sin(a+b)=4/5,tanb=4/3,0<a<b<pai,求tana
来源:百度知道 编辑:UC知道 时间:2024/06/09 07:39:22
因为0<a<b<π,并且tanb=4/3>0,y=tanx在0到π/2上大于0,在π/2到π上小于0,所以0<b<π/2,所以0<a<b<π/2.
tanb=4/3,tanb=(2tanb/2)/[1-(tanb/2)^2]
记tanb/2=x则2x/(1-x^2)=4/3,x=1/2或x=-2(舍)
sinb=(2tanb/2)/[1+(tanb/2)^2]=(2x)/1+x^2=4/5
cosb=[1-(tanb/2)^2]/[1+(tanb/2)^2]=(1-x^2)/(1+x^2)=3/5
sin(a+b)=sinacosb+cosasinb=4/5,该等式可化为
(3/5)sina+(4/5)cosa=4/5,即
3sina+4cosa=4
设tana/2=t则上式可化为6t/(1+t^2)+4(1-t^2)/(1+t^2)=4,解得t=3/4或t=0(舍)
tana=2t/(1-t^2)=24/7
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