maths

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2013-08-21 09:19:56 補充：
CORRECTION

Q1 Line 3

(cos²Acos²B - sin²Acos²B) + ...

Let tan A = a, so tan B = sqrt(a^2 - 1)/sqrt 2.
By Pythagoras thm.,
cos A = 1/sqrt ( a^2 + 1) and sin A = a/sqrt ( a^2 + 1).
cos B = sqrt 2/sqrt ( a^2 + 1) and sin B = sqrt ( a^2 - 1)/sqrt ( a^2 + 1).
cos 2A = cos^2 A - sin^2 A = 1/(a^2 + 1) - a^2/(a^2 + 1) = (1 - a^2)/(a^2 + 1)
sin^2 B = (a^2 - 1)/(a^2 + 1)
so cos 2A + sin^2 B = (1 - a^2)/(a^2 + 1) + (a^2 - 1)/(a^2 + 1) = 0.

2013-08-20 21:35:58 補充：
cos4A = cos2(2A) = 1 - 2sin^2 ( 2A) = 1 - 2 ( 2 sin A cos A)^2 = 1 - 8 sin^2 A cos^2 A
= 1 - 8 sin^2A ( 1 - sin^2 A) = 1 - 8 sin^2 A + 8 sin^4 A.
4cos2A = 4(1 - 2 sin^2 A) = 4 - 8 sin^2 A
so cos4A - 4cos2A + 3 = 1 - 8 sin^2 A + 8 sin^4 A - 4 + 8 sin^2 A + 3 = 8 sin^4 A.

2013-08-20 21:40:27 補充：
Using the above result and put x = 40 degree.
8 sin^4 40 = cos4(40) - 4 cos2(40) + 3 = cos 160 - 4 cos 80 + 3
= cos (180 - 20) - 4 cos 80 + 3 = - cos 20 - 4 cos 80 + 3
8 sin^4 40 + cos 20 + 4 cos 80 = 3
so 4 sin^4 40 + (cos 20)/2 + 2 cos 80 = 3/2.