Question

数学题。求学霸

数学。已知函数f(x)=sin(2x-π/6),x∈R
求函数的最小正周期
求f(0)的值
设a是第一象限角，且f(a+π/3)=3/5,求sina的值

Answer 1

解答
（1）∵最小正周期T=2π/|w|
|w|=2
∴T=π
（2）f(0)=sin(-π/6)=-1/2

(3)f(a+π/3)=sin(2(a+π/3)-π/6)
=sin(2a+2π/3-π/6)
=sin(2a+π/2)
=sin2acosπ/2+cos2a
=cos2a=3/5
∴
1-2sin的平方a=cos2a=3/5
解sina=√5/5或（-√5/5舍去）∵a是第一象限的角大于0

Answer 2

1.
最小正周期为2π/2=π

f(0)=sin(0-π/6)=-1/2

2.

f(a+π/3)=sin(2a+2π/3-π/6)=sin(2a+π/2)=cos2a=1-2sin^2a=3/5

sina=根号5/5

a=arcsin根号5/5

望采纳

Answer 3

答：

f(x)=sin(2x-π/6)
最小正周期T=2π/w=2π/2=π
f(0)=sin(0-π/6)=-sin(π/6)=-1/2
所以：最小正周期T=π，f(0)=-1/2

f(a+π/3)=sin[2(a+π/3)-π/6]=3/5
所以：
sin(2a+4π/6-π/6)=3/5
sin(2a+π/2)=3/5
cos(2a)=3/5
所以：cos(2a)=1-2(sina)^2=3/5
因为：a是第一象限角
所以：sina>0
解得：sina=√(1/5)=√5/5
所以：sina=√5/5

Answer 4

f(x)=sin(2x-π/6)
T=2π/2=π
f(0)=sin(2*0-π/6)
=sinπ/6
=1/2
f(a+π/3)=3/5
sin(2(a+π/3)-π/6)=3/5
sin(2a+π/2)=3/5
cos2a=3/5
1-2sin^2a=3/5
2sin^2a=2/5
sin^2a=4/5
sina=2√5/5 (a是第一象限角)

Answer 5

最小正周期T=π
f(0)=-0.5
sin143°=3/5，a+60°=143°，a=83°