不一回事。例如：1/cosx可以大于1。
三角函数问题,tanx+cotx=?sinx+cosx=?sin(π/4 + 2/x)和cos(π/4 - 2/x)相等吗? 答案 tanx+cotx=sinx/cosx+cosx/sinx=(sin^2x+cos^2x)/sinxcosx=1/sinxcosxsinx+cosx=√2(√2/2sinx+√2/2cosx)=√2sin(x+π/4 )sin(π/4 + 2/x)=sin[π/2-(π/4 - x/2)]=cos(π/4 -...
cos2x=(cosx)^2-(sinx)^2=2(cosx)^2-1=1-2(sinx)^2
答案解析 查看更多优质解析 解答一 举报 tanx+cotx=sinx/cosx+cosx/sinx=(sin^2x+cos^2x)/sinxcosx=1/sinxcosxsinx+cosx=√2(√2/2sinx+√2/2cosx)=√2sin(x+π/4 )sin(π/4 + 2/x)=sin[π/2-(π/4 - x/2)]=cos(π/4 - x/2) 解析看不懂?免费查看同类题视频解析查看解答 ...
三角函数问题,tanx+cotx=?sinx+cosx=?sin(π/4 + 2/x)和cos(π/4 - 2/x)相等吗? 答案 tanx+cotx=sinx/cosx+cosx/sinx=(sin^2x+cos^2x)/sinxcosx=1/sinxcosxsinx+cosx=√2(√2/2sinx+√2/2cosx)=√2sin(x+π/4 )sin(π/4 + 2/x)=sin[π/2-(π/4 - x/2)]=cos(π/4 -...
tanx+cotx=sinx/cosx+cosx/sinx =(sin^2x+cos^2x)/sinxcosx =1/sinxcosx sinx+cosx=√2(√2/2sinx+√2/2cosx)=√2sin(x+π/4 )sin(π/4 + 2/x)=sin[π/2-(π/4 - x/2)]=cos(π/4 - x/2)