请问sin(x+π/2)=?cos(x+π/2)=?tan(x+π/2)=?cot(x+π/2)=?sec(x+π/2)=?csc(x+π/2)=?
还有sin(x+π)=?cos(x+π)=?tan(x+π)=?cot(x+π)=?sec(x+π)=?csc(x+π)=?
请问sin(x+π/2)=?cos(x+π/2)=?tan(x+π/2)=?cot(x+π/2)=?sec(x+π/2)
答案:2 悬赏:60 手机版
解决时间 2021-12-19 22:41
- 提问者网友:火车头
- 2021-12-19 12:59
最佳答案
- 五星知识达人网友:千杯敬自由
- 2021-12-19 13:55
sin(x+π/2)=cos x
cos(x+π/2)=-sinx
tan(x+π/2)=sin(x+π/2)÷cos(x+π/2)=cos x÷(-sinx) = -cotx
cot(x+π/2)=1÷tan(x+π/2)=1÷(-cotx)= - tan x
sec(x+π/2)=1÷cos(x+π/2)=1÷(-sinx)=(-1/sinx)
csc(x+π/2)=1÷sin(x+π/2)=1÷(cos x)=1/cosx
sin(x+π)=-sinx
cos(x+π)=-cosx
tan(x+π)=sin(x+π)÷cos(x+π)=(-sinx)÷(-cosx)=tan x
cot(x+π)=1÷tan(x+π)=1÷tan x = cot x
sec(x+π)=1÷cos(x+π)=1÷(-cosx)=(-1/cos x)
csc(x+π)=1÷sin(x+π)=1÷(-sinx)=(-1/sin x)
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- 1楼网友:零点过十分
- 2021-12-19 15:11
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