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Homework Help: A cos theta + b sin theta

  1. Sep 8, 2006 #1
    Hey,
    I’ve got a test in one week’s time and was studying through my text book of geometry and trigonometry. I came across a “rule” which shows how to simplify expressions in the form of [tex]a \cos \theta + b \sin \theta[/tex] but I do not understand how this “rule” works.
    The simplify rule:

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

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

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

    where [tex]\alpha[/tex] is an acute angle

    In the text book this is explained using a right angle triangle where [tex]\alpha[/tex] is the unknown angle being measured, side a is the adjacent side and side b is the opposite, therefore the hypotenuse is equal to [tex]sqrt{a^2+b^2}[/tex]. However I do not understand this method which the book uses to explain, and was wondering if somebody out there knew how to explain/prove how/where this “rule” has come from. Thank you for any legitimate reply,
    Pavadrin

    EDIT: those latex code isn't wokring properly so ive attached this image link:
    PF1.JPG
     
    Last edited by a moderator: Sep 8, 2006
  2. jcsd
  3. Sep 8, 2006 #2

    J77

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  4. Sep 8, 2006 #3
    okay thanks for the link J77
     
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