请问sin(x+π/2)=?cos(x+π/2)=?tan(x+π/2)=?cot(x+π/2)=?sec(x+π/2)

学习时间:2026-03-30 08:50:05 阅读：4173

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)=?

提问回复

最佳回答

酷酷的电源

2026-03-30 08:50:05



sin(x+π/2)=cos xcos(x+π/2)=-sinxtan(x+π/2)=sin(x+π/2)÷cos(x+π/2)=cos x÷(-sinx) = -cotxcot(x+π/2)=1÷tan(x+π/2)=1÷（-cotx）= - tan xsec(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/cosxsin(x+π)=-sinxcos(x+π)=-cosxtan(x+π)=sin(x+π)÷cos(x+π)=（-sinx）÷（-cosx）=tan xcot(x+π)=1÷tan(x+π)=1÷tan x = cot xsec(x+π)=1÷cos(x+π)=1÷(-cosx)=(-1/cos x)csc(x+π)=1÷sin(x+π)=1÷(-sinx)=(-1/sin x)

请问sin(x+π/2)=?cos(x+π/2)=?tan(x+π/2)=?cot(x+π/2)=?sec(x+π/2)

最佳回答

最新回答共有2条回答

欢呼的巨人

热门文章

猜你喜欢