Find the value of sinBsinCsin2B+sin2C−sin2A given that sinCsinB=ABAC, sinBsinC=ACAB, and sinBsinCsinA=AC⋅ABBC.
A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.
Solution
Using the Law of Sines, we have sinBsinCsin2B+sin2C−sin2A=sinCsinB+sinBsinC−sinBsinAsinCsinA=ABAC+ACAB−ACBCABBC=8083
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