<s>[INST] 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.
 [/INST] If you want to prove it as if you didn't know what it actually should be equal to, then here what people usually do. Assume, you have the expression $$a \cos x + b \sin x$$ and you want to find what it is equal to. The very first step is to notice the if you pull some factor outside of the brackets, so the whole expression becomes $$c (a_1 \cos x + b_1 \sin x)$$, so that $$a_1^2 + b_1^2 = 1$$ and $$0 \le |a_1|, |b_1| \le 1,$$ i.e. you can substitute $a_1 = \cos t$ and $b_1 = \sin t$.
$$
a_1 = \frac ac; \qquad b_1 = \frac bc
$$
and now just use condition above
$$
\left ( \frac ac \right )^2 + \left (\frac bc \right )^2 = 1
$$
from which you can find that the factor is
$$
c = \sqrt{a^2 + b^2}
$$
and therefore
$$
a \cos x + b \sin x = \sqrt{a^2 + b^2} \left ( \frac a{\sqrt{a^2 + b^2}} \cos x + \frac b{\sqrt{a^2 + b^2}} \sin x\right ) = \\
= \sqrt{a^2 + b^2} \left ( \cos t \cos x + \sin t \sin x \right ) = \sqrt{a^2 + b^2} \cos (x - t)
$$
where
$$
\cos t = \frac a{\sqrt{a^2 + b^2}} \\
\sin t = \frac b{\sqrt{a^2 + b^2}}
$$
Now, just substitute your numbers.
$$
c = \sqrt{\frac 12 + \frac 12} = 1
$$
so no need to factor anything out, since coefficients already make unity. So, all you need to do is finding some angle that corresponds to those coefficients, namely
$$
\sin t = \frac {\sqrt 2}2 \\
\cos t = -\frac {\sqrt 2}2
$$
you can easily see, that $t = -\frac \pi 4$ is one of those angles. At the same time, you could've taken any angle of the form $-\frac \pi 4 + 2 \pi n$.
So, angle is found, then substitute it to the final answer
$$
\frac {\sqrt 2}2 \cos \omega t - \frac {\sqrt 2}2 \sin \omega t = \cos \left ( \omega t - \left (-\frac \pi 4 \right )\right ) = \cos \left ( \omega t + \frac \pi 4 \right )
$$
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