www4048jcom:高中数学人教A版必修四教师用书:第3章+3.2+简单的三角恒等变换

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3.2 简单的三角恒等变换 学习目标:1.能用二倍角公式导出半角公式,体会其中的三角恒等变换的基 本思想方法,以及进行简单的应用.(重点)2.了解三角恒等变换的特点、变换技 巧, 掌握三角恒等变换的基本思想方法, 能利用三角恒等变换对三角函数式化简、 求值以及三角恒等式的证明和一些简单的应用.(难点、易错点) [自 主 预 习· 探 新 知] 半角公式 α (1)sin2=± α (2)cos2=± α (3)tan2=± α (4)tan2= 1-cos 2 α , 1+cos α , 2 1-cos α , 1+cos α α α α sin 2 sin2· 2cos2 sin α = = α α α 1+cos α, cos2 cos2· 2cos2 α α sin2· 2sin2 α tan2= 1-cos α α= α α= sin α . cos2 cos2· 2sin2 [基础自测] α sin2 1.思考辨析 α (1)cos 2= 1+cos α .( 2 ) ) ) ) α 1 (2)存在 α∈R,使得 cos 2=2cos α.( α 1 (3)对于任意 α∈R,sin 2=2sin α 都不成立.( α (4)若 α 是第一象限角,则 tan 2= [ 解析 ] 1-cos α .( 1+cos α π α π (1)×. 只有当- 2 + 2kπ≤ 2 ≤ 2 + 2kπ(k ∈ Z) ,即- π + 4kπ≤α≤π + α 4kπ(k∈Z)时,cos 2= 1+cos α . 2 (2)√.当 cos α=- 3+1 时,上式成立,但一般情况下不成立. (3)×.当 α=2kπ(k∈Z)时,上式成立,但一般情况下不成立. α α (4)√.若 α 是第一象限角,则2是第一、三象限角,此时 tan 2= 成立. [答案] (1)× (2)√ (3)× (4)√ ) 1-cos α 1+cos α α 2.已知 180° <α<360° ,则 cos2的值等于( A.- C.- 1-cos α 2 1+cos α 2 B. D. 1-cos α 2 1+cos α 2 α C [∵180° <α<360° ,∴90° <2<180° , α 1+cos α 又 cos22= ,∴cos α=- 2 1+cos α .] 2 3 θ 3.已知 2π<θ<4π,且 sin θ=-5,cos θ<0,则 tan2的值等于________. 3 4 [由 sin θ=-5,cos θ<0 得 cos θ=-5, θ sin2 θ θ 2sin2cos2 sin θ = θ 1+cos θ 2cos22 -3 θ ∴tan2= 3 5 cos2 θ= - = =-3.] ? 4? 1+?-5? ? ? [合 作 探 究· 攻 重 难] 化简求值问题 θ θ (1)设 5π<θ<6π,cos2=a,则 sin4等于( ) A. 1+a 2 1+a 2 B. 1-a 2 1-a 2 C.- D.- 3π (2)已知 π<α< 2 ,化简: 1+sin α 1-sin α + . 1+cos α- 1-cos α 1+cos α+ 1-cos α 【导学号:84352339】 θ θ [思路探究] (1)先确定4的范围,再由 sin24= θ 1-cos2 2 得算式求值. α α α (2)1+cos θ=2cos22,1-cos α=2sin22,去根号,确定2的范围,化简. θ ?5π ? θ ?5π 3π? (1)D [(1)∵5π<θ<6π,∴2∈? 2 ,3π?,4∈? 4 , 2 ?. ? ? ? ? θ 又 cos2=a, θ ∴sin4=- θ 1-cos2 2 1-a 2 . =- α? α? ? α ? α ?sin2+cos2?2 ?sin2-cos2?2 ? ? ? ? (2)原式= + . α α α ? ? ? ? ? ? ? α? cos sin cos sin 2? 2?- 2? 2? 2? 2?+ 2? 2? ? ? ? ? ? ? ? ? 3π π α 3π α α ∵π<α< 2 ,∴2<2< 4 ,∴cos2<0,sin2>0, α? ? α ?sin2+cos2?2 ? ? ∴原式= α?+ ? α - 2?sin2+cos2? ? ? α α sin2+cos2 2 α α sin2-cos2 2 α? ? α ?sin2-cos2?2 ? ? α? ? α 2?sin2-cos2? ? ? =- + α =- 2cos2.] [规律方法] 1.化简问题中的“三变” (1)变角:三角变换时通常先寻找式子中各角之间的联系,通过拆、凑等手 段消除角之间的差异,合理选择联系它们的公式. (2)变名:观察三角函数种类的差异,尽量统一函数的名称,如统一为弦或 统一为切. (3)变式:观察式子的结构形式的差异,选择适当的变形途径,如升幂、降 幂、配方、开方等. 2.利用半角公式求值的思路 (1)看角:看已知角与待求角的 2 倍关系. (2)明范围:求出相应半角的范围为定符号作准备. 1-cos α α sin α (3)选公式:涉及半角公式的正切值时,常用 tan2= = sin α ,涉 1+cos α α 1-cos α 2α 1+cos α 及半角公式的正、余弦值时,常利用 sin22= , cos 计算. 2 2= 2 (4)下结论:结合(2)求值. α 提醒:已知 cos α 的值可求2的正弦、余弦、正切值,要注意确定其符号. [跟踪训练] 3 θ 1.已知 cos θ=-5,且 180° <θ<270° ,求 tan 2. θ θ θ [解] 法一:∵180° <θ<270° ,∴90° <2<135° ,即2是第二象限角,∴tan 2<0, θ ∴tan 2=- 1-

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