这道题里的πcosπ-sinπ=-π怎么求出来的?
1个回答
三角函数值呀!
cosπ=-1,sinπ=0
则πcosπ-sinπ=-π.
相关问题
请问这道题怎么求(a sin〖π/5〗+b cos〖π/5〗)/(a cos〖π/5〗-b sin〖π/5〗 )
(πcosπ-sinπ)/π2答案是1/π.求详细过程我算出来时-1/π
sinπ,sinπ/2,sin3π/2,cosπ/2,cos-π/2,cos/2π/3 等等一系列怎么算
求值【sin(2π-α)sin(π+α)cos(-π-α)】/【sin(3π-α)-cos(π-α)】
求sin2π,cosπ,tanπ的值?
求SINπ/64 COSπ/64 COSπ/32 COSπ/16 COSπ/8!急救急救
若sin(α-π)=2cos(2π-α),求[sin(π-α)+5cos(2π-α)]/[3cos(π-α)-sin(-
若sin(x-π)=2cos(2π-x),求(sin(π-x)+5cos(2π-x))/(3cos(π-x)-sin(-
sin(π/2+α)·cos(π/2-α)/cos(π+α)+sin(π-α)·cos(π/2+α)/sin(π+α)=
(cos[π/12-sinπ12])(cos[π/12+sinπ12])= ___ .