已知a为钝角,sin(a+π/4)=1/3,则cos(a+π/4)=?已知a为钝角,sin(a+π/4)=1/3,则cos(a+π/4)=?sin{π/2-(a+π/4)}=?sin(π/4-a)=?

已知a为钝角,sin(a+π/4)=1/3,则cos(a+π/4)=?已知a为钝角,sin(a+π/4)=1/3,则cos(a+π/4)=?sin{π/2-(a+π/4)}=?sin(π/4-a)=?
已知a为钝角,sin(a+π/4)=1/3,则cos(a+π/4)=?
已知a为钝角,sin(a+π/4)=1/3,则cos(a+π/4)=?sin{π/2-(a+π/4)}=?sin(π/4-a)=?

已知a为钝角,sin(a+π/4)=1/3,则cos(a+π/4)=?已知a为钝角,sin(a+π/4)=1/3,则cos(a+π/4)=?sin{π/2-(a+π/4)}=?sin(π/4-a)=?
因为a为钝角
而sin(a+π/4）>0
所以a+π/4还是钝角
所以cos(a+π/4)