例1在△ABC中,已知cosA =,sinB =
,则cosC的值为…………(A)
A
B
C
D
解:∵C = p - (A + B) ∴cosC = - cos(A + B)
又∵AÎ(0, p) ∴sinA = 而sinB =
显然sinA > sinB
∴A
> B 即B必为锐角 ∴ cosB =
∴cosC
= - cos(A + B) = sinAsinB -
cosAcosB =
例2在△ABC中,ÐC>90°,则tanAtanB与1的关系适合………………(B)
A tanAtanB>1 B
tanAtanB>1 C
tanAtanB =1 D
不确定
解:在△ABC中 ∵ÐC>90° ∴A, B为锐角 即tanA>0, tanB>0
又:tanC<0 于是:tanC
= -tan(A+B) = <0
∴1 - tanAtanB>0 即:tanAtanB<1
又解:在△ABC中 ∵ÐC>90° ∴C必在以AB为直径的⊙O内(如图)
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设CD = h,C’D = h’,AD = p,BD = q,
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例3已知,
,
,
,
求sin(a + b)的值
解:∵ ∴
又
∴
∵
∴
又
∴
∴sin(a + b) = -sin[p + (a + b)] =
例4已知sina +
sinb = ,求cosa + cosb的范围
解:设cosa + cosb = t,
则(sina +
sinb)2 + (cosa + cosb)2 = + t2
∴2
+ 2cos(a - b) = + t2
即
cos(a - b) = t2 -
又∵-1≤cos(a - b)≤1 ∴-1≤t2 -
≤1
∴≤t≤
例5设a,bÎ(,
),tana、tanb是一元二次方程
的两个根,求 a + b
解:由韦达定理:
∴
又由a,bÎ(,
)且tana,tanb < 0 (∵tana+tanb<0,
tanatanb >0)
得a + bÎ (-p, 0) ∴a + b =
例6 已知sin(p - a) - cos(p + a) =(0<a<p),求sin(p + a) + cos(2p - a)的值
解:∵sin(p - a) - cos(p + a) = 即:sin a + cos a =
①
又∵0<<1,0<a<p
∴sina>0, cosa<0
令a = sin(p + a) + cos(2p - a) = - sina + cosa 则 a<0
由①得:2sinacosa =
例7 已知2sin(p - a) - cos(p + a) = 1 (0<a<p),求cos(2p - a) + sin(p + a)的值
解:将已知条件化简得:2sin a + cos a = 1 ①
设cos(2p - a) + sin(p + a) = a , 则 a = cos a - sin a ②
①②联立得:
∵sin2a + cos2a =
1 ∴
∴5a2 + 2a - 7 = 0,
解之得:a1 = , a2 = 1(舍去)(否则sina = 0,
与0<a<p不符)
∴cos(2p - a) + sin(p + a) =