# Find the phase angle between two functions

Apologies if this is better suited for a mathematics forum, it's from my circuits textbook so I've decided to post here. Below is a practice problem from the textbook, with the provided solution.

If I try to convert i1 to a positive cosine function, I get 4cos(377t+140) by shifting -90 (converting to cos) and then shifting +180 (negating the amplitude). So I am then comparing i1=4cos(377t+140) and i2=5cos(377t-65), so a 205 degree phase angle between the functions, where 210 is the stated answer. Is this a mistake in the textbook? I spend hundreds for textbooks and I've spent more time verifying errors than reading... :( oh well

Additionally, it seems to me which function leads which is rather relative as I can shift any function 360 deg, any easy reconciliation?

• You don't need to add "Not homework" to your title. Homework is fine as long as you post your work, which you have.
– Drew
Jan 23, 2021 at 20:40
• Which textbook? So that others can avoid it...
– user16324
Jan 23, 2021 at 20:41
• Fundamentals of Electric Circuits 7e, Alexander and Sadiku. I should say that I have just started reading my first chapter, which included only 2 errors, but the result has been spending over an hour trying to verify the errors vs 20 min reading. It's very possible the text is otherwise well written Jan 23, 2021 at 20:47
• I would suggest learning LaTeX, which enables clean formatting for mathematical notation. Jan 24, 2021 at 5:08
• I'll definitely check it out. I was wondering how to format properly as I was typing this out. Jan 24, 2021 at 15:18

Additionally, it seems to me which function leads which is rather relative as I can shift any function 360 deg, any easy reconciliation?

Absolutely. What is leading or lagging is up for interpretation/definition. Personally, the only definition that would make sense to me would be that any lag by more than 180° is actually a lead. Maybe that book defines it otherwise, or it's plain inconsistent/underdefined there.

Re: your angle: Well, no matter what happens, the result needs to have a "last digit" of 5°, so we can directly conclude that the book's solution must be incorrect.

Let me try your calculation: $$\-\sin(x)=\cos(x+90°)\implies i_1=-\sin(337t+50°)=\cos(377t +140°)\$$, and $$\140 - (-65) = 205\$$, so I'd agree with you.

By the way, you say you've been fighting a lot of mistakes in this book. Note them down as errata (even in rough form, just scans/readable photos of paper or similar), and post them in your year's forum (or however you and your coursemates communicate).
You will a) earn eternal gratitude of peers through little extra work, b) if you did actually do something wrong but the book got it right, you get free corrections, and c) at the end of the whole circus, you drop an email to your prof and tell her or him that hey, you went through a bit of effort, whether they might be willing to share these errata which the course looked through with next semester's students. That might lead to you being asked to type them out for money, or to the professor more intensely reviewing course material.