= cos nθ
解
cos θ 1 D n = 2 cos θ D n 1 + ( 1) 2 n 1 ... 0
= 2 cos θ cos( n 1)θ [cos θ cos( n 1)θ + sin θ sin( n 1)θ ] = cos θ cos( n 1)θ sin θ sin( n 1)θ = cos n θ
= 2 cos θ D n 1 D n 2 = 2 cos θ cos( n 1)θ cos( n 2 )θ
练习题一、填空题
1
x 3 2
2 x 1
3 1 = x
———
2
x1 a1 D4 = a2 a3 x2 x2 x3
=
3
设
x 3 1 y 0 1 =1 z 2 11 1
,则
x 3 5 1
y 3 z 3 2 1 4 1 =
1
x 1 1 = 1 1
1 1 x +1 4 D4 = 1 1 x 1 x +1 1 1
5
3 1
3
2A12 + A22 + A32 + A42 =
3 3 1 1 D4 = 3 0 1 2 3 1A11 + A21 =
2
2
二
计算题
a1 1. b40 1 1 0 2 3
b1 a2 b3 b2 a3 a43 2 4 0
2. 2 3 0 2 1 3 2 2 0 5 4 0 1 5 0
3
0 1 0L M M 0 8 7a1 b1 c1
0 O
2 7 L 1a3 b3 = 5 c3
La2 b2 c2
3a1则 3a 2
b1 c1 b2 c2 b3 c3
2c1 2c2 = 2c3
4
设
3a3
λ=
x 5 求 f ( x) = 2 x
x 1 x 4 2 x 1 2 x 3 = 0 的根的个数是多少？
3x 3x 3 3x 5
6
λx1 + x 2 + x 3 = 0 齐次线性方程组 x1 + x 2 + x 3 = 0 只有零解, x + λx + x = 0 2 3 1λ=
则