Question

已知∏/2<α<∏, tanα-1/tanα=3/2 1 求tanα的值.

（2）求 5sin^2 α/2+8sinα/2cosα/2+11cos^2 α/2-8
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√2sin(α-2/∏

Answer 1

(1) 2(tana)^2-2=3tana
2(tana)^2-3tana-2=0
(2tana+1)(tana-2)=0
因∏/2<α<∏,
所以tana=-1/2
(2) 原式=[5+4sina+6cos^2(a/2)-8]/[-√2sin(π/2-a)]
=[4sina+3(cosa+1)-3]/(-√2cosa)
=(4sina+3cosa)/(-√2cosa)
=(-1/√2)(4tana+3)
=(-√2/2)[4*(-1/2)+3]
=-√2/2

Answer 2

2(tana)^2-3tana-1=0
(2tana+1)(tana-1)=0
tana=-1/2 cosa=-2/√5,sina=1/√5
tana=1(舍去）
5(sin(a/2))^2=5(1-cosa)/2
8sin(a/2)cos(a/2)=4sina
11(cos(a/2))^2=11(5+cosa)/2
√2sin(a-90)=-√2cosa
原式=(5/2+√5+4/√5+55/2-22/√5-8) / (2√2/√5)
=（ 11√10-9√2 ）/2

Answer 3

5∫≠