Question

已知sina+sinb=1,cosa+cosb=0,求cos(a+b),sin(a-b),sina(a+b)

Answer 1

sina+sinb=1
(sina)^2=(1-sinb)^2
cosa+cosb=0
(cosa)^2=(-cosb)^2
两式相加得
(sina)^2+(cosa)^2=(1-sinb)^2+(-cosb)^2
1=(sinb)^2+(cosb)^2+1-2sinb
2sinb=1
sinb=1/2
所以(cosb)^2=1-1/4=3/4
sina=1-sinb=1/2

cos(a+b)=cosacosb-sinasinb=-(cosb)^2-3/2*(-1/2)=1/2
sin(a-b)=sinacosb-cosasinb=cosb(sina+sinb)=±√3/2*1=±√3/2
sin(a+b)=sinacosb+cosasinb=cosb(sina-sinb)=±√3/2*0=0

Answer 2

为(sinc)^2=(sina+sinb)^2=(sina)^2+2sina*sinb+(sinb)^2,
同理(cosc)^2=(cosa)^2+2cosa*cosb+(cosb)^2,
所以相加得1=1+2(cosa*cosb+sina*sinb)+1,
公式cos(a-b)=cosa*cosb+sina*sinb
∴所以相加得1=1+2cos(a-b)+1,
2cos(a-b)=-1
所以cos(a-b)=-1/2。

Answer 3

解：由sina+sinb=1
cosa+cosb=0
(sina)^2+(cosa)^2=1
(sinb)^2+(cosb)^2=1

联立以上方程解得：sina=sinb=1/2
当cosa=√3/2时，cosb=-√3/2
当cosa=-√3/2时cosb=√3/2

cos(a+b)=cosa*cosb-sina*sinb

=(√3/2)*(-√3/2)-(1/2)^2=1/2
或
cos(a+b)=cosa*cosb-sina*sinb
=(-√3/2)*(√3/2)-(1/2)^2=1/2

sin(a-b)=sinacosb-cosasinb=
(1/2)*(√3/2)-(-√3/2)*1/2=√3/2
或
sin(a-b)=sinacosb-cosasinb
=(1/2)*(-√3/2)-(√3/2)*1/2=-√3/2

sina(a+b)=sinacosb+cosasinb
=1/2*(√3/2)+(-√3/2)*1/2=0
或
sina(a+b)=sinacosb+cosasinb
=1/2*(-√3/2)+(√3/2)*1/2=0
)

Answer 4

晕！你高一怎么学的，这都不会？