Show that b cos B + c cos C = a cos (B-C).
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i don't know had a few days back after school tomorrow
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It’s not hard to suppose that this identity pertains to a triangle. The Sine Law allows us to say,
asinA=bsinB=csinC=α ,
from which
a=αsinA , b=αsinB and c=αsinC .
Then the left side becomes
αsinBcosB+αsinCcosC
=12α[sin2B+sin2C]
=α[sin(B+C)cos(B−C)]
=αsin(180−A)cos(B−C)
=αsinAcos(B−C)
=acos(B−C
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