Did I actually say that? :)  The general result related to this is:

$$-\sqrt{a^2+b^2}\leq a\cos\theta + b\sin\theta \leq \sqrt{a^2+b^2}.$$

To show this is true, draw a right triangle with $a$ and $b$ as its two legs of the right angle, and then the hypotenuse is $\sqrt{a^2+b^2}$.  Let $\phi$ be the angle opposite to $a$, then $\sin\phi =\dfrac{a}{\sqrt{a^2+b^2}}$ and $\cos\phi=\dfrac{b}{\sqrt{a^2+b^2}}$.  Thus

$$a\cos\theta + b\sin\theta = \sqrt{a^2+b^2}(\sin\phi\cos\theta+\cos\phi\sin\theta) = \sqrt{a^2+b^2}\sin(\phi+\theta).$$

So the inequality you showed above should have $-\sqrt{20}$ and $\sqrt{20}$ as bounds.  If I really said $-20$ and $20$, it was wrong (technically not wrong, just not helpful to solving the problem).

I am going to answer the other questions, just need to find the time.  Sorry about the delay.