Question

在△ABC中，C-A=π/2，sinB=1/3 （1）求sinA的值 （2）设AC=根号6，求△ABC的面积

Answer 1

本题用三角函数来解比较麻烦，不如直接用数值解法简单：
∵A+B+C=π，C-A=π/2
∴A=（π/2-B）/2
∵sinB=1/3
∴B=0.339837
∴A=0.61548
sinA=0.57735

AC=√6，sinB=1/3
sinA/BC=sinB/AC
BC=sinA/sinB×AC=0.57735/0.33333×√6=4.242641
则△ABC的面积为
AC×BC×sinC=√6×0.471405×sin（0.61548+π/2）/2
=4.242641

当然，也可以用三角函数计算，由sinA=sin(（π/2-B）/2),根据和差化积等公式计算。

Answer 2

sinA=sin(B+C) = sin(pi/2+B+A) = cos(B+A)
sinA = cosB cosA -sinAsinB = cosA sqrt(8)/3 - sinA/3
sinA /cos A = sqrt(8)/4
sinA = 1/sqrt(3)

b = sqrt(6)
sinB/b = sinA/a
a = b sinA/sinB = sqrt(6)/sqrt(3) * 3 = 3sqrt(2)
sinC = sin(pi/2+A) = cosA = sqrt(6)/3
面积=1/2absinC = 3sqrt(12)*sqrt(6)/6 = 3sqrt(2)