If track was 360 deg and wind was 030/30 & TAS = 100 kts
Sine 30 = .5 Co sine = .86
Is that ground speed 86 knots?
No it is not.
The track is 360 and the wind is coming from the right (030 degrees). So the aircraft must be heading to the right such that the Easterly component of 100 kts TAS is equal to the westerly component of a 30 knots wind coming from 030.
If we use X for the aircraft heading we get.
100 kts x Sin X (East) = 30 kts x Sin 30 (West)
So
Sin X = 15 / 100 = 0.15
So X = 8.627 degrees.
The northerly component of TAS = 100 Cos 8.627
The southerly component of the wind = 30 Cos 30
So the northerly ground speed = 100 Cos 8.627 - 30 Cos 30
Northely ground speed = 98.87 kts - 25.98 kts = 72.89 kts.
I have done all of this pretty quickly so there will probably be some errors in it.
But with a bit of practice most people will find this kind of thing much easier with the CRP5.