如果三角形三内角满足sin2A+sin2B=5sin2C,求证:sinC≤3/5
问题描述:
如果三角形三内角满足sin2A+sin2B=5sin2C,求证:sinC≤3/5
答
(sinA)^2+(sinB)^2=5(sinC)^2
由正弦定理得:
a^2+b^2=5c^2≥2ab
ab≤5c^2/2
由余弦定理:
cosC=(a^2+b^2-c^2)/2ab
= 4c^2/2ab ≥4c^2/5c^2=4/5
(sinC)^2=1-(cosC)^2≤9/25
sinC≤3/5