Question

设f(x)=sinx(sinx+cosx)求f(x)最大的值及相应的x的值

设f(x)=sinx(sinx+cosx)求f(x)最大的值及相应的x的值
我快结束了

Answer 1

1。设f(x)=sinx(sinx+cosx)求f(x)最大的值及相应的x的值
解：f(x)=sin²x+sinxcosx=(1/2)(1-cos2x)+(1/2)sin2x=(1/2)(sin2x-cos2x)+1/2=(√2/2)sin(2x-π/4)+1/2
故当2x-π/4=π/2+2kπ，2x=π/4+π/2+2kπ=3π/4+2kπ，即当x=3π/8+kπ时f(x)获最大值(1/2)(1+√2)
2.锐角三角ABC中，满足f(A)=1 求sin(2B+C)的取值范围
解：f(A)=(√2/2)sin(2A-π/4)+1/2=1，故sin(2A-π/4)=1/√2=√2/2，故2A-π/4=π/4；A=π/4；
故B+C=π-π/4=3π/4；0<B<π/2，∴3π/4<2B+C<5π/4；
由于sin(5π/4)=sin(π+π/4)=-sin(π/4)=-√2/2；sin(3π/4)=sin(π-π/4)=sin(π/4)=√2/2；
故-√2/2<sin(2B+c)<√2/2.

Answer 2

f(x)=sinx(sinx+cosx)
=sin^2x+sinxcosx
=1/2(1-cos2x)+1/2sin2x
=1/2sin2x-1/2cos2x+1/2
=√2/2sin(2x-π/4)+1/2
sin(2x-π/4)最大值=1
f(x)=sinx(sinx+cosx)最大值=(1+√2)/2
2x-π/4=2Kπ+π/2
x=Kπ+3π/4
2,f(A)=1
f(A)=√2/2sin(2A-π/4)+1/2=1
sin(2A-π/4)=√2/2
2A-π/4=π/4 或 2A-π/4=3π/4
A=π/4 或 A=π/2(舍去)
B+C=3π/4,C=3π/4-B
sin(2B+C)=sin(B+3π/4)=sin[π/2+(B+π/4)]
=cos(B+π/4)
0<B<π/2,π/4<B+π/4<3π/4
-√2/2<sin(2B+C)<√2/2

Answer 3

f(x)=1,x=π/4