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2011届高三数学一轮复习 简单的三角恒等变换巩固与练习

巩固 1 1.已知 cos2α=4,则 sin2α=( 1 A.2 5 C.8 3 ∴sin2α=8,故选 D. 3 B.4 3 D.8 )

1 解析:选 D.cos2α=1-2sin2α,∴4=1-2sin2α, 2sin2α cos2α 2. · 等于( 1+cos2α cos2α A.tanα

)

B.tan2α 1 C.1 D.2 1+cos2α 2 2sin2α sin2α 解析:选 B.原式= · cos2α =cos2α=tan2α. 1+cos2α π 3. 已知 α, β, γ∈(0, 且 sinα+sinγ=sinβ, cosβ+cosγ=cosα, 2 ), 则 α-β 的值等于( ) π π A.3 B.-3 π π C.± D . ± 3 6 解析:选 B.sinβ-sinα=sinγ>0,cosα-cosβ=cosγ>0,则(sinβ 1 π -sinα)2+(cosα-cosβ)2=1, 且 β>α, 即 cos(α-β)=2(0<α<β<2), π 则 α-β=-3,故选 B. 4. 定义运算 a b=ab2+a2b, 则 sin15° cos15° 的值是________. 解析:依题意,可知 sin15° cos15° =sin15° cos215° +sin215° cos15° 2 6 = 2 sin30° sin(15° +45° )= 8 . 6 答案: 8 4 5 . ( 原创题 ) 已知 sinθ = 5 ,且 cosθ - sinθ + 1<0 ,则 sin2θ =

________. 4 1 解析:∵sinθ=5,cosθ-sinθ+1<0,即 cosθ<-5, 3 24 ∴cosθ=-5.∴sin2θ=2sinθcosθ=-25. 24 答案:-25 6.化简: . π 2 π 2tan(4-x)· sin (4+x) 1 4 2 2(4cos x-4cos x+1) 解:原式= π sin(4-x) π 2· π · cos2(4-x) cos(4-x) (2cos2x-1)2 = π π 4sin(4-x)cos(4-x) cos22x cos22x 1 = =2cos2x=2cos2x. π 2sin(2-2x) 1 2cos4x-2cos2x+2

练习

π 4 π π 1.若 α∈(2,π),且 sinα=5,则 sin(α+4)+cos(α+4)=( 4 2 A. 5 3 2 C. 5 3 ∴cosα=-5, π π π ∴sin(α+4)+cos(α+4)= 2sin(α+2) 4 2 B.- 5 3 2 D.- 5

)

4 π 解析:选 D.∵sinα=5,2<α<π,

3 2 = 2cosα=- 5 . 2.化简 2+cos2-sin21的结果是( ) A.-cos1 B.cos1 C. 3cos1 D.- 3cos1 2 解析:选 C. 2+cos2-sin 1= 1+2cos21-sin21= 3cos1. 2 2 3.已知 sinx-siny=-3, cosx-cosy=3, 且 x, y 为锐角, 则 sin(x +y)的值是( ) A.1 B.-1 1 1 C.3 D.2 2 2 解析:选 A.∵sinx-siny=-3,cosx-cosy=3,两式相加得: sinx+cosx=siny+cosy, ∴sin2x=sin2y.又∵x、y 均为锐角, π ∴2x=π-2y,∴x+y=2,∴sin(x+y)=1. 1 10 π π π 4.若 tanα+tanα= 3 ,α∈(4,2),则 sin(2α+4)的值为( ) 2 2 A.- 10 B. 10 5 2 7 2 C. 10 D. 10 1 10 解析:选 A.由 tanα+tanα= 3 ?(tanα-3)(3tanα-1)=0 得 tanα 1 π π 1 =3 或 tanα=3,由 α∈(4,2)得 tanα>1,故 tanα=3舍去,而 sin(2α 2 2 π 2 sin2α+cos2α 2 2sinαcosα+cos α-sin α +4 )= 2 × =2× ,将分式分 1 cos2α+sin2α 2 π 2 2tanα+1-tan α 2 2 子与分母同除以 cos α 得 sin(2α+4)= 2 × =- 10 . 2 1+tan α 7 5. 已知 cosA+sinA=-13, A 为第四象限角, 则 tanA 等于( ) 12 5 A. 5 B.12 12 5 C.- 5 D.-12 7 解析:选 C.cosA+sinA=-13<0,

∵cosA>0,sinA<0,∴|sinA|>cosA, ∴|tanA|>1.又∵tanA<0,故选 C. π 1 2π 6.若 sin(6-α)=3,则 cos( 3 +2α)=( ) 1 7 A.-3 B.-9 7 1 C.9 D.3 π 1 解析:选 B.∵sin(6-α)=3, π π ∴cos(3-2α)=cos[2(6-α)] π 1 7 =1-2sin2(6-α)=1-2×(3)2=9, 2 π ∴cos(3π+2α)=cos[π-(3-2α)] π 7 =-cos(3-2α)=-9. 7.化简 2sin2x· sinx+cos3x 的结果为________. 解析:原式=2sin2xsinx+cos(2x+x) =2sin2x· sinx+cos2xcosx-sin2x· sinx =cos2x· cosx+sin2x· sinx =cos(2x-x)=cosx. 答案:cosx sinα+cosα 8.若 =3,tan(α-β)=2,则 tan(β-2α)=________. sinα-cosα sinα+cosα tanα+1 解析:∵ = =3,∴tanα=2. sinα-cosα tanα-1 又 tan(α-β)=2, ∴tan(β-2α)=tan[(β-α)-α] =-tan[(α-β)+α] tan(α-β)+tanα 4 =- = . 1-tan(α-β)· tanα 3 4 答案:3 π 3 9. 在△ABC 中, 已知 cos(4+A)=5, 则 cos2A 的值为________. π π π 解析:cos(4+A)=cos4cosA-sin4sinA

2 3 = 2 (cosA-sinA)=5, 3 2 ∴cosA-sinA= 5 >0.① π π ∴0<A<4,∴0<2A<2, 18 7 ①2 得 1-sin2A=25,∴sin2A=25. 24 ∴cos2A= 1-sin22A=25. 24 答案:25 1 5 10.已知 tanα=-3,cosβ= 5 ,α,β∈(0,π). (1)求 tan(α+β)的值; (2)求函数 f(x)= 2sin(x-α)+cos(x+β)的最大值. 5 解:(1)由 cosβ= 5 ,β∈(0,π), 2 5 得 sinβ= 5 ,tanβ=2, tanα+tanβ 所以 tan(α+β)= =1. 1-tanαtanβ 1 (2)因为 tanα=-3,α∈(0,π), 1 3 所以 sinα= ,cosα=- , 10 10 3 5 5 5 2 5 f(x)=- 5 sinx- 5 cosx+ 5 cosx- 5 sinx =- 5sinx, 所以 f(x)的最大值为 5. π π 1 4 11.已知:0<α<2<β<π,cos(β-4)=3,sin(α+β)=5. (1)求 sin2β 的值; π (2)求 cos(α+4)的值. π π π 解:(1)法一:∵cos(β-4)=cos4cosβ+sin4sinβ 2 2 1 = 2 cosβ+ 2 sinβ=3.

2 ∴cosβ+sinβ= 3 . 2 7 ∴1+sin2β=9,∴sin2β=-9. π 法二:sin2β=cos(2-2β) π 7 =2cos2(β-4)-1=-9. π π π 3π π 3π (2)∵0<α<2<β<π,∴4<β-4< 4 ,2<α+β< 2 . π ∴sin(β-4)>0,cos(α+β)<0. π 1 4 ∵cos(β-4)=3,sin(α+β)=5, π 2 2 3 ∴sin(β-4)= 3 ,cos(α+β)=-5. π π ∴cos(α+4)=cos[(α+β)-(β-4)] π π =cos(α+β)cos(β-4)+sin(α+β)sin(β-4) 3 1 4 2 2 8 2-3 =-5×3+5× 3 = 15 .

12.如图,点 P 在以 AB 为直径的半 圆上移动,且 AB=1,过点 P 作圆的切 线 PC,使 PC=1.连结 BC,当点 P 在什 1 么位置时, 四边形 ABCP 的面积等于2? 解:设∠PAB=α,连结 PB. ∵AB 是直径,∴∠APB=90° . 又 AB=1,∴PA=cosα,PB=sinα. ∵PC 是切线,∴∠BPC=α.又 PC=1, ∴S 四边形 ABCP=S△ APB+S△ BPC

[例 1]求经过两点 P1(2,1)和 P2(m,2) (m∈R)的直线 l 的斜率,并且求出 l 的 倾斜角α 及其取值范围. 选题意图:考查倾斜角与斜率之间的关系及斜率公式. 解:(1)当 m=2 时,x1=x2=2,∴直线 l 垂直于 x 轴,因此直线的斜率不存在,倾斜角 α =

? 2

(2)当 m≠2 时,直线 l 的斜率 k= ∴α =arctan

1 ? ,α ∈(0, ) , m?2 2

1 ∵m>2 时,k>0. m?2

∵当 m<2 时,k<0 ∴α =π +arctan

1 ? ,α ∈( ,π ). m?2 2 1 ,m)共线,求 m 的值. 2

说明:利用斜率公式时,应注意斜率公式的应用范围. [例 2]若三点 A(-2,3) ,B(3,-2) ,C(

选题意图:考查利用斜率相等求点的坐标的方法. 解:∵A、B、C 三点共线,

∴kAB=kAC,

?2?3 m?3 ? . 1 3? 2 ?2 2

解得 m=

1 . 2

说明:若三点共线,则任意两点的斜率都相等,此题也可用距离公式来解. [例 3]已知两点 A(-1,-5),B(3,-2),直线 l 的倾斜角是直线 AB 倾斜角的一半,求直 线 l 的斜率. 选题意图:强化斜率公式. 解:设直线 l 的倾斜角α ,则由题得直线 AB 的倾斜角为 2α . ∵tan2α =kAB=

? 2 ? (?5) 3 ? . 3 ? (?1) 4

?

2 tan ? 3 ? 2 1 ? tan ? 4 1 或 tanα =-3. 3

即 3tan2α +8tanα -3=0, 解得 tanα = ∵tan2α =

3 >0,∴0°<2α <90°, 4

0°<α <45°, ∴tanα =

1 . 3 1 3

因此,直线 l 的斜率是

说明:由 2α 的正切值确定α 的范围及由α 的范围求α 的正切值是本例解法中易忽略 的地方.

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