已知向量a=(2cosx/2,1),b=(√2sin(x/2+∏/4),-1),令f(x)=a*b1.化简f(x)2.求f(∏/12)的值
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/14 15:47:43
![已知向量a=(2cosx/2,1),b=(√2sin(x/2+∏/4),-1),令f(x)=a*b1.化简f(x)2.求f(∏/12)的值](/uploads/image/z/5387199-15-9.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%282cosx%2F2%2C1%29%2Cb%3D%EF%BC%88%E2%88%9A2sin%28x%2F2%2B%E2%88%8F%2F4%29%2C-1%29%2C%E4%BB%A4f%28x%29%3Da%2Ab1.%E5%8C%96%E7%AE%80f%28x%292.%E6%B1%82f%28%E2%88%8F%2F12%29%E7%9A%84%E5%80%BC)
x){}KN?V(9BHPS'G34bڏ:M4utrOv/IӨдMJ2{3HƦ4
BC#Z6I*ҧ)Vڧ_`gC/ꁤlՔ-̬le 5A5,TV8 lHM!2-
C2 ]
T
T' !!g|zNy} #Y"t")U@Q
4
J
Hj"(P%l@1 ]
已知向量a=(2cosx/2,1),b=(√2sin(x/2+∏/4),-1),令f(x)=a*b1.化简f(x)2.求f(∏/12)的值
已知向量a=(2cosx/2,1),b=(√2sin(x/2+∏/4),-1),令f(x)=a*b
1.化简f(x)
2.求f(∏/12)的值
已知向量a=(2cosx/2,1),b=(√2sin(x/2+∏/4),-1),令f(x)=a*b1.化简f(x)2.求f(∏/12)的值
1.f(x) = a•b = (2cos(x/2),1)•(√2sin(x/2 + π/4),-1)
= 2√2cos(x/2)sin(x/2 + π/4) - 1
= 2√2 * (1/2) * {sin[(x/2) + (x/2 + π/4)] - sin[(x/2) - (x/2 + π/4)] - 1 (积化和差)
= √2[sin(x + π/4) - sin(-π/4)] - 1
= √2 sin(x + π/4)
2.f(π/12) = √2 sin(π/12 + π/4)
= √2 sin(π/3)
= √6/2