已知f(x)=sinx+2sin(π/4+x/2)cos(π/4+x/2).若f(a)=√2/2,a∈(-π/2,0),
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已知f(x)=sinx+2sin(π/4+x/2)cos(π/4+x/2).若f(a)=√2/2,a∈(-π/2,0),求a的值
请告诉我答案及解题过程!谢谢!
若sin(x/2)=4/5,x∈(π/2,π),f(x)=?
请告诉我答案及解题过程!谢谢!
若sin(x/2)=4/5,x∈(π/2,π),f(x)=?
利用2倍角公式
f(x)=sinx+sin[2(π/4+x/2)]
=sinx+sin(π/2+x)
=sinx+cosx
=√2(√2/2sinx+√2/2cosx)
=√2(cosπ/4sinx+sinπ/4cosx)
=√2sin(x+π/4)
f(a)=√2/2
即sin(x+π/4)=1/2
a∈(-π/2,0)
a+π/4(-π/4,π/4)
所以a+π/4=π/6 x=-π/12
x∈(π/2,π) x∈(π/4,π/2) cos(x/2)>0
cos(x/2)=√[1-(sin(x/2))^2]=3/5
2倍角公式
sinx=2sin(x/2)cos(x/2)=2*(4/5)*(3/5)=24/25
cosx=1-2(sinx/2)^2=-7/25
f(x)=sinx+2sin(π/4+x/2)cos(π/4+x/2)
=sinx+cosx
=17/25
f(x)=sinx+sin[2(π/4+x/2)]
=sinx+sin(π/2+x)
=sinx+cosx
=√2(√2/2sinx+√2/2cosx)
=√2(cosπ/4sinx+sinπ/4cosx)
=√2sin(x+π/4)
f(a)=√2/2
即sin(x+π/4)=1/2
a∈(-π/2,0)
a+π/4(-π/4,π/4)
所以a+π/4=π/6 x=-π/12
x∈(π/2,π) x∈(π/4,π/2) cos(x/2)>0
cos(x/2)=√[1-(sin(x/2))^2]=3/5
2倍角公式
sinx=2sin(x/2)cos(x/2)=2*(4/5)*(3/5)=24/25
cosx=1-2(sinx/2)^2=-7/25
f(x)=sinx+2sin(π/4+x/2)cos(π/4+x/2)
=sinx+cosx
=17/25
已知f(x)=sinx+2sin(π/4+x/2)cos(π/4+x/2).若f(a)=√2/2,a∈(-π/2,0),
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