2018版高中数学人教B版必修四学案:第三单元+3.3+三角函数的积化和差与和差化积+Word版含答案

学习目标 1.了解利用两角和与差的正弦、余弦公式导出积化和差、和差化积两组公式的过 程.2.理解在推导积化和差、和差化积公式中方程思想、换元思想所起的作用. 知识点一 积化和差公式 思考 根据两角和与差的正、余弦公式把下列等式补充完整. ①sin(α+β)+sin(α-β)=________________; ②sin(α+β)-sin(α-β)=________________; ③cos(α+β)+cos(α-β)=________________; ④cos(α+β)-cos(α-β)=________________. 1 在上述四个等式两边同乘以 ,等号两端互换,就可以得出四个相应的积化和差公式. 2 梳理 积化和差公式 (1)sin αcos β=________________________________. (2)cos αsin β=________________________________. (3)cos αcos β=________________________________. (4)sin αsin β=________________________________. 知识点二 和差化积公式 思考 在四个积化和差公式中, 如果我们令 α+β=θ, α-β=φ, 则 α=________, β=________, 由此可以得出四个相应的和差化积公式,请你试一试写出这四个公式: sin θ+sin φ=________________; sin θ-sin φ=________________; cos θ+cos φ=________________________; cos θ-cos φ=________________________. 梳理 和差化积公式 (1)sin x+sin y=2sin x+y x-y cos , 2 2 (2)sin x-sin y=2cos x+y x-y sin , 2 2 x+y x-y (3)cos x+cos y=2cos cos , 2 2 x+y x-y (4)cos x-cos y=-2sin sin . 2 2 类型一 利用积化和差与和差化积公式化简求值 例 1 求值:sin 20° cos 70° +sin 10° sin 50° . 反思与感悟 套用和差化积公式的关键是记准、记牢公式,为了能够把三角函数式化为积的 形式,有时需要把常数首先化为某个角的三角函数,然后再化积,有时函数不同名,要先化 为同名再化积,化积的结果能求值则尽量求出值来. 跟踪训练 1 求值:cos 20° +cos 60° +cos 100° +cos 140° . 类型二 三角恒等式的证明 例 2 在△ABC 中,求证:sin 2A+sin 2B+sin 2C=4sin Asin Bsin C. 反思与感悟 数间的关系. 在运用积化和差求值时,尽量出现特殊角,同时注意互余角、互补角的三角函 A B C 跟踪训练 2 已知 A+B+C=π,求证:sin A+sin B-sin C=4sin sin cos . 2 2 2 1.sin 75° -sin 15° 的值为( 1 2 3 1 A. B. C. D.- 2 2 2 2 2.sin 15° cos 165° 的值是( 1 1 1 1 A. B. C.- D.- 4 2 4 2 3.sin 105° +sin 15° 等于( A. 3 2 B. 2 6 6 C. D. 2 2 4 ) ) ) ) 4.sin 37.5°cos 7.5° 等于( A. 2 2 B. 2+1 2+2 2 C. D. 4 4 4 5.在△ABC 中,若 B=30° ,求 cos Asin C 的取值范围. 1.本节学习了积化和差公式、和差化积公式,一定要清楚这些公式的形式特征,理解公式间的 关系. 2.和差化积、积化和差公式不要求记忆,但要注意公式推导中应用的数学思想方法,同时注意 这些公式与两角和与差公式的联系. 答案精析 问题导学 知识点一 思考 ①2sin αcos β ②2cos αsin β ③2cos αcos β ④-2sin αsin β 1 梳理 (1) [sin(α+β)+sin(α-β)] 2 1 (2) [sin(α+β)-sin(α-β)] 2 1 (3) [cos(α+β)+cos(α-β)] 2 1 (4)- [cos(α+β)-cos(α-β)] 2 知识点二 思考 2cos 2cos θ+φ θ-φ θ+φ θ-φ 2sin cos 2 2 2 2 θ+φ θ-φ sin 2 2 θ+φ θ-φ cos 2 2 θ+φ θ-φ sin 2 2 -2sin 题型探究 例1 解 sin 20° cos 70° +sin 10° sin 50° 1 1 = (sin 90° -sin 50° )- (cos 60° -cos 40° ) 2 2 1 1 1 = - sin 50° + cos 40° 4 2 2 1 1 1 1 = - sin 50° + sin 50° = . 4 2 2 4 1 跟踪训练 1 解 原式=cos 20° + +(cos 100° +cos 140° ) 2 1 =cos 20° + +2cos 120° cos 20° 2 1 1 =cos 20° + -cos 20° = . 2 2 例 2 证明 左边=sin 2A+sin 2B+sin 2C 2A+2B 2A-2B =2sin cos +sin 2C 2 2 =2sin(A+B)cos(A-B)-2sin(A+B)· cos(A+B) =2sin C[cos(A-B)-cos(A+B)] ?A-B?+?A+B? ?A-B?-?A+B? =2sin C· (-2)sin · sin 2 2 =4sin Asin Bsin C=右边. 所以原等

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