已知AB为锐角,且COS(A+B)=12⼀13,COS(2A+B)3⼀5,则cosA=

要解法
2025-02-02 21:44:37
推荐回答(2个)
回答1:

∵A,B为锐角,则0∴0∵COS(2A+B)=3/5>0,
∴0<2A+B<90°.
∴sin(A+B)>0,sin(2A+B)>0.
∵cos(A+B)=12/13,cos(2A+B)=3/5,
∴sin(A+B)=5/13,sin(2A+B)=4/5.
故cosA=cos[(2A+B)-(A+B)]
=cos(2A+B)cos(A+B)+sin(2A+B)sin(A+B)
=(3/5)(12/13)+(4/5)(5/13)
=56/65.

回答2:

sin(A+B)=√[1-(12/13)²]=5/13
sin(2A+B)=√[1-(3/5)²]=4/5
cosA=cos[(2A+B)-(A+B)]=(3/5)(12/13)+(4/5)(5/13)=56/65