题目
已知sin a=2cos a 计算⑴(sinα+2cosα)/(5cosα-sinα) ⑵tan(α+π/4)
提问时间:2020-11-15
答案
⑴(sinα+2cosα)/(5cosα-sinα)
=(2cosα+2cosα)/(5cosα-2cosα)
=(4cosα)/(3cosα)
=4/3
⑵tan(α+π/4)
=sin(α+π/4)/cos(α+π/4)
=[sinacos(π/4)+cosαsin(π/4)]/[cosαsin(π/4)-sinαsin(π/4)]
=(sinα+cosα)/(cosα-sinα)
=(2cosα+cosα)/(cosα-2cosα)
=(3cosα)/(-cosα)
=-3
=(2cosα+2cosα)/(5cosα-2cosα)
=(4cosα)/(3cosα)
=4/3
⑵tan(α+π/4)
=sin(α+π/4)/cos(α+π/4)
=[sinacos(π/4)+cosαsin(π/4)]/[cosαsin(π/4)-sinαsin(π/4)]
=(sinα+cosα)/(cosα-sinα)
=(2cosα+cosα)/(cosα-2cosα)
=(3cosα)/(-cosα)
=-3
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