sin(a-b)sin(b-r)-cos(a-b)cos(r-b)
sin(a-b)sin(b-r)-cos(a-b)cos(r-b)
貌似不难,sin(a+b)cos(c-b)-cos(b+a)sin(b-c)sin(a-b)sin(b-c)-cos(a
已知a=(1,cosΘ),b=(1,sinΘ),0∈R
cos(a-b)cosb-sin(a-b)sinb
为什么sin(a+b)-sina=2sin(b/2)cos(a+b/2)
向量a等于(cos阿尔法,sin阿尔法)向量b等于(cos贝塔,sin贝塔)
设向量a=(cos(a+b),sin(a+b)),b=(cos(a-b),sin(a-b)),(括号里的为阿尔法,贝塔)
已知向量a=(sinθ,cosθ)(θ∈R),向量b=(根号下3,3)
若cosθ/a+sinθ/b=1(θ∈R)成立,则一定有
cos(a+b)=cos(a)*cos(b)-sin(a)*sin(b) 用三角形证明
sin(A+B/2)=cos(C/2)
由cos(a+b)=cos a cos b-sin a sin b cos(a-b)=cos a cos b+sin a