Recall that sin(2θ) = 2sinθcosθ
So, sin(Bxi)cos(Bxi) = ½sin(2Bxi)
Does that help at all?

Mark M.
tutor
Let θ = Bxi for any real number,B.
Then, since sin(2θ) = 2sinθcosθ, we have sin(2Bxi) = 2sinBxicosBxi
So, sinBxicosBxi = ½sin(2Bxi)
Mark M (Bayport, NY)
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10/15/16

Mark M.
Mark M (Bayport, NY)
True, yet the constant B remains as a coefficient of x.
Hubert wanted to perform some type of factoring, i.e., moving it from a coefficient of the angles to a coefficient of the function.
Mark M (Carson, CA)
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10/15/16
Mark M.
tutor
OK. Of course "factoring out" B doesn't work. As far as I can tell, the best that can be done is to use the double angle identity so that there is only one trig function to contend with instead of two of them.
Mark M (Bayport, NY)
Report
10/15/16
Mark M.
10/15/16