已知a,b是直角三角形ABC的两条直角边，c为斜边，arcsin(1/a)+arcsin(1/b)= π/2.
求证：lgc=lga+lgb

解：∵arcsin(1/a)+arcsin(1/b)= π/2.
∴ sin[arcsin(1/a)+arcsin(1/b)]=sinπ/2.
= sin[arcsin(1/a)] cos[arcsin(1/b)]+ cos[arcsin(1/a)] sin[arcsin(1/b)]
=sin[arcsin(1/a)]cos[π/2-arcsin(1/a)]+cos[π /2-arcsin(1/b)]sin[arcsin(1/b)]
==>(1/a)²+(1/b)²=1
==>(a²+b²)/a²b² =1
==>a²+b²=a²b²【∵a²+b²=c²】
==>c²=a²b²
==>c=ab
==>lgc=lgab
==>lgc=lga+lgb