MHB Confused about a couple of trig. identities.

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The discussion centers on proving the trigonometric identity Ccos(ω₀t - φ) = Asin(ω₀t) + Bcos(ω₀t), which the original poster finds challenging. They reference the cosine addition formula as a potential tool but struggle to see how to apply it effectively. A participant suggests that the constants A and B can be defined as A = Csin(φ) and B = Ccos(φ), indicating that the goal is to demonstrate the existence of such constants rather than to evaluate them. The conversation also touches on the Auxiliary Angle formula, hinting at alternative methods for solving the problem. Overall, the thread highlights the complexities of applying trigonometric identities in proofs.
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