- #1
Lunat1c
- 66
- 0
Hi,
Can someone please show me how the 2nd equation shown below is arrived to from the first one?
http://img263.imageshack.us/img263/126/voltages.jpg
I started with:
[tex] V_R = \frac{\sqrt(2)E}{\pi} \Bigg[\bigg(-cos(\omega t)\bigg)_{\alpha}^{\beta} - sin(\beta)\omega CR\bigg(exp\bigg({-\frac{\omega t - \beta}{\omega CR}}\bigg)_{\beta}^{\pi+\alpha}\Bigg] [/tex]
[tex] V_R = \frac{\sqrt(2)E}{\pi} \Bigg[-cos(\beta)+cos(\alpha) - sin(\beta)\omega CR\bigg(exp\bigg({-\frac{\pi+\alpha-\beta}{\omega CR}}\bigg) - 1\bigg)\Bigg] [/tex]
I can't figure out how to continue from there.. there must be some kind of identity that I can use which I'm not familiar with
Can someone please show me how the 2nd equation shown below is arrived to from the first one?
http://img263.imageshack.us/img263/126/voltages.jpg
I started with:
[tex] V_R = \frac{\sqrt(2)E}{\pi} \Bigg[\bigg(-cos(\omega t)\bigg)_{\alpha}^{\beta} - sin(\beta)\omega CR\bigg(exp\bigg({-\frac{\omega t - \beta}{\omega CR}}\bigg)_{\beta}^{\pi+\alpha}\Bigg] [/tex]
[tex] V_R = \frac{\sqrt(2)E}{\pi} \Bigg[-cos(\beta)+cos(\alpha) - sin(\beta)\omega CR\bigg(exp\bigg({-\frac{\pi+\alpha-\beta}{\omega CR}}\bigg) - 1\bigg)\Bigg] [/tex]
I can't figure out how to continue from there.. there must be some kind of identity that I can use which I'm not familiar with
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