在△ABC中,求证：S△ABC=a^2/[2(cotB+cotC)]
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S△ABC=1/2absinC
=1/2a^2*(b/a)*sinC
=1/2a^2*(sinB/sinA)*sinC
=1/2a^2*sinB*sinC/sinA
=1/2a^2*sinB*sinC/sin(B+C)
=1/2a^2*sinB*sinC/(sinBcosC+cosBsinC)
=1/2a^2/(sinBcosC/sinB*sinC+cosBsinC/sinB*sinC)
=1/2a^2/(cosC/sinC+cosB/sinB)
=1/2*a^2/(cotB+cotC)
=a^2/[2(cotB+cotC)]
你从下往上看就知道了这证明得推算步骤.
=1/2a^2*(b/a)*sinC
=1/2a^2*(sinB/sinA)*sinC
=1/2a^2*sinB*sinC/sinA
=1/2a^2*sinB*sinC/sin(B+C)
=1/2a^2*sinB*sinC/(sinBcosC+cosBsinC)
=1/2a^2/(sinBcosC/sinB*sinC+cosBsinC/sinB*sinC)
=1/2a^2/(cosC/sinC+cosB/sinB)
=1/2*a^2/(cotB+cotC)
=a^2/[2(cotB+cotC)]
你从下往上看就知道了这证明得推算步骤.
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