A,B滿足acosx+bsinx=c,且a^2+B^2<>0,A,B終邊也不重合求證;4cos^2(A/2)cos^2(B/2)=(a+c)^2/(a^2+b^2)

熱心網(wǎng)友

acosx+bsinx=c == a*{[cos(x/2)]^2-[sin(x/2)]^2}+b*2*sin(x/2)*cos(x/2)=c*{[cos(x/2)]^2+[sin(x/2)]^2}== (c+a)[tan(x/2)]^2 -2b*tan(x/2) +(c-a) = 0 。。。(1)== A、B為(1)的兩個根== tan(A/2)+tan(B/2) = 2b/(c+a) 。。。(2)tan(A/2)*tan(B/2) = (c-a)/(c+a) 。。。(3)因此：4cos^2(A/2)cos^2(B/2) = 4/{1 +[tan(A/2)]^2}{1 +[tan(B/2)]^2}= 4/{[tan(A/2)+tan(B/2)]^2 +[tan(A/2)*tan(B/2) -1]^2}= 4/{[2b/(c+a)]^2 + [(c-a)/(c+a) - 1]^2}= (a+c)^2/(a^2+b^2)證畢。

熱心網(wǎng)友

有高2 高3 的題嗎?這種題高考不會考的!

熱心網(wǎng)友

a^2+B^20　？