Finding Phase Shift in Trigonometric Equations

AI Thread Summary
To find the phase shift in the trigonometric equations, the in-phase term should be divided by the out-of-phase term. The phase shift can be determined using the arctan function of the ratio of these terms. The user is struggling to manipulate the formula to achieve the arctan representation. A proper approach involves expressing the equations in a form that highlights the sine and cosine components. Clarification on these manipulations is needed to progress towards finding the phase shift.
Fluidman117
Messages
34
Reaction score
0
Hello,

Probably a simple problem, but I am not able to figure it out.

a \cos (\epsilon) - b \sin (\epsilon) = c in-phase part
a \sin (\epsilon) - b \cos (\epsilon) = d out-of-phase part

In order to find the phase shift, the in-phase term has to be divided by the out-of-phase term?

\frac{a \cos (\epsilon) - b \sin (\epsilon) = c}{a \sin (\epsilon) - b \cos (\epsilon) = d}

And the phase shift is the arctan of the out of phase and the in-phase term to my knowledge. But I am not able to manipulate the formula in a way that I'll get to an arctan. Can someone point me in the right direction?

Thanks
 
Mathematics news on Phys.org
$$ a \cos x + b \sin x = \sqrt {a^2 + b^2 } ( \frac a {\sqrt {a^2 + b^2} } \cos x + b \frac b {\sqrt {a^2 + b^2} } \sin x)
\\ = \sqrt {a^2 + b^2 } ( \sin c \cos x + \cos c \sin x) = \ ...
$$
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Back
Top