已知α为锐角,且sin^2α-sinαcosα-2cos^2=0.(1)求tanα的值.(2)求sin[α-(π/3)]
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/02 19:51:12
xQN@~v)EdXJDxb bK3hZx|n"@,_Zdy{XW`Ǔ1֕ZWG'%e+ۻf2
ޜ}mj{(=<{m,`,*E h1>މ$B-6@<$
veH(dE
Ed-yxi#`ppdx]إmY$]
已知α为锐角,且sin^2α-sinαcosα-2cos^2=0.(1)求tanα的值.(2)求sin[α-(π/3)]
已知α为锐角,且sin^2α-sinαcosα-2cos^2=0.(1)求tanα的值.(2)求sin[α-(π/3)]
已知α为锐角,且sin^2α-sinαcosα-2cos^2=0.(1)求tanα的值.(2)求sin[α-(π/3)]
1.sin^2α-sinαcosα-2cos^2= sin^2 a-cos^2 a-1/2(sin2a)-cos^2a+1-1
=-cos2a-1/2(sin2a)-1=0
cos2a+1/2(sin2a)=-1
(1-tan^2a)/(1+tan^2a)-0.5[ 2tana/(1+tan^2a)]=-1
1-tan^2a-tana=-1-tan^2a
tana=2
2.已知α为锐角,则sina=2/根号5 cosa=1/根号5
sina[a-(TT/3)]=sinacos60-sin60cosa=(1-根号3)/2根号5