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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?

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    $\begingroup$ 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 $270$? $\endgroup$
    – Gerry Myerson
    Commented Jan 19 at 9:08

1 Answer 1

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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 $360°-150°=210°$.

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