Expressing a combination of sine and cosine as a single cosine Prove that:
$$\dfrac{\sqrt2}2 \cos \omega t - \dfrac{\sqrt2}2 \sin \omega t = \cos \left(\omega t + \dfrac\pi4\right)$$
Obviously, if we are evaluating the right side of the equation, it would a easier. It would only take the following steps:
$cos \left(\omega t + \dfrac\pi4\right) = \cos \omega t\cos\dfrac\pi4 - \sin \omega t\sin\dfrac\pi4 = \dfrac{\sqrt2}2 \cos \omega t - \dfrac{\sqrt2}2 \sin \omega t$ 
Thus, it was proven.
But when I was evaluating the left side of the equation, I was having a hard time. Any idea on how or what approach I should do?
Thanks a lot.

Getting too long for a comment:
For proving an identity such as yours, $$A(t) = B(t)$$it is sufficient to show that  


*

*The left-hand side can be manipulated to equate with the right-hand side, $$A(t) = A^*(t) = A^{**}(t) = \cdots = B(t)$$ 

*OR the right-hand side can be manipulated until it equates with the left-hand side: $$B(t) = B^{*}(t) = B^{**}(t) = \cdots = A(T)$$

*Or each side of the proposed equality can be manipulated until each expression is equivalent: $$A(t) = A^*(t) = A^{**}(t) = B^{**}(t) = B^*(t) = B(t)$$ 


So your work is just fine to show that each side is indeed equivalent. You show equivalence by showing the right-hand side can be manipulated to equate with the left-hand side.
Indeed, as your steps show, you can simply apply them in "reverse order"...i.e., invert the process.
