已知cos[x+(π/4)]=3/5,7π/12<x<7π/4,
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sin
2x
=
2sinxcosx
(sin
2x+2sin^2
x)/(1-tan
x)
=
2sin
x(cos
x+sin
x)/(1-tan
x)
=
2sin
x(cos
x+sin
x)/(1-(sin
x/cos
x))
=
2sin
x(cos
x+sin
x)×cos
x/(cos
x-sin
x)
=
sin2x(cos
x+sin
x)/(cos
x-sin
x)
(中间你可以随意省略!)
cos[x+(π/4)]=3/5
=>
cosxsin(π/4)-sinxcos(π/4)=3/5
=>
(cos
x-sin
x)×√2/2=3/5
(√2/2是二分之根号二)
=>
cos
x-sin
x
=
3√2/5
①
(3√2/5
是五分之三倍根号二)
=>
(cos
x-sin
x)^2
=
18/25
=>
1
-
2sinxcosx
=
18/25
=>
2sinxcosx
=7/25
=>
sin2x
=
7/25
②
然后再回到
(sin
2x+2sin^2
x)/(1-tan
x)
=
sin2x(cos
x+sin
x)/(cos
x-sin
x)
现在就只有
(cos
x+
sinx)不知道
等于多少了
。
(cos
x+sin
x)^2=1+2sinxcosx=1+sin2x
所以
cos
x
+sin
x
=
√(1+7/25)
=
√(32/25)
③
由
①
②
③
原式等于=
28/75
2x
=
2sinxcosx
(sin
2x+2sin^2
x)/(1-tan
x)
=
2sin
x(cos
x+sin
x)/(1-tan
x)
=
2sin
x(cos
x+sin
x)/(1-(sin
x/cos
x))
=
2sin
x(cos
x+sin
x)×cos
x/(cos
x-sin
x)
=
sin2x(cos
x+sin
x)/(cos
x-sin
x)
(中间你可以随意省略!)
cos[x+(π/4)]=3/5
=>
cosxsin(π/4)-sinxcos(π/4)=3/5
=>
(cos
x-sin
x)×√2/2=3/5
(√2/2是二分之根号二)
=>
cos
x-sin
x
=
3√2/5
①
(3√2/5
是五分之三倍根号二)
=>
(cos
x-sin
x)^2
=
18/25
=>
1
-
2sinxcosx
=
18/25
=>
2sinxcosx
=7/25
=>
sin2x
=
7/25
②
然后再回到
(sin
2x+2sin^2
x)/(1-tan
x)
=
sin2x(cos
x+sin
x)/(cos
x-sin
x)
现在就只有
(cos
x+
sinx)不知道
等于多少了
。
(cos
x+sin
x)^2=1+2sinxcosx=1+sin2x
所以
cos
x
+sin
x
=
√(1+7/25)
=
√(32/25)
③
由
①
②
③
原式等于=
28/75
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