两道高一数学题,详过程
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两道高一数学题,详过程
求下列函数的值域:
y=sinxcos(x+π/3)
已知sina/(1+cosa)=3,则cosa-sina=_____
答案是 -7/5.
求下列函数的值域:
y=sinxcos(x+π/3)
已知sina/(1+cosa)=3,则cosa-sina=_____
答案是 -7/5.
![两道高一数学题,详过程](/uploads/image/z/18399960-0-0.jpg?t=%E4%B8%A4%E9%81%93%E9%AB%98%E4%B8%80%E6%95%B0%E5%AD%A6%E9%A2%98%2C%E8%AF%A6%E8%BF%87%E7%A8%8B)
[-sin1,sin1]
2sin(a/2)cos(a/2)/(1+2cos^2(a/2)-1)=tan(a/2)=3
cosa-sina=2cos^2(a/2)-1-2sin(a/2)cos(a/2)
=2cos^2(a/2)/(sin^2(a/2)+cos^2(a/2))-1-2sin(a/2)cos(a/2)/(sin^2(a/2)+cos^2(a/2))
=2/(tan^2(a/2)+1)-1-2tan(a/2)/(tan^2(a/2)+1)
=2/(3^2+1)-1-2*3/(3^2+1)
=1/5-1-3/5
=-7/5
2sin(a/2)cos(a/2)/(1+2cos^2(a/2)-1)=tan(a/2)=3
cosa-sina=2cos^2(a/2)-1-2sin(a/2)cos(a/2)
=2cos^2(a/2)/(sin^2(a/2)+cos^2(a/2))-1-2sin(a/2)cos(a/2)/(sin^2(a/2)+cos^2(a/2))
=2/(tan^2(a/2)+1)-1-2tan(a/2)/(tan^2(a/2)+1)
=2/(3^2+1)-1-2*3/(3^2+1)
=1/5-1-3/5
=-7/5