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17-18版 第5章 第22课 同角三角函数的基本关系及诱导公式

第 22 课 同角三角函数的基本关系及 诱导公式 [最新考纲] 内容 A 同角三角函数的基本关系 三角函数的诱导公式 要求 B √ √ C 1.同角三角函数的基本关系式 (1)平方关系 sin2α+cos2α=1; (2)商数关系 sin α tan α=cos α. 2.诱导公式 组序 角 正弦 余弦 正切 口诀 记忆规律 一 2kπ+α(k∈Z) sin α cos α tan α 二 π+α -sin α -cos α tan α 三 -α -sin α cos α -tan α 四 π-α sin α -cos_α -tan_α 函数名改变符 号看象限 五 π 2-α cos α sin α 六 π 2+α cos_α -sin α 函数名不变,符号看象限 奇变偶不变,符号看象限 1 1.(思考辨析)判断下列结论的正误.(正确的打“√”,错误的打“×”) (1)若 α,β 为锐角,则 sin2α+cos2β=1.( sin α (2)若 α∈R,则 tan α=cos α恒成立.( ) ) ) (3)sin(π+α)=-sin α 成立的条件是 α 为锐角.( (4)诱导公式的记忆口诀中“奇变偶不变,符号看象限”,其中的奇、偶是 π 指2的奇数倍、偶数倍,变与不变指函数名称是否变化.( [答案] (1)× (2)× (3)× (4)√ ) 5 2.(教材改编)已知 α 是第二象限角,sin α=13,则 cos α 等于________. 12 -13 5 [∵sin α=13,α 是第二象限角, 12 1-sin2α=-13.] ∴cos α=- 1 3.若 tan α=2,则 sin4α-cos4α 的值为________. 3 -5 3 =-5.] 4.(2016· 四川高考)sin 750° =________. 1 2 1 [sin 750° =sin(750° -360° ×2)=sin 30° =2.] [sin α - cos α = (sin α - cos α)(sin α + cos α) = 4 4 2 2 2 2 sin2α-cos2α sin2α+cos2α = tan2α-1 tan2α+1 π? ?π ? 3 ? 5.已知 sin?2+α?=5,α∈?0,2?,则 sin(π+α)=________. ? ? ? ? - 4 5 π? 3 ?π ? ? [因为 sin?2+α?=cos α= ,α∈?0,2?,所以 sin α= 5 ? ? ? ? 4 1-cos2α= , 5 4 所以 sin(π+α)=-sin α=-5.] 同角三角函数基本关系式的应用 2 1 5π 3π (1)已知 sin αcos α=8, 且 4 <α< 2 , 则 cos α-sin α 的值为________. 3 (2)(2016· 全国卷Ⅲ改编)若 tan α=4,则 cos2α+2sin 2α=________. 3 (1) 2 64 (2)25 5π 3π [(1)∵ 4 <α< 2 , ∴cos α<0,sin α<0 且 cos α>sin α, ∴cos α-sin α>0. 1 3 又(cos α-sin α)2=1-2sin αcos α=1-2×8=4, 3 ∴cos α-sin α= 2 . 3 (2)∵tan α=4, 则 cos2α+2sin 2α= 64 25.] [规律方法] 1.利用 sin2α+cos2α=1 可以实现角 α 的正弦、余弦的互化,利 sin α 用cos α=tan α 可以实现角 α 的弦切互化. 2.应用公式时要注意方程思想的应用:对于 sin α+cos α,sin αcos α,sin α -cos α 这三个式子,利用(sin α± cos α)2=1± 2sin αcos α,可以知一求二. 3.注意公式逆用及变形应用:1=sin2α+cos2α,sin2α=1-cos2α,cos2α=1 -sin2α. [变式训练 1] (1)已知 sin α-cos α= 2,α∈(0,π),则 tan α 等于________. 【导学号:62172123】 cos2α+4sin αcos α sin2α+cos2α 1+4tan α tan2α+1 3 1+4×4 = = 3 = ? ?2 ?4? +1 ? ? 3 -1 ? ?sin α-cos α= 2, [由? 2 2 ? ?sin α+cos α=1, 消去 sin α 得:2cos2α+2 2cos α+1=0, 即( 2cos α+1)2=0, 2 ∴cos α=- 2 . 3π 又 α∈(0,π),∴α= 4 , 3π ∴tan α=tan 4 =-1.] ? π? 1 (2)设 θ 为第二象限角,若 tan?θ+4?=2,则 sin θ+cos θ=________. ? ? 10 - 5 1+tan θ 1 1 ? π? 1 [∵tan?θ+4?=2,∴ =2,解得 tan θ=-3. ? ? 1-tan θ 2 ∴(sin θ+cos θ) = sin2θ+cos2θ+2sin θ· cos θ sin2θ+cos2θ 1 2 9-3+1 2 = 1 =5. 9+1 = tan2θ+2tan θ+1 tan2θ+1 1 ∵θ 为第二象限角,tan θ=-3, 3π ∴2kπ+ 4 <θ<2kπ+π,∴sin θ+cos θ<0, 10 ∴sin θ+cos θ=- 5 .] 诱导公式的应用 (1)已知 A= ________. ? π? 1 ? 5π? ?π ? (2)(2017· 南通一模 ) 已知 sin ?x+6? = 3 ,则 sin ?x- 6 ? + sin2 ?3-x? 的值是 ? ? ? ? ? ? 4 sin?kπ+α? cos?kπ+α? sin α + cos α (k∈Z

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