Confused about a couple of trig. identities.

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SUMMARY

The discussion centers on proving the trigonometric identity \(C\cos(\omega_{o}t-\phi)=A\sin(\omega_{o}t)+B\cos(\omega_{o}t)\). Participants reference the cosine addition formula \( \cos(a \pm b) = \cos(a)\cos(b) \mp \sin(a)\sin(b) \) as a key tool for the proof. The constants \(A\) and \(B\) are defined as \(A = C\sin(\phi)\) and \(B = C\cos(\phi)\), indicating that the identity can be satisfied by appropriate choices of \(A\) and \(B\). The discussion emphasizes the importance of understanding these relationships in trigonometric identities.

PREREQUISITES
  • Understanding of trigonometric identities, specifically the cosine addition formula.
  • Familiarity with the concepts of sine and cosine functions.
  • Knowledge of the Auxiliary Angle method in trigonometry.
  • Basic algebraic manipulation skills to rearrange and equate trigonometric expressions.
NEXT STEPS
  • Study the Auxiliary Angle formula and its applications in trigonometric proofs.
  • Practice deriving and manipulating trigonometric identities using the cosine addition formula.
  • Explore the relationship between sine and cosine functions through graphical representations.
  • Learn how to apply trigonometric identities in physics, particularly in classical mechanics.
USEFUL FOR

Students of mathematics, particularly those studying trigonometry and its applications in physics, as well as educators looking for insights into teaching trigonometric identities effectively.

skate_nerd
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I've got this problem right now, which asks me to prove that
$$Ccos(\omega_{o}t-\phi)=Asin(\omega_{o}t)+Bcos(\omega_{o}t)$$
This proved to be a bit more difficult than I expected, so I looked up a complete list of trig identities.
$$cos(a\pm{b})=cos(a)cos(b)\mp{sin(a)sin(b)}$$
seems like the only one that could be helpful in my situation, however when I try to think of a way where the original equation makes sense, I am really not able to convince myself.
Wouldn't the only way for the original equation to make sense be if
\(sin(\phi)=cos(\phi)=1\)? As far as I know this is only possible for \(\frac{-3\pi}{4}\).
Kind of stuck here, any help would be very appreciated
 
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Re: confused about a couple trig identities.

skatenerd said:
I've got this problem right now, which asks me to prove that
$$Ccos(\omega_{o}t-\phi)=Asin(\omega_{o}t)+Bcos(\omega_{o}t)$$
This proved to be a bit more difficult than I expected, so I looked up a complete list of trig identities.
$$cos(a\pm{b})=cos(a)cos(b)\mp{sin(a)sin(b)}$$
seems like the only one that could be helpful in my situation, however when I try to think of a way where the original equation makes sense, I am really not able to convince myself.
Wouldn't the only way for the original equation to make sense be if
\(sin(\phi)=cos(\phi)=1\)? As far as I know this is only possible for \(\frac{-3\pi}{4}\).
Kind of stuck here, any help would be very appreciated

I am 90% positive I have shown this in my classical mechanics notes in the forum's notes section. They aren't completed so it should be easy to find.
 
Re: confused about a couple trig identities.

skatenerd said:
I've got this problem right now, which asks me to prove that
$$Ccos(\omega_{o}t-\phi)=Asin(\omega_{o}t)+Bcos(\omega_{o}t)$$
This proved to be a bit more difficult than I expected, so I looked up a complete list of trig identities.
$$cos(a\pm{b})=cos(a)cos(b)\mp{sin(a)sin(b)}$$
seems like the only one that could be helpful in my situation, however when I try to think of a way where the original equation makes sense, I am really not able to convince myself.
Wouldn't the only way for the original equation to make sense be if
\(sin(\phi)=cos(\phi)=1\)? As far as I know this is only possible for \(\frac{-3\pi}{4}\).
Kind of stuck here, any help would be very appreciated

I agree with what you are thinking. Using the identity you have been given

\displaystyle \begin{align*} C\cos{ \left( \omega_ot - \phi \right)} &= C \left[ \cos{ \left( \omega_ot \right)}\cos{ \left( \phi \right) } + \sin{\left( \omega_ot \right) } \sin{\left( \phi \right) } \right] \\ &= C\sin{\left( \phi \right) } \sin{\left( \omega_ot \right) } + C\cos{\left( \phi \right)} \cos{\left( \omega_ot \right) } \\ &= A\sin{\left( \omega_ot \right) } + B \sin{ \left( \omega_ot \right) } \end{align*}

where \displaystyle \begin{align*} A = C\sin{(\phi)} \end{align*} and \displaystyle \begin{align*} B = C\cos{(\phi)} \end{align*}. Here you're not expected to be able to evaluate these constants, you're just expected to show that there ARE some constants that exist which would satisfy your identity.
 
Re: confused about a couple trig identities.

$\LaTeX$ tip:

Precede trigonometric/logarithmic functions with a backslash so that their names are not italicized as if they are strings of variables. For example:

sin(\theta) produces $sin(\theta)$

\sin(\theta) produces $\sin(\theta)$
 
Re: Confused about a couple trig identities.

Hi Skatenerd! :D

Have you come across the Auxiliary Angle formulae - from trigonometry - before...?

$$A\cos x+ B\sin x=\sqrt{A^2+B^2}\cos(x\pm\varphi)\quad ; \, \varphi=\tan^{-1}\left(\mp \frac{B}{A}\right) $$$$A\cos x+ B\sin x=\sqrt{A^2+B^2}\sin(x\pm\varphi)\quad ; \, \varphi=\tan^{-1}\left(\pm \frac{A}{B}\right) $$

Incidentally, I just posted a thread in the Puzzles Board that might be of interest...

http://mathhelpboards.com/challenge-questions-puzzles-28/auxiliary-angle-proof-6759-new.html
 

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