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【高考领航】高考数学(理科)新一轮总复习考点突破课件3.5两角和与差的正弦、余弦和正切公式_图文

第5课时 两角和与差的正弦、余弦和 正切公式 ? (一)考纲点击 ? 1.会用向量的数量积推导出两角差的余弦 公式. ? 2.能利用两角差的余弦公式推导出两角差 的正弦、正切公式. ? 3.能利用两角差的余弦公式推导出两角和 的正弦、余弦、正切公式,推导出二倍角 的正弦、余弦、正切公式,了解它们的内 在联系. ? (二)命题趋势 ? 通 过近三年的高考试题分析,以两角和与 差的三角函数公式及二倍角为基础,求三 角函数的值是考查的重点,选择题、填空 题、解答题均有可能,难度不大. ? 1.两角和与差的正弦、余弦、正切公式 cos ? (1)C(α-β):cos (α-β ) αcosβ+sinαsinβ = ; cosαcos β-sinαsinβ sin ? (2)C(α+β):cos(α+β ) αcosβ+cosαsinβ = ; sinαcos β-cosαsinβ tan α+ tan β ? (5)T (3)S : sin( α + β ) tan( ; (α +βα ) +β)= (α β): 1-tan αtan β = ; tan α-tan β -β )= α-β) . (α β):tan(α ? (6)T (4)S : sin( 1+tan αtan β (α-β) = ; + - 对点演练 (1)化简:sin 200° cos 140° -cos 160° sin 40° =________. 3 解析:原式=sin 20° cos 40° +cos 20° sin 40° =sin 60° = . 2 3 答案: 2 ? π? 3 3 (2)若 sin α= ,cos β= ,其中 α,β∈?0,2?,则 α+β=________. 5 5 ? ? ? π? 解析:∵α、β∈?0,2?,∴α+β∈(0,π), ? ? 4 4 ∴cos α=5,sin β=5, 4 3 3 4 π ∴cos(α+β)= × - × =0,∴α+β= . 5 5 5 5 2 π 答案: 2 2.二倍角的正弦、余弦、正切公式 (1)S2α:sin 2α=2sin αcos α ; (2)C2α:cos 2α= cos2α-sin2α 2tan α (3)T2α:tan 2α= . 1-tan2α = 1-2sin2α =2cos2α-1 ; 对点演练 2 (1)已知 sin α=3,则 cos(π-2α)等于 5 A.- 3 1 C. 9 4 1 2×9-1=-9. 答案:B 1 B.- 9 5 D. 3 ( ) 解析:cos(π-2α)=-cos 2α=-(1-2sin2α)=2sin2α-1= 1 (2)若 tan θ+ =4,则 sin 2θ 等于 tan θ ( 1 A.5 1 C.3 1 B.4 1 D.2 ) 1 sin θ cos θ 1 解析:由 tan θ+ = + = =4, tan θ cos θ sin θ sin θcos θ 1 得 sin θcos θ= , 4 1 1 则 sin 2θ=2sin θcos θ=2× = . 4 2 答案:D ? 3.函数f(α)=acos α+bsin α(a,b为常数), 可以化为 ? f(α)=sin(α+φ)或f(α)=cos(α-φ),其中 φ可由a,b的值唯一确定. 对点演练 4m-6 要使 sin α- 3cos α= 有意义,则应 4-m 7 A.m≤3 7 C.m≤-1 或 m≥ 3 B.≥-1 7 D.-1≤m≤ 3 ( ) ? ? π? 4m-6 π? 2m-3 解析:2sin?α-3?= ?sin?α-3?= . 4-m 4-m ? ? ? ? 2m-3 7 由-1≤ ≤1?-1≤m≤ . 3 4-m 答案:D ? 三角变换中的“三变” ? (1) 变角:目的是沟通题设条件与结论中所 “配凑” 涉及的角,其手法通常是 . ? (2) 变名:通过变换函数名称达到减少函数 种类的目的,其手法通常有 “切化弦” “升幂与降幂” 、 等. 常值代换 ? (3) 变式:根据式子的结构特征进行变形, 使其更贴近某个公式或某个期待的目标, “逆用变用公式” 其手法通常有“ ”、 ? 、“通分约分”、“分解 与组合”、“配方与平方”等. 题型一 三角函数式的化简 化简下列各式: (1) 1 1 2-2 ? ?3π ?? 1 1 ?α∈? ,2π??=______. + cos 2 α 2 2 ? ?2 ?? cos2α-sin2α (2) ?π ? ?π ?=________. 2tan?4-α?cos2?4-α? ? ? ? ? 【解析】 所以 3π (1)因为 <α<2π, 2 1 1 2+2cos 2α=|cos α|=cos α, 3π α 又因为 4 <2<π, 所以 1 1 α α - cos α=|sin |=sin , 2 2 2 2 α 所以,原式=sin 2. (2)原式= cos 2α ?π ? ? ? 2 π 2tan?4-α?cos ?4-α? ? ? ? ? cos 2α cos 2α = =cos 2α=1. ?π ? ?π ?= π 2sin?4-α?cos?4-α? sin ?2-2α? ? ? ? ? 【答案】 α (1)sin 2 (2)1 cos2α ? 【归纳提升】 (1) 三角函数式的化简要遵 循 “ 三看 ” 原则,一看角,二看名,三看 式子结构与特征. ? (2) 对于给角求值问题,往往所给角都是非 特殊角,解决这类问题的基本思路有 ? ①化为特殊角的三角函数值; ? ②化为正、负相消的项,消去求值; ? ③化分子、分母出现公约数进行约分求 值. 针对训练 1 cos α- 2 1.化简: ?π ? ?π ?=________. 2tan?4-α?sin2?4+α? ? ? ? ? 2 1

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