一组方程①sinx+cosy=1 ②cosx+siny=√3 求这个方程.
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一组方程①sinx+cosy=1 ②cosx+siny=√3 求这个方程.
是不是还是用到sin’2 θ+sin‘2 θ=1 这个知识点.
是不是还是用到sin’2 θ+sin‘2 θ=1 这个知识点.
(1)+(2):(sinx+siny) + (cosx+cosy) = 1+√3
sin(x+y)/2cos(x-y)/2 + cos(x+y)/2cos(x-y)/2 = (1+√3)/2 (3)
(1)-(2):cos(x+y)/2sin(x-y)/2 + sin(x+y)/2sin(x-y)/2=(1-√3)/2 (4)
cos(x-y)/2 [sin(x+y)/2+cos(x+y)/2]=(1+√3)/2 (3)
sin(x-y)/2 [cos(x+y)/2+sin(x+y)/2]=(1-√3)/2 (4)
(4)/(3):tan(x-y)/2 = (1-√3)/(1+√3) (5)
解出:x-y = arcTAN[(1-√3)/(1+√3)] (6)
(1)^2+(2)^2:sin(x+y)=(√3-1)/2
解出:x+y = arcSIN(√3-1)/2 (7)
联立(6)、(7)立马解出:
x = {arcTAN[(1-√3)/(1+√3)]+arcSIN(√3-1)/2}/2 (8)
y = {arcTAN[(1-√3)/(1+√3)]-arcSIN(√3-1)/2}/2 (9)
请再检查一遍.
sin(x+y)/2cos(x-y)/2 + cos(x+y)/2cos(x-y)/2 = (1+√3)/2 (3)
(1)-(2):cos(x+y)/2sin(x-y)/2 + sin(x+y)/2sin(x-y)/2=(1-√3)/2 (4)
cos(x-y)/2 [sin(x+y)/2+cos(x+y)/2]=(1+√3)/2 (3)
sin(x-y)/2 [cos(x+y)/2+sin(x+y)/2]=(1-√3)/2 (4)
(4)/(3):tan(x-y)/2 = (1-√3)/(1+√3) (5)
解出:x-y = arcTAN[(1-√3)/(1+√3)] (6)
(1)^2+(2)^2:sin(x+y)=(√3-1)/2
解出:x+y = arcSIN(√3-1)/2 (7)
联立(6)、(7)立马解出:
x = {arcTAN[(1-√3)/(1+√3)]+arcSIN(√3-1)/2}/2 (8)
y = {arcTAN[(1-√3)/(1+√3)]-arcSIN(√3-1)/2}/2 (9)
请再检查一遍.
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