依題意,tan(a)*tan(b)=sin(a)*sin(b)/cos(a)*cos(b)=1/3^(1\2);
兩邊平方,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?yàn)閏os(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];
將sin(a)^2+sin(b)^2=1-2[sin(a)^2]*[sin(b)^2]帶入上式則所求式為1+2{1-2[sin(a)^2]*[sin(b)^2]}+4[sin(a)^2]*[sin(b)^2]=3;
即所求.
注意:^是冪的符號(hào).
打起來(lái)真困難..其實(shí)不很麻煩,你用紙寫(xiě)一寫(xiě),加油啊.