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what trig identity is used? (1 Viewer)
- Thread starter bos1234
- Start date
P
pLuvia
Guest
Are you sure its not
bsin(@-$)
=b.2sin((@-$)/2)cos((@-$)/2)?
bsin(@-$)
=b.2sin((@-$)/2)cos((@-$)/2)?
ellen.louise
Member
I think it should be = 2bsin((theta-y)/2)cos((theta-y)/2)
cause then you just use the sin2x = 2sinxcosx
or, in this case, sinx = 2sin(x/2)cos(x/2)
cause then you just use the sin2x = 2sinxcosx
or, in this case, sinx = 2sin(x/2)cos(x/2)