f(x)=cos2x/sin(x+π/4) 若f(x)=4/3,求sin2x的值

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/29 00:12:45
f(x)=cos2x/sin(x+π/4) 若f(x)=4/3,求sin2x的值
xSJ@~AjYvS4Ť/RZPARXD%9YB Uj4 m{+8;TC2;*5G+ܡ椾4-"jlθ}"{DWPrZ5۬@2A`kl&N/#Lm(S&b?K'H e~MxѕHªQ6ۤB حLcRBOR)eu㺝,\)@dD>&<QWuJRd"j`4@t6* 8Y]"q

f(x)=cos2x/sin(x+π/4) 若f(x)=4/3,求sin2x的值
f(x)=cos2x/sin(x+π/4) 若f(x)=4/3,求sin2x的值

f(x)=cos2x/sin(x+π/4) 若f(x)=4/3,求sin2x的值
cos2x/sin(x+π/4) =4/3
cos2x/sin(x+π/4) =4/3
(cos² x-sin² x)/sin(x+π/4) =4/3
(cosx-sinx)(cosx+sinx)/sin(x+π/4) =4/3
(cosx-sinx)[√2(√2/2cosx+√2/2sinx)]/sin(x+π/4) =4/3
(cosx-sinx)[√2(sinπ/4cosx+cosπ/4sinx)]/sin(x+π/4) =4/3
(cosx-sinx)[√2sin(x+π/4)]/sin(x+π/4) =4/3
√2(cosx-sinx)=4/3(平方)
2(cosx-sinx)²=16/9
(cosx-sinx)²=8/9
cos²x+sin²x-2sinxcosx=8/9
1-2sinxcosx=8/9
2sinxcosx=1/9
sin2x=1/9

cos2x=sin(2x+π/2)=2sin(x+π/4)cos(x+π/4)
∴f(x)=2cos(x+π/4)=4/3 ∴cos(x+π/4)=2/3
又∵sin2x=﹣cos(2x+π/2)=﹣[2cos(x+π/4)^2-1]= ﹣(2·4/9-1)=1/9

∵ f(x)=cos2x/sin(x+π/4) ,f(x)=4/3, ∴ cos2x/sin(x+π/4)=4/3
(cos^2x-sin^2x)/[根2/2(sinx+cosx)=4/3 分子平方差分解后,约去公因式得:
cosx-sinx=3倍根2/8, 两边平方得1-sin2x=9/32 即sin2x=23/32