数学
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
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 arcsin(1/a)+arcsin(1/b)= π/2. ==>(1/a)^2+(1/b)^2=1 ==>(a^2+b^2)/a^2b^2=1 ==>a^2+b^2=a^2b^2【∵a^2+b^2=c^2】 ==>c^2=a^2b^2 ==>c=ab ==>lgc=lgab ==>lgc=lga+lgb
答:解:∵arcsin(1/a)+arcsin(1/b)= π/2. ∴ sin[arcsin(1/a)+arcsin(1/b)]=sinπ/2. = sin[ar...详情>>
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