1987 CE AMATH PAPER2 #12b
α and β are two acute angles satisfying the equation
6cos²θ– 5cosθ+ 1 = 0
Without solving the equation, show that
cos ( α+β/2 ) + cos ( α-β/2 ) = √2
ANS:
α+β =5/6 , αβ=1/6
cos ( α+β/2 ) + cos ( α-β/2 )
= 2cosα/2 cosβ/2
= 2[1/2( 1+ cosα)]^0.5[1/2(1+ cosβ)]^0.5
= [( 1+ cosα) (1+ cosβ)]^0.5
= [1+( cosα+ cosβ )+ cosαcosβ ]^0.5
= (1+5/6+1/6)^0.5 ← 點解cosα+ cosβ=5/6 同 cosαcosβ=1/6 ?
= 2^0.5
Q.2
1995 CE AMATH PAPER2 #9a
Show that cos²A - cos²B = sin(A+B)sIn(B-A)
ANS:
cos²A - cos²B
= (cosA + cosB) (cosA - cosB)
= [ 2cos(A+B/2) cos(A-B/2)][-2sin(A+B/2)sin(A-B/2)]
= [2sin(A+B/2)cos(A+B/2)][-2 sin(A-B/2) cos(A-B/2)]
= -sin(A+B) sin(A-B) ← 點樣轉成依步 ?
= sin(A+B) sin(B-A) ← 點解個負冇左 就變成咁 ?
更新1:
alpha and beta are two acute angles satisfying the equation
更新2:
alphaand beta are two acute angles satisfying the equation
更新3:
alpha and beta are two acute angles satisfying the equation
更新4:
點解 cos α 及 cos β 是其兩根 ?
