一道反三角题:sin[2 arc sin(-3/5)]的值??

设a=arc sin(-3/5)
则sin a=-3/5,cos a=4/5
sin 2a=2*sin a*cos a=-24/25

sin[2 arc sin(-3/5)]=2 sin arcsin(-3/5) *cos arc sin (-3/5)
因为 sin arc sin(-3/5)=-3/5
所以cos arc sin(-3/5)=4/5
原式=2*4/5*(-3/5)=-24/25

因为sin[ arc sin(-3/5)]=-3/5
所以sin[2 arc sin(-3/5)]
=2*(-3/5)*cos[ arc sin(-3/5)]
=(-6/5)*[根号开始1-(-3/5)^2根号结束]
=-24/25