题目

已知tan(a)*tan(b)=1/根号3,求[2-cos(2a)]*[2-cos(2b)]的值,

提问时间：2020-10-12

答案

依题意,tan(a)*tan(b)=sin(a)*sin(b)/cos(a)*cos(b)=1/3^(12);
两边平方,有[sin(a)^2]*[sin(b)^2]/[cos(a)^2]*[cos(b)^2]=1/3;
所以3[sin(a)^2]*[sin(b)^2]=[cos(a)^2]*[cos(b)^2];
3[sin(a)^2]*[sin(b)^2]=[1-sin(a)^2]*[1-sin(b)^2];
3[sin(a)^2]*[sin(b)^2]=1+[sin(a)^2]*[sin(b)^2]-[sin(a)^2+sin(b)^2];
即有sin(a)^2+sin(b)^2=1-2[sin(a)^2]*[sin(b)^2];
又因为cos(2a)=1-2sin(a)^2;
cos(2b)=1-2sin(b)^2;
所求式[2-cos(2a)]*[2-cos(2b)]
=[1+2sin(a)^2][1+2sin(b)^2]
=1+2[sin(a)^2+sin(b)^2]+4[sin(a)^2]*[sin(b)^2];
将sin(a)^2+sin(b)^2=1-2[sin(a)^2]*[sin(b)^2]带入上式则所求式为1+2{1-2[sin(a)^2]*[sin(b)^2]}+4[sin(a)^2]*[sin(b)^2]=3;
即所求.
注意:^是幂的符号.
打起来真困难..其实不很麻烦,你用纸写一写,加油啊.