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Homework Help: Compound Angle Formulas

  1. Apr 24, 2007 #1
    It's a previous exam question using compound angle forumlas. Ive tried everything to try and get the answer involving rearrangments!

    I know the forumlas i just cant get them to work to show this.

    I assume its something to do with adding or subtracting the forumla away from each other, but ive tried it. The furthest i got was this:

    cos (alpha) + (root 3)sin (alpha)

    but i cant get it into the format of what they want

    Here's the question :)

    http://img251.imageshack.us/my.php?image=mathelpsk7.jpg

    Thanks!
     
  2. jcsd
  3. Apr 24, 2007 #2

    VietDao29

    User Avatar
    Homework Helper

    Yup, so far so good. :)

    It goes like this. If you want to combine the expression:
    [tex]a \sin \alpha + b \cos \alpha[/tex] to get some expression with only one sine, or one cos function, you should pull out the factor: [tex]\sqrt{a ^ 2 + b ^ 2}[/tex]

    [tex]a \sin \alpha + b \cos \alpha = \sqrt{a ^ 2 + b ^ 2} \left( \frac{a}{\sqrt{a ^ 2 + b ^ 2}} \sin \alpha + \frac{b}{\sqrt{a ^ 2 + b ^ 2}} \sin \alpha \right)[/tex]

    Now, let [tex]\beta[/tex] be some angle such that:
    [tex]\left\{ \begin{array}{l} \sin \beta = \frac{a}{\sqrt{a ^ 2 + b ^ 2}} \\ \cos \beta = \frac{b}{\sqrt{a ^ 2 + b ^ 2}} \end{array} \right. \quad \mbox{or} \quad \left\{ \begin{array}{l} \sin \beta = \frac{b}{\sqrt{a ^ 2 + b ^ 2}} \\ \cos \beta = \frac{a}{\sqrt{a ^ 2 + b ^ 2}} \end{array} \right.[/tex]

    There will definitely be an angle [tex]\beta[/tex] like that, since, we have:
    [tex]\left| \sin \beta \right| = \left| \frac{a}{\sqrt{a ^ 2 + b ^ 2}} \right| \leq 1[/tex]
    [tex]\left| \cos \beta \right| = \left| \frac{b}{\sqrt{a ^ 2 + b ^ 2}} \right| \leq 1[/tex]
    and
    [tex]\sin ^ 2 \beta + \cos ^ 2 \beta = \frac{a ^ 2}{a ^ 2 + b ^ 2} + \frac{b ^ 2}{a ^ 2 + b ^ 2} = \frac{a ^ 2 + b ^ 2}{a ^ 2 + b ^ 2} = 1[/tex]

    So, we have:
    [tex]a \sin \alpha + b \cos \alpha = \sqrt{a ^ 2 + b ^ 2} \left( \frac{a}{\sqrt{a ^ 2 + b ^ 2}} \sin \alpha + \frac{b}{\sqrt{a ^ 2 + b ^ 2}} \cos \alpha \right) = \left[ \begin{array}{l} \sqrt{a ^ 2 + b ^ 2} (\sin \beta \sin \alpha + \cos \beta \sin \alpha) \\ \sqrt{a ^ 2 + b ^ 2} (\cos \beta \sin \alpha + \sin \beta \cos \alpha) \end{array} \right.[/tex]

    [tex]= \left[ \begin{array}{l} \sqrt{a ^ 2 + b ^ 2} \cos (\alpha - \beta) \\ \sqrt{a ^ 2 + b ^ 2} \sin (\alpha + \beta) \end{array} \right.[/tex]


    -------------------------

    Applying this to your problem, we have:
    [tex]\cos \alpha + \sqrt{3} \sin \alpha[/tex]
    Pull out [tex]\sqrt{1 ^ 2 + (\sqrt{3}) ^ 2} = \sqrt{4} = 2[/tex], we have:
    [tex]... = 2 \left( \frac{1}{2} \cos \alpha + \frac{\sqrt{3}}{2} \sin \alpha \right)[/tex]
    Now, we will try to find such angle [tex]\beta[/tex], we have:
    [tex]\sin \left( \frac{\pi}{6} \right) = \frac{1}{2}[/tex], and
    [tex]\cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}[/tex], so: [tex]\beta = \frac{\pi}{6}[/tex]. Substitute [tex]\beta[/tex] into the expression, yielding:

    [tex]...= 2 \left( \sin \left( \frac{\pi}{6} \right) \cos \alpha + \cos \left( \frac{\pi}{6} \right) \sin \alpha \right) = 2 \sin \left( \alpha + \frac{\pi}{6} \right)[/tex] (Q.E.D)
    Yay, it's done.
    Is it clear?
    Can you get it? :)
     
    Last edited: Apr 24, 2007
  4. Apr 24, 2007 #3
    yes! thankyou :D
     
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