Trig question regarding electrical engineering

[mrjoet]

Hi,

I'm self-studying with an electrical engineering textbook and the authors sometimes jump math steps. Most of the time I can work out the steps in-between but this time I'm really struggling as my knowledge of trig isn't good enough.

my equation so far is:

V_S cos (ωt) = I_S (R cos (ωt + φ) - X_L sin (ωt + φ))

which can apparently be converted to:

V_S cos (ωt) = I_S ([MATH]\sqrt{ }([/MATH]R² + X_L²) cos (ωt + φ + ψ)) where ψ = tan^-1 (X_L / R)

But I have no idea how to do this. Is anyone able to show the steps for this?

Thanks in advance!

 

[Deleted member 4993]

Hi,

I'm self-studying with an electrical engineering textbook and the authors sometimes jump math steps. Most of the time I can work out the steps in-between but this time I'm really struggling as my knowledge of trig isn't good enough.

my equation so far is:

V_S cos (ωt) = I_S (R cos (ωt + φ) - X_L sin (ωt + φ))

which can apparently be converted to:

V_S cos (ωt) = I_S ([MATH]\sqrt{ }([/MATH]R² + X_L²) cos (ωt + φ + ψ)) where ψ = tan^-1 (X_L / R)

But I have no idea how to do this. Is anyone able to show the steps for this?

Thanks in advance!

V_S cos (ωt) = I_S (R cos (ωt + φ) - X_L sin (ωt + φ))

\(\displaystyle \frac{V_s}{I_s}cos (ωt) \ = \ R * cos (ωt + φ) - X_s * sin (ωt + φ)\)

\(\displaystyle \frac{V_s}{I_s}cos (ωt) \ = \ \sqrt{(R^2+X_s^2)} * \left( \frac{R}{\sqrt{(R^2+X_s^2)}} * cos (ωt + φ) - \frac{X_s}{\sqrt{(R^2+X_s^2)}} * sin (ωt + φ)\right)\)

Now, let:

\(\displaystyle \frac{R}{\sqrt{(R^2+X_s^2)}} \ = \cos(u)\) and

\(\displaystyle \frac{X_s}{\sqrt{(R^2+X_s^2)}} \ = \sin(u)\)

continue...

 

[mrjoet]

V_S cos (ωt) = I_S (R cos (ωt + φ) - X_L sin (ωt + φ))

\(\displaystyle \frac{V_s}{I_s}cos (ωt) \ = \ R * cos (ωt + φ) - X_s * sin (ωt + φ)\)

\(\displaystyle \frac{V_s}{I_s}cos (ωt) \ = \ \sqrt{(R^2+X_s^2)} * \left( \frac{R}{\sqrt{(R^2+X_s^2)}} * cos (ωt + φ) - \frac{X_s}{\sqrt{(R^2+X_s^2)}} * sin (ωt + φ)\right)\)

Now, let:

\(\displaystyle \frac{R}{\sqrt{(R^2+X_s^2)}} \ = \cos(u)\) and

\(\displaystyle \frac{X_s}{\sqrt{(R^2+X_s^2)}} \ = \sin(u)\)

continue...

Thanks for your helpful reply (and showing me how to code the symbols correctly )

continued...

\(\displaystyle \frac{V_s}{I_s}cos (ωt) \ = \ \sqrt{(R^2+X_L^2)} * \left( cos (ωt + φ) * cos (u) - sin (ωt + φ) * sin (u)\right)\)

\(\displaystyle \frac{V_s}{I_s}cos (ωt) \ = \ \sqrt{(R^2+X_L^2)} * \left( \frac{cos(ωt + φ + u) + cos(ωt + φ - u)}{2} - \frac{cos(ωt + φ - u) - cos(ωt + φ + u)}{2}\right)\)

\(\displaystyle \frac{V_s}{I_s}cos (ωt) \ = \ \sqrt{(R^2+X_L^2)} * \frac{1}{2} * \left(cos(ωt + φ + u) + cos(ωt + φ - u) - cos(ωt + φ - u) + cos(ωt + φ + u)\right)\)

\(\displaystyle \frac{V_s}{I_s}cos (ωt) \ = \ \sqrt{(R^2+X_L^2)} * \frac{1}{2} * \left(2* cos(ωt + φ + u) \right)\)

\(\displaystyle \frac{V_s}{I_s}cos (ωt) \ = \ \sqrt{(R^2+X_L^2)} * cos(ωt + φ + u)\)

\(\displaystyle \frac{X_L}{\sqrt{(R^2+X_L^2)}} \ = \sin(u)\) and \(\displaystyle \frac{R}{\sqrt{(R^2+X_L^2)}} \ = \cos(u)\)

\(\displaystyle tan(u) \ = \ \frac{sin(u)}{cos(u)} \ = \ \frac{\frac{X_L}{\sqrt{(R^2+X_L^2)}}}{\frac{R}{\sqrt{(R^2+X_L^2)}}} \ = \ \frac{X_L}{R}\)

\(\displaystyle u \ = \ tan^-1 \left(\frac{X_L}{R}\right)\)

\(\displaystyle \frac{V_s}{I_s}cos (ωt) \ = \ \sqrt{(R^2+X_L^2)} * cos(ωt + φ + u)\) where \(\displaystyle u \ = \ tan^-1 \left(\frac{X_L}{R}\right)\)

Have I done that right?

 

[Deleted member 4993]

Thanks for your helpful reply (and showing me how to code the symbols correctly )

continued...

\(\displaystyle \frac{V_s}{I_s}cos (ωt) \ = \ \sqrt{(R^2+X_L^2)} * \left( cos (ωt + φ) * cos (u) - sin (ωt + φ) * sin (u)\right)\)

\(\displaystyle \frac{V_s}{I_s}cos (ωt) \ = \ \sqrt{(R^2+X_L^2)} * \left( \frac{cos(ωt + φ + u) + cos(ωt + φ - u)}{2} - \frac{cos(ωt + φ - u) - cos(ωt + φ + u)}{2}\right)\)

\(\displaystyle \frac{V_s}{I_s}cos (ωt) \ = \ \sqrt{(R^2+X_L^2)} * \frac{1}{2} * \left(cos(ωt + φ + u) + cos(ωt + φ - u) - cos(ωt + φ - u) + cos(ωt + φ + u)\right)\)

\(\displaystyle \frac{V_s}{I_s}cos (ωt) \ = \ \sqrt{(R^2+X_L^2)} * \frac{1}{2} * \left(2* cos(ωt + φ + u) \right)\)

\(\displaystyle \frac{V_s}{I_s}cos (ωt) \ = \ \sqrt{(R^2+X_L^2)} * cos(ωt + φ + u)\)

\(\displaystyle \frac{X_L}{\sqrt{(R^2+X_L^2)}} \ = \sin(u)\) and \(\displaystyle \frac{R}{\sqrt{(R^2+X_L^2)}} \ = \cos(u)\)

\(\displaystyle tan(u) \ = \ \frac{sin(u)}{cos(u)} \ = \ \frac{\frac{X_L}{\sqrt{(R^2+X_L^2)}}}{\frac{R}{\sqrt{(R^2+X_L^2)}}} \ = \ \frac{X_L}{R}\)

\(\displaystyle u \ = \ tan^-1 \left(\frac{X_L}{R}\right)\)

\(\displaystyle \frac{V_s}{I_s}cos (ωt) \ = \ \sqrt{(R^2+X_L^2)} * cos(ωt + φ + u)\) where \(\displaystyle u \ = \ tan^-1 \left(\frac{X_L}{R}\right)\)

Have I done that right?

Looks good to me!