需要詳解，我想要理解自己盲點

1.tan+cot=4
=>sin/cos + cos/sin =4
=> 1/sin*cos= 4
=> sin*cos =1/4

sin^4+cos^4
=(sin^2+ cos^2)^2 -2sin^2*cos^2
= 1^2- 2(1/4)^2
= 1 - 1/8 = 7/8

2.
n
Σ a =n平方-99，
k=1
-----------------------------
10
Σa =
k=6
------------------------------
10
Σa 減掉
k=1
5
Σa = (10^2-99)-(5^2-99) =100-25=75
k=1

3.3球至少一球是白球的機率
=1 -(隨機取3球，3球都不是白的機率)=1-1/12=11/12
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3球都不是白的機率=(紅球2顆黃球3顆中任取3顆)/(10顆球中任取3顆)
=C(5,3)/C(10,3)=10/120=1/12

tan(x)+cot(x)=4
令 y=tan(x), 則 y+1/y=4, 故 y=2±√3.
即: tan(x) = 2±√3
sec^2(x) = 1/cos^2(x) = 1+tan^2(x) = 8±4√3
故 cos^2(x) = 1/(8±4√3) = (1/2)±(1/4)√3
sin^4(x)+cos^4(x) = (sin^2(x)+cos^2(x))^2-2sin^2(x)cos^2(x)
= 1-2(1-cos^2(x))cos^2(x) = 1-2[(1/2)^2-(√3/4)^2] = 7/8.

2012-12-30 01:06:35 補充：
紅球2顆 黃球3顆 白球5顆 隨機取3球 3球至少一球是白球的機率為?

P(至少一白球) = 1-P(0白球) = 1-C(5,0)C(2+3,3)/C(2+3+5,3)
= 1-C(5,3)/C(10,3) = 11/12.