# TrigonometryConfused about a couple of trig. identities.

#### skatenerd

##### Active member
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

#### dwsmith

##### Well-known member
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

##### Well-known member
MHB Math Helper
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

\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.

#### MarkFL

Staff member
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

##### Well-known member
Re: Confused about a couple trig identities.

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

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

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

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

Auxiliary angle proof