(sint+cost)/(cost-sint)怎么算的，等于tan(t+π/4)
答案:3 悬赏:50 手机版
解决时间 2021-02-25 02:46
- 提问者网友：别再叽里呱啦
- 2021-02-24 10:33
(sint+cost)/(cost-sint)怎么算的，等于tan(t+π/4)
最佳答案
- 五星知识达人网友：躲不过心动
- 2021-02-24 10:45
(sint + cost) / (cost-sint)
= [(sint + cost)/√2] / [(cost - sint)/√2]
= (sint cosπ/4 + cost sinπ/4) / (cost cosπ/4 - sint sinπ/4)
= sin(t+π/4) / cos(t+π/4)
= tan(t+π/4)
= [(sint + cost)/√2] / [(cost - sint)/√2]
= (sint cosπ/4 + cost sinπ/4) / (cost cosπ/4 - sint sinπ/4)
= sin(t+π/4) / cos(t+π/4)
= tan(t+π/4)
全部回答
- 1楼网友：三千妖杀
- 2021-02-24 12:29
三角函数?只有计算器能算。。。追问就是（sint+cost)/(cost-sint)继续化简 然后就变成了tan(t+π/4)追答。。。就是答案咯
- 2楼网友：街头电车
- 2021-02-24 11:51
(sint+cost)/(cost-sint)分子分母同时除以cost
=(sint/cost+cost/cost)/(cost/cost-sint/cost)
=(tant+1)/(1-tant)
=(tant+tanπ/4)/(1-tanttanπ/4)
=tan(t+π/4)
=(sint/cost+cost/cost)/(cost/cost-sint/cost)
=(tant+1)/(1-tant)
=(tant+tanπ/4)/(1-tanttanπ/4)
=tan(t+π/4)
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