专题02 三角恒等变换与解三角形
【要点提炼】
1.三角函数公式
(1)两角和与差的正弦、余弦、正切公式:
sin(α±β)=sin αcos β±cos αsin β;
cos(α±β)=cos αcos β∓sin αsin β;
tan(α±β)=.
(2)二倍角公式:sin 2α=2sin αcos α,cos 2α=cos2α-sin2α=2cos2α-1=1-2sin2α.
(3)辅助角公式:asin x+bcos x=sin(x+φ),其中tan φ=
.
2.正弦定理、余弦定理、三角形面积公式
(1)正弦定理
在△ABC中,=
=
=2R(R为△ABC的外接圆半径);
变形:a=2Rsin A,sin A=,
a∶b∶c=sin A∶sin B∶sin C等.
(2)余弦定理
在△ABC中,a2=b2+c2-2bccos A;
变形:b2+c2-a2=2bccos A,cos A=.
(3)三角形面积公式
S△ABC=absin C=
bcsin A=
acsin B.
考点一 三角恒等变换
考向一 三角恒等变换
【典例1】 (1)(2020·全国Ⅲ卷)已知2tan θ-tan=7,则tan θ=( )
A.-2 B.-1 C.1 D.2
(2)(2019·全国Ⅱ卷)已知α∈,2sin 2α=cos 2α+1,则sin α=( )
A. B.
C.
D.
解析 (1)2tan θ-tan=2tan θ-
=7,解得tan θ=2.故选D.
(2)由2sin 2α=cos 2α+1,得4sin αcos α=2cos2α.
由α∈知cos α≠0,
则2sin α=cos α,代入sin2α+cos2α=1,解得sin2α=,
又α∈,所以sin α=
.
答案 (1)D (2)B
探究提高 1.三角恒等变换的基本思路:找差异,化同角(名),化简求值.三角变换的关键在于对两角和与差的正弦、余弦、正切公式,二倍角公式,三角恒等变换公式的熟记和灵活应用,要善于观察各个角之间的联系,发现题目所给条件与恒等变换公式的联系.
2.求解三角函数中给值求角的问题时,要根据已知先求这个角的某种三角函数值,然后结合角的取值范围,求出角的大小.求解时,尽量缩小角的取值范围,避免产生增解.
【拓展练习1】 (1)(2020·深圳统测)已知tan α=-3,则sin=( )
A. B.-
C.
D.-
(2)(2020·江南名校联考)已知α,β均为锐角,且α+β≠,若sin(2α+β)=
sin β,则
=________.
解析 (1)由题意,得sin=sin
=cos 2α=cos2α-sin2α=
=
=
=-
.故选D.
(2)因为sin(2α+β)=sin β,
则2sin[(α+β)+α]=3sin[(α+β)-α]
∴2[sin(α+β)cos α+cos(α+β)sin α]=3[sin(α+β)cos α-cos(α+β)sin α]
从而sin(α+β)cos α=5cos(α+β)sin α.
∴tan(α+β)=5tan α,故=5.
答案 (1)D (2)5
考点二 解三角形
考向二 利用正(余)弦定理进行边角计算
【典例2】 (2020·青岛质检)在△ABC中,a,b,c分别为内角A,B,C的对边,2b2=(b2+c2-a2)(1-tan A).
(1)求角C;
(2)若c=2,D为BC的中点,在下列两个条件中任选一个,求AD的长度.
条件①:S△ABC=4且B>A;
条件②:cos B=.
(注:如果选择多个条件分别解答,按第一个解答计分)
解 (1)已知2b2=(b2+c2-a2)(1-tan A).
由余弦定理,得2b2=2bccos A·(1-tan A),
所以b=c(cos A-sin A).
由正弦定理,得sin B=sin C(cos A-sin A),
所以sin(A+C)=sin Ccos A-sin Csin A,
所以sin Acos C=-sin Csin A,
又sin A≠0,所以tan C=-1,
又C∈(0,π),所以C=π.
(2)若选择条件①:S△ABC=4且B>A.
因为S△ABC=4=absin C=
absin
,所以ab=8
.
由余弦定理,得c2=(2)2=40=a2+b2-2abcos
,
所以a2+b2+ab=40.
由解得
或
因为B>A,所以b>a,所以所以CD=
.
在△ACD中,AD2=CA2+CD2-2CA·CD·cos C=16+2-2×4×cos
=26,所以AD=
.
若选择条件②:cos B=.
因为cos B=,B∈(0,π),所以sin B=
.
因为sin A=sin(B+C)=sin Bcos C+sin Ccos B=,
所以结合正弦定理=
,得a=
=2
.
在△ABD中,由余弦定理,得AD2=AB2+BD2-2AB·BD·cos B=(2)2+(
)2-2×2
×
×
=26,解得AD=
.
探究提高 1.高考的考向是利用正、余弦定理求三角形的边、角、面积等基本计算,或将两个定理与三角恒等变换相结合综合解三角形.
2.关于解三角形问题,一般要用到三角形的内角和定理,正、余弦定理及有关三角形的性质,常见的三角变换方法和原则都适用,同时要注意“三统一”,即“统一角、统一函数、统一结构”,这是使问题获得解决的突破口.
【拓展练习2】 (2020·日照联考)在①3c2=16S+3(b2-a2),②5bcos C+4c=5a,这两个条件中任选一个,补充在下面横线处,然后解答问题.
在△ABC中,内角A,B,C的对边分别为a,b,c,设△ABC的面积为S,已知________.
(1)求tan B的值;
(2)若S=42,a=10,求b的值.
(注:如果选择多个条件分别解答,按第一个解答计分)
解 选择条件①:(1)由题意得8acsin B=3(a2+c2-b2),
即4sin B=3·,整理可得3cos B-4sin B=0.
又sin B>0,所以cos B>0,所以tan B==
.
(2)由tan B=,得sin B=
.
又S=42,a=10,
所以S=acsin B=
×10c×
=42,解得c=14.
将S=42,a=10,c=14代入3c2=16S+3(b2-a2),
得3×142=16×42+3(b2-102),解得b=6.
选择条件②:(1)已知5bcos C+4c=5a,
由正弦定理,得5sin Bcos C+4sin C=5sin A,
即5sin Bcos C+4sin C=5sin(B+C),
即sin C(4-5cos B)=0.
在△ABC中,因为sin C≠0,所以cos B=.
所以sin B==
,所以tan B=
.
(2)由S=acsin B=
×10c×
=42,解得c=14.
又a=10,所以b2=100+196-2×140×=72,所以b=6
.
考向三 正、余弦定理与其它知识的交汇问题
角度1 正、余弦定理与三角函数的结合命题
【典例3】 已知m=(2cos x+2sin x,1),n=(cos x,-y),且满足m·n=0.
(1)将y表示为x的函数f(x),并求f(x)的最小正周期;
(2)已知a,b,c分别为△ABC的三个内角A,B,C对应的边长,f(x)(x∈R)的最大值是f,且a=2,求b+c的取值范围.
解 (1)由m·n=0,得2cos2 x+2sin xcos x-y=0,
即y=2cos2 x+2sin xcos x=cos 2x+
sin 2x+1
=2sin+1,
所以f(x)=2sin+1,其最小正周期为π.
(2)由题意得f=3,
所以A+=2kπ+
(k∈Z),
因为0<A<π,所以A=.
由正弦定理,得b=sin B,c=
sin C,
则b+c=sin B+
sin C
=sin B+
sin
=4sin
,
又因为B∈,
所以sin∈
,
所以b+c∈(2,4],所以b+c的取值范围是(2,4].
角度2 正、余弦定理与向量的结合命题
【典例4】 (2020·潍坊模拟)△ABC的内角A,B,C的对边分别为a,b,c,已知向量m=(c-a,sin B),n=(b-a,sin A+sin C),且m∥n.
(1)求C;
(2)若c+3b=3a,求sin A.
解 (1)因为m∥n,
所以(c-a)(sin A+sin C)=(b-a)sin B.
由正弦定理得(c-a)(a+c)=(b-a)b,所以a2+b2-c2=ab,
所以cos C==
=
.
因为C∈(0,π),所以C=.
(2)由(1)知B=-A.
由题设及正弦定理得sin C+3sin
=3sin A,
即+
cos A+
sin A=sin A,
可得sin=
.
由于0<A<,因此-
<A-
<
,
所以cos=
.
故sin A=sin=sin
cos
+cos
·sin
=
.
探究提高 1.该题求解的关键是利用向量的知识将条件“脱去向量外衣”,转化为三角函数的相关知识进行求解.
2.与解三角形有关的交汇问题的关注点
(1)根据条件恰当选择正弦、余弦定理完成边角互化.
(2)结合三角形内角和定理、面积公式等,灵活运用三角恒等变换公式.
【拓展练习3】 已知向量a=,b=(-sin x,
sin x),f(x)=a·b.
(1)求函数f(x)的最小正周期及f(x)的最大值;
(2)在锐角△ABC中,角A,B,C的对边分别为a,b,c,若f=1,a=2
,求△ABC面积的最大值并说明此时△ABC的形状.
解 (1)由已知得a=(-sin x,cos x),又b=(-sin x,sin x),
则f(x)=a·b=sin2x+sin xcos x
=(1-cos 2x)+
sin 2x=sin
+
,
∴f(x)的最小正周期T==π,
当2x-=
+2kπ(k∈Z),即x=
+kπ(k∈Z)时,
f(x)取得最大值.
(2)锐角△ABC中,因为f=sin
+
=1,
∴sin=
,∴A=
.
因为a2=b2+c2-2bccos A,所以12=b2+c2-bc,
所以b2+c2=bc+12≥2bc,
所以bc≤12(当且仅当b=c=2时等号成立),此时△ABC为等边三角形.
S△ABC=bcsin A=
bc≤3
.
所以当△ABC为等边三角形时面积取最大值3.
【专题拓展练习】
一、单选题
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A.![]() | B.![]() | C.![]() | D.![]() |
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A.![]() | B.![]() | C.![]() | D.![]() |
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A.![]() | B.![]() | C.![]() | D.![]() |
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A.![]() | B.![]() | C.![]() | D.![]() |
【知识点】 正弦定理边角互化的应用解读 余弦定理解三角形解读 三角函数与解三角形
二、多选题
A.半径为![]() ![]() ![]() |
B.若![]() ![]() ![]() ![]() ![]() |
C.若![]() ![]() ![]() ![]() ![]() |
D.若![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() |
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A.![]() |
B.![]() |
C.若![]() ![]() ![]() |
D.若![]() ![]() ![]() |
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A.![]() ![]() |
B.当![]() ![]() ![]() |
C.当![]() ![]() ![]() ![]() ![]() |
D.当![]() ![]() ![]() ![]() ![]() ![]() ![]() |
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A.![]() | B.![]() |
C.![]() | D.![]() |
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A.![]() ![]() | B.![]() ![]() |
C.![]() ![]() | D.![]() ![]() |
三、解答题
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(1)求
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(2)若
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(Ⅰ)求角
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(Ⅱ)求
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(1)求
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(2)若
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