sqrt{A^2+ B^2}}.
     
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