1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A cos theta + b sin theta

  1. Sep 8, 2006 #1
    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,

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


    User Avatar

  4. Sep 8, 2006 #3
    okay thanks for the link J77
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: A cos theta + b sin theta
  1. Cos B (Replies: 19)