1. Limited time only! Sign up for a free 30min personal 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!

Alternative deduction of sum of sine and cosine

  1. May 26, 2014 #1
    Hi!
    Many students know that [itex]A\sin(x) + B\cos(x) =\sqrt{A^2+B^2} \sin{(x+\arctan \frac{B}{A})}[/itex]. I have seen just one deduction of that relation, showed by set up a system of two equations, solving for amplitude and phase shift.

    Is it possible to deduce the relation in a vectorial way, or in any way including complex numbers? The cosine part on the x-axis and the sine-part on the y-axis, the resultant vector should be the sum of the given trigonometric functions. But the magnitude of that resultant vector becomes [itex]\sqrt{(A\sin{x})^2+(B\cos{x})^2}[/itex], which is not equal to the actual magnitude of the sum.

    What am I missing in this kind of deduction? Or maybe it's not possible?
     
  2. jcsd
  3. May 26, 2014 #2
    I can prove it backward using complex numbers. The Euler formula says [itex]\sin x=\frac{e^{ix}-e^{-ix}}{2i}[/itex] and [itex]\cos x=\frac{e^{ix}+e^{-ix}}{2}[/itex]. Moreover you have [itex]\arctan x=\frac{1}{2}i\left[\log(1-ix)-\log(1+ix)\right][/itex]. Then in your case:
    \begin{align*}
    \sin(x+\arctan(B/A))=&\frac{e^{ix}e^{-\frac{1}{2}\log(1-B/A)}e^{\frac{1}{2}\log(1+iB/A)}-e^{-ix}e^{\frac{1}{2}\log(1-B/A)}e^{-\frac{1}{2}\log(1+iB/A)}}{2i} \\
    =&\frac{e^{ix}\sqrt{\frac{1+iB/A}{1-iB/A}}-e^{-ix}\sqrt{\frac{1-iB/A}{1+iB/A}}}{2i}=\frac{e^{ix}\left(\frac{1+iB/A}{\sqrt{1+B^2/A^2}}\right)-e^{-ix}\left(\frac{1-iB/A}{\sqrt{1+B^2/A^2}}\right)}{2i} \\
    =&\frac{e^{ix}(1+iB/A)-e^{-ix}(1-iB/A)}{2i}\frac{A}{\sqrt{A^2+B^2}}=\frac{A(e^{ix}-e^{-ix})+iB(e^{ix}+e^{-ix})}{2i}\frac{1}{\sqrt{A^2+B^2}} \\
    =&\left(A\sin x+B\cos x\right)\frac{1}{\sqrt{A^2+B^2}}.
    \end{align*}

    I hope this is clear.
     
  4. May 27, 2014 #3

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    You can show it geometrically on an Argand diagram. ##\sin x = \cos(x - \pi/2)##, so start with the equivalent expression, the real part of ##Ae^{i(x-\pi/2)} + Be^{ix}##.

    If you draw ##Ae^{i(x-\pi/2)}## and ##Be^{ix}## on an Argand diagram, for an arbitrary choice of ##x##, you have a right angled triangle with sides ##A## and ##B##. The result follows from Pythagoras's theorem and simple trig.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Alternative deduction of sum of sine and cosine
  1. Sine and cosine (Replies: 4)

  2. Sine and cosine (Replies: 5)

Loading...