Question

高一数学 2.3.4题

Answer 1

Answer 2

2.
cosa=-3/5,a∈(pi/2,pi)求cos(pi/4-a)
因为cos(a+b)=cosacosb-sinasinb
∴cos(pi/4-a)=cos(a-pi/4)=cosacos(-pi/4)-sinasin(-pi/4)
cos(-pi/4)=-sin(-pi/4)=√2/2
cosa=-3/5,sina=√1-cos²a(第二象限取正)=√1-9/25=4/5
代入得cos(pi/4-a)=(-3/5)√2/3-4/5*(-√2/2)=-√2/5+2√2/5=√2/5

3.
sina=15/17.
cos(a-pi/3)=cosacos(-pi/3)-sinasin(-pi/3)=cosa/2+√3sina/2
sina=15/17,cosa=-√1-15²/17²=-8/17
代入得cos(a-pi/3)=-4/17+√3*15/17*2=(15√3-8)/34

4.
sina=-2/3,a∈(pi,.3pi/2),cosb=3/4,b∈(3pi/2,2pi)求cos(b-a)
cos(b-a)=cosbcosa+sinbsina
cosa=-√1-4/9=-√5/3,sinb=-√1-9/16=-√5/4
代入得cos(b-a)=3/4*(-√5/3)+(-√5/4)*(-2/3)
=-√5/4+√5/6
=-√5/12