Another trig question relate to retarded potential

[yungman]

This is from Griffiths page 446.

In radiating dipoles:

$$V(\vec r,t)=\frac 1 {4\pi \epsilon_0} \left [ \frac {q_0 cos [\omega(t- \frac {\eta_+} c )]}{\eta_+}- \frac {q_0 cos [\omega(t- \frac {\eta_- } c)]}{\eta_-} \right ]$$

Given conditions d<< $\eta\;$ and d<< $\frac c {\omega}$ :

$$V_{(\eta,\theta,t)} = \frac {q_0 d cos \theta}{4\pi \epsilon_0 r} \left [ -\frac {\omega}{c} sin[\omega(t-\frac {\eta}{c}]+\frac 1 {\eta} cos[\omega(t-\frac {\eta}{c}]\right ]$$

But then the book claimed if $\eta$ >> $\frac c {\omega}\;$, then:$$V_{(\eta,\theta,t)} = \frac {q_0 d cos \theta}{4\pi \epsilon_0 r} \left [ -\frac {\omega}{c} sin[ \omega(t-\frac {\eta}{c} ] \right ] = -\frac {q_0\; d\;\omega\; cos \theta}{4\pi \epsilon_0 c\; r} sin[ \omega(t-\frac {\eta}{c} ]$$I don't see why if $\eta$ >> $\frac c {\omega}\;$, then

$$\frac 1 {\eta} cos[\omega(t-\frac {\eta}{c}] = 0$$

It look so simple but I just don't see it. Please explain to me.

Thanks

Alan

 

[yungman]

I think I have the answer, it is very simple if I am correct:

$$V_{(\eta,\theta,t)} = \frac {q_0 d cos \theta}{4\pi \epsilon_0 r} \left [ -\frac {\omega}{c} sin[\omega(t-\frac {\eta}{c}]+\frac 1 {\eta} cos[\omega(t-\frac {\eta}{c}]\right ]$$Since both the sine and cosine max out at +/-1, so if $\eta$ >> $\frac c {\omega}\;$, then The first term with the sine function is much larger than the second term with cosine term. So the second term just disappeared. Tell me whether I am correct. It's just that simple!