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.
 A: 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.
A: 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
$$
\frac {\sqrt 2}2 \cos \omega t - \frac {\sqrt 2}2 \sin \omega t = \cos \left ( \omega t - \left (-\frac \pi 4 \right )\right ) = \cos \left ( \omega t + \frac \pi 4 \right )
$$
A: $$\dfrac{\sqrt2}2 \cos \omega t - \dfrac{\sqrt2}2 \sin \omega t =\frac1{\sqrt2}\left(\cos\omega t -\sin  \omega t \right)$$
Let $1=r\cos A, 1=r\sin A$ where $r\ge 0$
So that $$\cos\omega t -\sin  \omega t=r\cos A\cos\omega t -r\sin A\sin  \omega t=r\cos(\omega t+A)$$
Now, $(r\cos A)^2+(r\sin A)^2=1+1\implies r^2=2\implies r=\sqrt2$ as $r\ge0$
$\implies \cos A=\sin A=\frac1{\sqrt2}\implies A=2n\pi+\frac{\pi}4$ where $n$ is any integer
So, the principal value of $A$ will be $\frac{\pi}4$
Can you take it from here?
A: There is a more general statement:

PROP For any $x,y\in\Bbb R$, it holds that $$\tag 1\cos(x+y)=\cos x \cos y-\sin x\sin y$$

Then let $x=\omega t,y=\dfrac{\pi }4$ and note that $\cos y=\sin y=\dfrac{\sqrt 2}2$
To prove $(1)$, fix $y\in \Bbb R$ and consider the function $$f(x)=\cos(x+y)-\cos x \cos y+\sin x\sin y$$ 
Note that $f(0)=0$. Moreover, $$f'(x)= - \sin (x + y) + \sin x\cos y + \cos x\sin y$$
so $f'(0)=0$. Finally $$f''\left( x \right) =  - \cos (x + y) + \cos x\cos y - \sin x\sin y$$
so $f''+f=0$. But this means $$f'f''+ff'=0$$ $$(f')^2+f^2=\rm const$$
Since $f(0)=f'(0)=0$, we get $$(f')^2+f^2=0$$
But this is a sum of non-negative terms, which means $f$ vanishes identically. Since $y$ was arbitrary, we have proven the proposition.
A: Here is my answer to the question guys. This is when evaluating the left side of the equation.
We were given,
$$\dfrac{\sqrt2}2 \cos \omega t - \dfrac{\sqrt2}2 \sin \omega t = \cos \left(\omega t + \dfrac\pi4\right)$$
Then,
$$\dfrac{\sqrt2}2 (\cos \omega t - \sin \omega t) = \cos \left(\omega t + \dfrac\pi4\right)$$
Recall that the sum of the cosine and sine of the same angle is given by:
$$cos x \pm sin x = \sqrt2 sin (\dfrac\pi4 \pm x )$$
Thus,
$$\dfrac{\sqrt2}2(\sqrt2 sin (\dfrac\pi4 - wt )) = \cos \left(\omega t + \dfrac\pi4\right)$$
$$ sin (\dfrac\pi4 - wt ) = \cos \left(\omega t + \dfrac\pi4\right)$$
Recall that, $sin(A - B)$ can be alternatively represented as $cos (-A + b + \dfrac\pi2) $. Therefore,
$$cos (-\dfrac\pi4 + wt + \dfrac\pi2) = = \cos \left(\omega t + \dfrac\pi4\right)$$
$$\cos (\omega t + \dfrac\pi4)= \cos \left(\omega t + \dfrac\pi4\right)$$
QED
