www.22991277.com:2019年试题一轮优化探究理数 苏教版 第四章 第四节 两角和与差的正弦、余弦、正切公式

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一、填空题 3 π π 5π 1.若 sin α=5,α∈(-2,2),则 cos(α+ 4 )=________. π π 3 4 解析:∵α∈(-2,2),sin α=5,∴cos α=5, 5π 2 2 ∴cos(α+ 4 )=- 2 (cos α-sin α)=- 10 . 答案:- 2 10

1-cos 2α 1 2.已知sin αcos α =1,tan(β-α)=-3,则 tan(β-2α)=________. 1-cos 2α 解析:依题意由 sin αcos α =1 2sin2 α 1 得sin αcos α=1,则 tan α=2, 从而 tan(β-2α)=tan[(β-α)-α] tan?β-α?-tan α = 1+tan?β-α?· tan α 1 1 -3-2

=- 1 1=-1. 1+?-3?×2 答案:-1 π 3 π 2 3.已知 tan(α-6)=7,tan(6+β)=5,则 tan(α+β)的值为________. π π 解析:tan(α+β)=tan [(α-6)+(6+β)] 3 2 7+5 = = π π 3 2=1. 1-tan?α-6?· tan?6+β? 1-7×5 答案:1 4.在等式 cos(*)(1+ 3tan 10° )=1 的括号中,填写一个锐角,使得等式成立,
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π π tan?α-6?+tan?6+β?

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这个锐角的度数是________. + 3sin 10° 2sin?30° +10° ? 3sin 10° cos 10° 解 析 : 1 + 3 tan 10° = 1 + cos 10° = = cos 10° cos 10° = 2sin 40° ,所以填 40° . cos 10° 答案:40° 6 5.设 a=sin 14° +cos 14° ,b=sin 16° +cos 16° ,c= 2 ,则 a、b、c 的大小关系 是________. 3 解析:∵a2=1+2sin 14° cos 14° =1+sin 28° ∈(1,2),b2=1+2sin 16° cos 16° =1 3 3 +sin 32° ∈(2,2),c2=2,且 a>0,b>0,c>0,∴a<c<b. 答案:a<c<b 5 10 6.已知 A、B 均为钝角,且 sin A= 5 ,sin B= 10 ,则 A+B 等于________. 2 5 3 10 解析:由已知可得 cos A=- 5 ,cos B=- 10 , 2 ∴cos(A+B)=cos Acos B-sin Asin B= 2 , π π 又∵2<A<π,2<B<π, 7π ∴π<A+B<2π,∴A+B= 4 . 7π 答案: 4 2 π 1 π 7.若 tan(α+β)= , tan(β- )= ,则 tan (α+ )=______. 5 4 4 4 π π 解析:tan(α+4)=tan [(α+β)-(β-4)] 2 1 5-4 3 = π= 2 1=22. 1+tan?α+β?tan?β-4? 1+5×4 π tan?α+β?-tan?β-4?

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3 答案:22 3π 3 π 12 π 8.已知 α,β∈( 4 ,π),sin(α+β)=-5,sin(β-4)=13,则 cos(α+4)=________. 3π 3π π π 3π 4 解析: 由于 α, β∈ ( 4 , π), 所以 2 <α+β<2π, < β - < , 故 cos( α + β ) = cos(β 2 4 4 5, π 5 π π 4 5 3 12 -4)=-13,cos(α+4)=cos[(α+β)-(β-4)]=5×(-13)+(-5)×13 56 =-65. 56 答案:-65 π 9. 非零向量 a=(sin θ, 2), b=(cos θ, 1), 若 a 与 b 共线, 则 tan(θ-4)=________. 解析:因为非零向量 a,b 共线,所以 a=λb,即(sin θ,2)=λ(cos θ,1),所以 λ π tan θ-1 1 =2,sin θ=2cos θ,得 tan θ=2,所以 tan(θ-4)= = . 1+tan θ 3 1 答案:3 二、解答题 π 10.已知 α 为锐角,且 tan(4+α)=2. (1)求 tan α 的值; (2)求 sin 2αcos α-sin α 的值. cos 2α

1+tan α 1+tan α π 解析:(1)tan(4+α)= ,所以 =2, 1-tan α 1-tan α 1+tan α=2-2tan α, 1 所以 tan α=3. sin 2αcos α-sin α 2sin αcos2α-sin α (2) = cos 2α cos 2α

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sin α?2cos2α-1? sin αcos 2α = = cos 2α =sin α. cos 2α 1 因为 tan α=3,所以 cos α=3sin α,又 sin2α+cos2α=1, 1 所以 sin2 α=10, 10 又 α 为锐角,所以 sin α= 10 , sin 2αcos α-sin α 10 所以 = . cos 2α 10 11.如图所示,在平面直角坐标系 xOy 中,以 Ox 轴为始边作两 个锐角 α,β,它们的终边分别与单位圆交于 A、B 两点,已知 A、 2 2 5 B 的横坐标分别为 10 、 5 . (1)求 tan(α+β)的值; (2)求 α+2β 的值. 2 2 5 解析:由已知条件得 cos α= 10 ,cos β= 5 . ∵α、β 为锐角,∴sin α= sin β= 7 2 1-cos2α= 10 ,

5 1 1-cos2β= 5 ,因此 tan α=7,tan β=2. tan α+tan β 1-tan α· tan β 1 7+2

(1)tan(α+β)=



1=-3. 1-7×2

(2)∵tan 2β=

2tan β 4 = = , 2 1-tan β 1-?1?2 3 2 4 7+3

1 2×2

∴tan(α+2β)=

4=-1. 1-7×3

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3π ∵α,β 为锐角,∴0<α+2β< 2 , 3π ∴α+2β= 4 . → 12.已知向量OA=(cos α,sin α)(α∈[-π,0]).向量 m=(2,1),n=(0,- 5), → 且 m⊥(OA-n). (1)求 tan α 的值; 2 (2)若 cos(β-π)= 10 ,且 0<β<π,求 cos(2α-β). → 解析:(1)∵OA=(cos α,sin α), → ∴OA-n=(cos α,sin α+ 5), → → ∵m⊥(OA-n),∴m· (OA-n)=0, 即 2cos α+(sin α+ 5)=0,① 又 sin2α+cos2α=1,② 由①②联立方程组解得, 2 5 5 cos α=- 5 ,sin α=- 5 . sin α 1 ∴tan α=cos α=2. 2 (2)∵cos(β-π)= 10 , 2 即 cos β=- 10 ,0<β<π, 7 2 π ∴sin β= 10 ,2<β<π, 5 2 5 4 又∵sin 2α=2sin αcos α=2×(- 5 )×(- 5 )=5, 4 3 cos 2α=2cos2α-1=2×5-1=5,

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∴cos(2α-β)=cos 2αcos β+sin 2αsin β 3 2 4 7 2 2 =5×(- 10 )+5× 10 = 2 .

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