Confused about a couple of trig. identities.

[skate_nerd]

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

 

[Dustinsfl]

Re: confused about a couple trig identities.

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.

 

[Prove It]

Re: confused about a couple trig identities.

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

[tex]\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*}[/tex]

where [tex]\displaystyle \begin{align*} A = C\sin{(\phi)} \end{align*}[/tex] and [tex]\displaystyle \begin{align*} B = C\cos{(\phi)} \end{align*}[/tex]. 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.

 

[MarkFL]

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 $\sin(\theta)$

 

[DreamWeaver]

Re: Confused about a couple trig identities.

Hi Skatenerd! :D

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

Spoiler

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

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