AB=2,AC=(√2)BC,求三角形ABC面积的最大值?
c=AB=2,b=AC,a=BC,b=(√2)a；
cosC=(a²+b²-4)/2ab=(3a²-4)/[(2√2)a²]
sinC=√(1-cos²C)=√[1-(3a²-4)²/(8a⁴)]=√{[8a⁴-(9a⁴-24a²+16)]/8a⁴}
=√[(-a⁴+24a²-16)/8a⁴]=√[-(a⁴-24a²+16)/8a⁴]=√[-(a²-12)²+128]/8a⁴]
SΔABC=(1/2)absinC=(√2/2)a²sinC=(√2/2)a²√[-(a²-12)²+128]/8a⁴]
=(1/4)√[-(a²-12)²+128]≦(1/4)√128=2√2.
即ΔABC面积的最大值为2√2.此时a²=12,a=2√3,b=2√6；
cosC=32/(24√2)=4/(3√2)=(2/3)√2,
sinC=√(1-8/9)=1/3,
S=(1/2)×2√3×2√6×(1/3)=(2/3)√18=2√2.