Phases difference

Question

Recently, I learned in my class about harmonic motion and the difference in phase. According to the wikipedia and many other sources, you find that difference by subtracting the phases. Take a look at this:

$4cos(20t+10)$

$-4 \cdot 20sin(20t + 10)$

$-4 \cdot20^2cos(20t + 10)$

By changing the sine to cosine, the second one is $\pi /2$ different from the first, and the third one is a $\pi$ different. But it occurred to me that why would you need to change the sine to cosine, or minus cosine to cosine? Isn't only subtracting the phase enough, and whatever number before it doesn't matter?

Answer 1

$-\sin(x)=\sin(x \pm \pi)$, likewise with $\cos$.

Hence multiplying by $-1$ has the effect of introducing a phase difference of $\pi$

Answer 2

You don't really need to change $sin$ by $cosine$ or whatever. As you say, you can just write $\cos (\varphi \pm \pi)$. Sometimes it's just about lazyness, as writing $\sin$ is shorter than adding another factor to the phase. In many other cases, it is because derivatives are learned like that. If you calculate the derivative of those, sines and cosines are switched (with some minus signs). That's why both forms are used, and you must eb able to notice that the onl difference ni phase is a $\pi$ term.