At 10:15 you say that we can ignore the +/- factors because cosine is even. But the phase-shifted cosine function that we have assumed is NOT necessarily even. Have I missed something?
You're right that cos(ωt+φ) is not necessarily an even function of t, but it's still an even function of its argument as a whole. This means that cos(-ωt+φ) = cos(ωt-φ). You could then let ψ = -φ and write it as cos(ωt+ψ). Since the phase is arbitrary (to be determined by initial conditions), cos(ωt+φ) and cos(ωt+ψ) are completely equivalent solutions and the sign of ω therefore doesn't matter.
Great video sir keep uploading.
Thanks! Will do.
Thank you, sir.
At 10:15 you say that we can ignore the +/- factors because cosine is even. But the phase-shifted cosine function that we have assumed is NOT necessarily even. Have I missed something?
You're right that cos(ωt+φ) is not necessarily an even function of t, but it's still an even function of its argument as a whole. This means that cos(-ωt+φ) = cos(ωt-φ). You could then let ψ = -φ and write it as cos(ωt+ψ). Since the phase is arbitrary (to be determined by initial conditions), cos(ωt+φ) and cos(ωt+ψ) are completely equivalent solutions and the sign of ω therefore doesn't matter.
good video!!
Thanks for watching, I'm glad you enjoyed it!