（1）sin^4α-cos^4α=2sin^2α-1 （2）tan^2α-sin^2α=tan^2αsin^2α （3）cosα/1-sinα=1+sinα/cos（1）sin^4α-cos^4α=2sin^2α-1 （2）tan^2α-sin^2α=tan^2αsin^2α （3）cosα/1-sinα=1+sinα/cosα求证
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（1）sin^4α-cos^4α=2sin^2α-1 （2）tan^2α-sin^2α=tan^2αsin^2α （3）cosα/1-sinα=1+sinα/cos（1）sin^4α-cos^4α=2sin^2α-1 （2）tan^2α-sin^2α=tan^2αsin^2α （3）cosα/1-sinα=1+sinα/cosα求证
（1）sin^4α-cos^4α=2sin^2α-1 （2）tan^2α-sin^2α=tan^2αsin^2α （3）cosα/1-sinα=1+sinα/cos
（1）sin^4α-cos^4α=2sin^2α-1
（2）tan^2α-sin^2α=tan^2αsin^2α
（3）cosα/1-sinα=1+sinα/cosα
求证
（1）sin^4α-cos^4α=2sin^2α-1 （2）tan^2α-sin^2α=tan^2αsin^2α （3）cosα/1-sinα=1+sinα/cos（1）sin^4α-cos^4α=2sin^2α-1 （2）tan^2α-sin^2α=tan^2αsin^2α （3）cosα/1-sinα=1+sinα/cosα求证
(1) 设 x= (sin α)^2,
则 (cos α)^2 = 1-x.
所以 (sin α)^4 - (cos α)^4
= x^2 - (1-x)^2
= 2x -1
= 2(sin α)^2 -1.
(2) 设 x= (sin α)^2,
则 (cos α)^2 = 1-x,
(tan α)^2 = x / (1-x).
所以 (tan α)^2 -(sin α)^2 = x / (1-x) -x
= x^2 / (1-x).
=(tan α)^2 * (sin α)^2.
= = = = = = = = = =
换元法+暴力破解, sin 神马的都是浮云.
当sin, cos, tan等都带平方时可用, 因为不用考虑正负问题.
(3) 因为 (cos α)^2 = 1 -(sin α)^2
= (1 +sin α) (1 -sin α).
所以 cos α / (1 -sin α) = (1 +sin α) / cos α.
= = = = = = = = =
如果要证明 a /b = c /d,
可先证明 ad =bc.
不等式也类似, 不过要注意正负问题.