Question

两道简单的高一数学题

已知(1/cosA)-(1/sinA)=1则sin2A值 已知5cos(A-(B/2)+7cos(B/2)=0则tan(A/2)tan(A-B/2)=

Answer 1

1)
1/cosa-1/sina=1
(sina-cosa)/cosa*sina=1
sinacosa=sina-cosa
-----------(1)
另外sina^2
+
cosa^2
=
1
------(2)
由(2)-2*(1)
sina^2+cosa^2-2ainacosa=1-2(sina-cosa)
(sina-cosa)^2+2(sina-cosa)-1=0
sinacosa=sina-cosa
=
根号2-1
sin2a=2sina*cosa
=
2根号2-2
2)用积化差做,5cos(x/2)*cos(x-y)/2=-cos(y/2),tg化sin/cos形式,[sin(x/2)*sin(x-y)/2]/-cos(y/2),sin(x/2)*sin(x-y)/2]化cos形式(积化差)算答案-6即用积化差做,5cos(x/2)*cos(x-y)/2=-cos(y/2),tg化sin/cos形式,[sin(x/2)*sin(x-y)/2]/-cos(y/2),sin(x/2)*sin(x-y)/2]化cos形式(积化差)算答案-6即答案
x,y即A,B
我帮找
理解吧
2倍根号2-2
绝确啊