已知sin^4a/cos^2b+cos^4a/sin^2b=1,求证:sin^4 b/cos^2 a+cos^4b/si
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已知sin^4a/cos^2b+cos^4a/sin^2b=1,求证:sin^4 b/cos^2 a+cos^4b/sin^2a=1
证明:令 x = (sin a)^2,
y = (sin b)^2,
则 (cos a)^2 = 1-x,
(cos b)^2 = 1-y.
由已知,
(x^2) / (1-y) +[ (1-x)^2 ] / y =1.
即 (x^2) y + (1-y) (1-x)^2 = y (1-y).
所以 0= y (x^2) +(1-y) (1-2x+x^2) -y (1-y)
= x^2 -2 (1-y) x +(1-y)^2
= (x+y-1)^2.
所以 x+y-1 = 0.
即 y = 1-x,
x = 1-y.
所以 (sin a)^2 = (cos b)^2 = x,
(cos a)^2 =(sin b)^2 = y.
所以 (sin b)^4 / (cos a)^2 + (cos b)^4 / (sin a)^2
= y^2 / y + x^2 / x
= y+x
= 1.
= = = = = = = = =
换元法+暴力破解.
y = (sin b)^2,
则 (cos a)^2 = 1-x,
(cos b)^2 = 1-y.
由已知,
(x^2) / (1-y) +[ (1-x)^2 ] / y =1.
即 (x^2) y + (1-y) (1-x)^2 = y (1-y).
所以 0= y (x^2) +(1-y) (1-2x+x^2) -y (1-y)
= x^2 -2 (1-y) x +(1-y)^2
= (x+y-1)^2.
所以 x+y-1 = 0.
即 y = 1-x,
x = 1-y.
所以 (sin a)^2 = (cos b)^2 = x,
(cos a)^2 =(sin b)^2 = y.
所以 (sin b)^4 / (cos a)^2 + (cos b)^4 / (sin a)^2
= y^2 / y + x^2 / x
= y+x
= 1.
= = = = = = = = =
换元法+暴力破解.
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