Question

急求一道数学题

已知sin(q+π/3)=1/3,q∈(0,π)则cos(q+π/12)

Answer 1

q∈(0,π) q+π/3∈(π/3,4π/3)
0<sin(q+π/3)=1/3<1/2 所以 5π/6<q+π/3<π

cos(q+π/3)=-(2√2)/3
cos(q+π/12)=cos(q+π/3-π/4)=cos(q+π/3)cos(π/4)+sin(q+π/3)sin(π/4)
=-(2√2)/3 （√2/2）+(1/3)(√2/2)=-2/3+√2/6=(√2-4)/6

Answer 2

解：∵sin(q+π/3)=1/3,q∈(0,π)
　∴cos(q+π/12)
　　=cos(q+4π/12-3π/12)
　　=cos(q+π/3)cos(π/4)+sin(q+π/3)sin(π/4)
　∵sin(q+π/3)=1/3<1
　∴2/π<q+π/3<π
　∴cos(q+π/3)=-(2√2)/3
　∴cos(q+π/12)
　　=cos(q+π/3)cos(π/4)+sin(q+π/3)sin(π/4)
　　=-(2√2)/3×√2/2+1/3×√2/2
　　=-2/3+√2/6
　　=(√2-4)/6

Answer 3

q+π/3=arcsin(1/3) q=arcsin(1/3)-π/3<0 不满足q∈(0,π)
所以q+π/3=π-arcsin(1/3) q=π-arcsin(1/3)-π/3

cos(π-arcsin(1/3)-π/3+π/12)
= -cos(arcsin(1/3)+π/3-π/12)
= -cos(arcsin(1/3)+π/4)
= -[cos(arcsin(1/3)cos(π/4)-sin(arcsin1/3)sinπ/4]
= -（4-根号2）/6

Answer 4

cos（a-b）=cosa cosb+sina sinb
sin(q+π/3)=1/3,q∈(0,π)得cos(q+π/3)=（-？）

cos(q+π/12)=cos(q+π/3-π/4)=cos(q+π/3）cosπ/4+sin(q+π/3）sinπ/4=？
自己算哈啊，具体值我都忘了···

Answer 5

上半年平均数是（50+51+48+50+52+49）/6=50
设增长率X
50X2=72
X=1.2