Question

已知α，β均为锐角。cos（α+β）=12/13，cos（2α+β）=3/5,求tanα

Answer 1

已知α，β均为锐角
则0<α<π/2 0<β<π/2
0<2α+β<3π/2
因cos(2α+β)=3/5>0
所以0<2α+β<π/2
同理0<α+β<π
所以sin(α+β)=√(1-cos²(α+β)=5/13
sin(2α+β)=√(1-cos²(2α+β)=4/5
所以sinα=sin[(2α+β)-(α+β)]
=sin(2α+β)cos(α+β)-cos(2α+β)sin(α+β)
=(4/5)*(12/13)-(3/5)(5/13)
=33/65
cosα=√(1-sin²α)=56/65
所以tanα=sinα/cosα=(33/65)/(56/65)=33/56

Answer 2

解：
由题设可知
cos(a+b)=12/13. sin(a+b)=5/13
cos(2a+b)=3/5. sin(2a+b)=4/5
∴sina
=sin[(2a+b)-(a+b)]
=sin(2a+b)cos(a+b)-cos(2a+b)sin(a+b)
=(48/65)-(15/65)
=33/65
cosa
=cos[(2a+b)-(a+b)]
=cos(2a+b)cos(a+b)+sin(2a+b)sin(a+b)
=56/65
∴tana=sina/cosa
=33/56