数学题 不难 200分 来抢
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数学题 不难 200分 来抢
在△ABC中,求证:a cosA+b cosB+c cosC = 2a sinB sinC
谢谢大家~~小妹感激不尽-\(^o^)/~
在△ABC中,求证:a cosA+b cosB+c cosC = 2a sinB sinC
谢谢大家~~小妹感激不尽-\(^o^)/~
![数学题 不难 200分 来抢](/uploads/image/z/18860781-21-1.jpg?t=%E6%95%B0%E5%AD%A6%E9%A2%98+%E4%B8%8D%E9%9A%BE+200%E5%88%86+%E6%9D%A5%E6%8A%A2)
证明:左=a cosA+b·cosB+c·cosC
=a(cosA+b/a cosB+c/a cosC)
=a(cosA+sinB/sinA cosB+sinC/sinA cosC)
=a[cosA+(sinBcosB+sinCcosC)/sinA]
=a[cosA+sin(B+C)cos(B-C)/sinA]
=a[-cos(B+C)+cos(B-C)]
=2asinBsinC
=右
=a(cosA+b/a cosB+c/a cosC)
=a(cosA+sinB/sinA cosB+sinC/sinA cosC)
=a[cosA+(sinBcosB+sinCcosC)/sinA]
=a[cosA+sin(B+C)cos(B-C)/sinA]
=a[-cos(B+C)+cos(B-C)]
=2asinBsinC
=右