Sin2α=1/4,α属于(π/4,π/2),则cosα-sinα等于?

Question

Sin2α(不是平方!)=1/4,α属于(π/4,π/2),则cosα-sinα等于?

要非常详细!
(cosα-sinα)^2 怎么会出现这个?
我是小白,所以大家解答的清楚点,谢谢!

(cosα-sinα)^2=1-2sinαcosα=3/4 1是怎么来的? 我拆开变成cos^2α+sin^2α+2sinαcosα

Answer 1

令cosα-sinα=A
两边平方
A^2=1-2cosa*sina=1-sin2a=1-1/4=3/4
又在α属于(π/4,π/2),则cosα-sinα为负数
=-根号3/2

Answer 2

(cosa-sina)^2=cos^2a+sin^2a-2sinacosa=1-sin2a=3/4
又α属于(π/4,π/2), 所以sina＞cosa
所以cosa-sina=-√3/2

Answer 3

Sin2α=2sinαcosα
(cosα-sinα)^2=1-2sinαcosα=3/4
α属于(π/4,π/2),则cosα-sinα=负的二分之根号3

Answer 4

(cosα-sinα)^2 = 1 + 2cosαsinα = cosα^2 +sinα^2+ 2cosαsinα
cosα-sinα = (1.5)^(0.5)