又因为sin 2α=5/13,

　　所以cos 2α=-√(1"-" sin^2 2α)=-√(1"-(" 5/13 ")" ^2 )=-12/13.

　　于是sin 4α=2sin 2αcos 2α=2×5/13×(-12/13)=-120/169;

　　cos 4α=1-2sin22α=1-2×(5/13)2=119/169;

　　tan 4α=sin4α/cos4α=("-" 120/169)/(119/169)=-120/119.

　　【例2】解:方法一:在△ABC中,由cos A=4/5,0

　　sin A=√(1"-" cos^2 A)=√(1"-(" 4/5 ")" ^2 )=3/5,

　　所以tan A=sinA/cosA=3/5×5/4=3/4,

　　tan 2A=2tanA/(1"-" tan ^2 A)=(2×3/4)/(1"-(" 3/4 ")" ^2 )=24/7,

　　又tan B=2,

　　所以tan 2B=2tanB/(1"-" tan ^2 B)=(2×2)/(1"-" 2^2 )=-4/3.

　　于是tan(2A+2B)=(tan2A+tan2B)/(1"-" tan2Atan2B)=(24/7 "-" 4/3)/(1"-" 24/7×"(-" 4/3 ")" )=44/117.

　　方法二:在△ABC中,由cos A=4/5,0

　　sin A=√(1"-" cos ^2 A)=√(1"-(" 4/5 ")" ^2 )=3/5.

　　所以tan A=sinA/cosA=3/5×5/4=3/4.又tan B=2,

　　所以tan(A+B)=(tanA+tanB)/(1"-" tanAtanB)=(3/4+2)/(1"-" 3/4×2)=-11/2,

　　于是tan(2A+2B)=tan[2(A+B)]=(2tan"(" A+B")" )/(1"-" tan ^2 "(" A+B")" )=(2×"(-" 11/2 ")" )/(1"-(-" 11/2 ")" ^2 )=44/117.

　　四、变式演练,深化提高

　　1.解:原式=√("(" sin15"°" +cos15"°" ")" ^2 )=√(sin ^2 15"°" +2sin15"°" cos15"°" +cos ^2 15"°" )=√6/2.

　　2.解:原式=cos 80°cos 60°cos 40°cos 20°=(2^3 "·" sin20"°" cos20"°" cos40"°" cos80"°" )/(2^3 "·" 2sin20"°" )

　　=sin160"°" /16sin20"°" =sin20"°" /16sin20"°" =1/16.

　　3.解:(1)由cos α=1/7,0<α<π/2,得sin α=√(1"-" cos ^2 α)=√(1"-(" 1/7 ")" ^2 )=(4√3)/7.

∴tan α=sinα/cosα=(4√3)/7×7/1=4√3.于是tan 2α=2tanα/(1"-" tan ^2 α)=(2×4√3)/(1"-(" 4√3 ")" ^2 )=-(8√3)/47.