Question

悬赏20…全给了…急

已知函数f(x)=sin的平方+sinxcosx.求f(x)的最大值及取得最大值时对应的x的值

Answer 1

f(x)=sin的平方+sinxcosx=(1-cos2x)/2+1/2sin2x=1/2+1/2(sin2x-cos2x)=1/2+√2/2sin(2x-π/4)
因为-1≤sin(2x-π/4)≤1
所以f(x)的最大值是1/2+√2/2
此时sin(2x-π/4)=1,
则2x-π/4=2kπ+π/2
故x=kπ+3π/8

Answer 2

最大值1.5，当x=45度时取到

Answer 3

f(x)=(1-cos2x)/2+1/2*sin2x
=1/2*(sin2x-cos2x)+1/2
=√2/2(√2/2*sinx-√2/2cos2x)+1/2
=√2/2sin(2x-π/4)+1/2
所以
sin(2x-π/4)=1时最大
2x-π/4=2kπ+π/2
所以x=kπ+3π/8,f(x)最大=(√2+1)/2

Answer 4

f=(1/2)(1-cos(2x)) +(1/2)sin(2x)
=(1/2)(1+2^0.5 * sin(2x+45度))
f 最大值在sin(2x+45度)=1时,
即f=(1/2)(1+2^0.5),当x=22.5度+-180n度, n=0,1,2,..

Answer 5

f(x)=(1-cos2x)/2+sin2x/2=1/2+(sin2x-cos2x)/2=1/2+2?(sin2xcos45°-cos2xsin45°)/2=1/2+2?sin(2x-45°)/2
∴当sin(2x-45°)=1时取最大值（1+2?）/2
此时2x-π/4=2kπ+π/2，k∈N，即x=kπ+3π/4

Answer 6

f(x)=sin²x+sinxcosx
f(x)=[1-cos(2x)+sin(2x)]/2
f(x)=1/2+[sin(2x)cos(π/4)-sin(π/4)cos(2x)]/√2
f(x)=1/2+√2/2[sin(2x-π/4)]
当2x-π/4=π/2+2kπ ,k∈Z,即x=3π/8+kπ ,k∈Z,时,f(x)有最大值(1+√2)/2