)利用平面向量数量积证明不等式(x²+y²)(m²+n²)≥(xm+yn)
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)利用平面向量数量积证明不等式(x²+y²)(m²+n²)≥(xm+yn)²
)利用平面向量数量积证明不等式(x²+y²)(m²+n²) ≥ (xm+yn)²
(2)利用平面向量数量积证明cos(α-β)= cosαcosβ + sinαsinβ
)利用平面向量数量积证明不等式(x²+y²)(m²+n²) ≥ (xm+yn)²
(2)利用平面向量数量积证明cos(α-β)= cosαcosβ + sinαsinβ
平面向量u = (x,y)
平面向量v = (m,n)
数量积 u*v = |u||v|cos
u*v |u|^2 * |v|^2 (x²+y²)(m²+n²) ≥ (xm+yn)²
(2) u = (cos a,sin a),a = the angle between x-axis and u
v = (cos b,sin b),b = the angle between x-axis and v.
angle is the angle between u,v.==> = b - a.
数量积u*v = cos a * cos b + sin a * sin b
|u| = (cos^2 a + sin^2 a)^(1/2) = 1
|v| = 1.
u*v = |u||v| cos = cos = cos (b-a)
So,cos(b-a) = cos a cos b + sin a sin
平面向量v = (m,n)
数量积 u*v = |u||v|cos
u*v |u|^2 * |v|^2 (x²+y²)(m²+n²) ≥ (xm+yn)²
(2) u = (cos a,sin a),a = the angle between x-axis and u
v = (cos b,sin b),b = the angle between x-axis and v.
angle is the angle between u,v.==> = b - a.
数量积u*v = cos a * cos b + sin a * sin b
|u| = (cos^2 a + sin^2 a)^(1/2) = 1
|v| = 1.
u*v = |u||v| cos = cos = cos (b-a)
So,cos(b-a) = cos a cos b + sin a sin
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