51问答网 > 化简 √2⼀4sin(π⼀4-x)+√6⼀4cos(π⼀4-x)

化简 √2⼀4sin(π⼀4-x)+√6⼀4cos(π⼀4-x)

2024-12-29 16:56:55

推荐回答（3个）

回答1：

设直线是y=kx+b
则0=3k+b
-3=0+b
m=k+b
则b=-3
k=-b/3=1
m=k+b=-2通分原式=[(2x+2)(x-1)+5]/(x-1)
=(2x²-2+5)/(x-1)
=(2x²+3)/(x-1)

回答2：

原式=√2/2(1/2sin(π/4-x)+√3/2cos(π/4-x))
=√2/2(cosπ/3sin(π/4-x)+√sinπ/3cos(π/4-x))
=√2/2sin(7π/12-x)

回答3：

√2/4sin(π/4-x)+√6/4cos(π/4-x)
=√2/2[sin(π/4-x)*1/2+cos(π/4-x)*√3/2]
=√2/2[sin(π/4-x)cosπ/3+cos(π/4-x)sinπ/3]
=√2/2sin(π/4-x+π/3)
=√2/2sin(7π/12-x)
=√2/2sin(π/2+π/12-x)
=√2/2cos(π/12-x)
=√2/2cos(x-π/12)