# Confusion in finding phase difference of two sinusoids.

If two sinusoids are given that,

$$v_1(t)=-a\text{sin}(\omega t +55°)$$

$$v_2(t)=b\text{cos}(\omega t -65°)$$

Then what is the phase difference between them?

Method -1

$$v_1(t)=-a\text{sin}(\omega t +55°) \\ =a\text{sin}((\omega t +55°)-180°) \\ =a\text{sin}(\omega t -125°) \\$$ $$v_2(t)=b\text{cos}(\omega t -65°) \\ =b\text{sin}(90°+(\omega t -65°)) \\ =b\text{sin}(\omega t +25°) \\$$

So, phase difference $$= 25°-(-125°)= \bbox[5px,border:2px solid black]{150°}$$

Method -2

$$v_1(t)=-a\text{sin}(\omega t +55°) \\ =a\text{cos}((\omega t +55°)+90°) \\ =a\text{cos}(\omega t +145°) \\$$ $$v_2(t)=b\text{cos}(\omega t -65°)$$

So, phase difference $$= 145°-(-65°)= \bbox[5px,border:2px solid black]{210°}$$

Which one one is correct and why, which one should be granted and what is the actual process that should be followed to get the correct answer?

• The phase difference between $v_1$ and $v_2$ is $150$. The phase difference between $v_2$ and $v_1$ is $210$. Both correct, just interpreting "phase difference between them" two different ways. If not convinced, think of simplest case, $\sin x$ and $\cos x$. Is the phase difference $90$, or $270$? Commented Jan 19 at 9:08

As noticed in the comments there is not difference indeed $$150° + 210° = 360°$$ and in the first case you are considering the phase difference between $$v_2$$ and $$v_1$$ that is $$150°$$ while in the second case you are considering the phase difference between $$v_1$$ and $$v_2$$ that is $$210°$$.
Note also that for the method $$1$$, taking the phase difference between $$v_1$$ and $$v_2$$ we have: $$-125°-25° = -150°$$ and $$360°-150°=210°$$.