关于高一数学新教材的利用三角函数证明题目 高中数学三角函数关于诱导公式方面的例题，越多越好，我会加分的
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S=t absinC=2S=2t ab=4
cc=aa+bb-2abcosC=aa+bb-ab=4 (a+b)^2=4+3ab=16 (a-b)^2=4-ab=0 a=b=2
\u4ee4P=\u5706\u5468\u7387 A+B=P-C=2P/3 B-A=2P/3-2A
t/2=sinC=2sin2A-sin(B-A)=2sin2A-sin(2P/3-2A)=2sin2A-[tcos2A+sin2A]/2
t/2=3sin2A/2-tcos2A/2 t=3sin2A-tcos2A 1=tsin2A-cos2A=2sin(2A-30)
sin(2A-30)=1/2 2A-30=30, 150 A=30, 90
S=t/2
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A. B. C. 2 D. 4
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1\u3001tan600\u00b0\u7684\u503c\u662f\uff08 \uff09
\uff08A\uff09 \uff08B\uff09 \uff08C\uff09 (D)
2\u3001\u51fd\u6570y\uff1dsin(2x\uff0b )\u7684\u6700\u5c0f\u6b63\u5468\u671f\u662f( )
(A) (B) (C) 2 (D) 4
3\u3001\u201c\u7b49\u5f0f \u6210\u7acb\u201d\u662f\u201c \u6210\u7b49\u5dee\u6570\u5217\u201d\u7684\uff08 \uff09
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4\u3001\u5f53 \u7684\u503c\u57df\u662f\uff08 \uff09
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5\u3001\u82e5 \u7684\u5947\u51fd\u6570\uff0c\u5219 \u53ef\u4ee5\u662f\uff08 \uff09
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6\u3001\u5c06\u51fd\u6570 \u7684\u56fe\u8c61\u6309\u5411\u91cf \u5e73\u79fb\uff0c\u5e73\u79fb\u540e\u7684\u56fe\u8c61\u5982\u56fe\u6240\u793a\uff0c\u5219\u5e73\u79fb\u540e\u7684\u56fe\u8c61\u6240\u5bf9\u5e94\u51fd\u6570\u7684\u89e3\u6790\u5f0f\u662f\uff08 \uff09
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7\u3001\u82e5\u25b3ABC\u9762\u79efS= \u5219\u2220C=\uff08 \uff09
\uff08A\uff09 \uff08B\uff09 \uff08C\uff09 \uff08D\uff09
8\u3001\u5728 \u4e2d\uff0c\u82e5 \uff0c\u5219 \u4e00\u5b9a\u662f\uff08 \uff09
\uff08A\uff09\u7b49\u8170\u4e09\u89d2\u5f62 \uff08B\uff09\u76f4\u89d2\u4e09\u89d2\u5f62 \uff08C\uff09\u7b49\u8170\u76f4\u89d2\u4e09\u89d2\u5f62 \uff08D\uff09\u7b49\u8170\u6216\u76f4\u89d2\u4e09\u89d2\u5f62
9\u3001 \u5219 \u7684\u6700\u5927\u503c\u548c\u6700\u5c0f\u503c\u5206\u522b\u662f\uff08 \uff09
\uff08A\uff097\u30015 \uff08B\uff097\u3001\uff0d \uff08C\uff095\u3001\uff0d \uff08D\uff097\u3001\uff0d5
10\uff0e\u5df2\u77e5\u5411\u91cf \u5219 \u4e0e \u7684\u5939\u89d2\u4e3a\uff08 \uff09
\uff08A\uff09 \uff08B\uff09 \uff08C\uff09 \uff08D\uff09
\u4e8c\u3001\u586b\u7a7a\u9898\uff08\u672c\u5927\u9898\u51714\u5c0f\u9898\uff0c\u6bcf\u5c0f\u98985\u5206\uff0c\u517120\u5206\uff09
11. \u5982\u679c \uff1d \uff0c\u4e14 \u662f\u7b2c\u56db\u8c61\u9650\u7684\u89d2\uff0c\u90a3\u4e48 \uff1d \uff1b
12. \u5df2\u77e5 \uff0c\u4e14 \uff0c\u5219 __________ \uff1b
13. \u5df2\u77e5 \u7684\u4e09\u4e2a\u5185\u89d2A\u3001B\u3001C\u6210\u7b49\u5dee\u6570\u5217\uff0c\u4e14 \u5219\u8fb9BC\u4e0a\u7684\u4e2d\u7ebfAD\u957f\u4e3a \uff1b
14. \u82e5 \u662f\u4ee55\u4e3a\u5468\u671f\u7684\u5947\u51fd\u6570\uff0c =4,\u4e14cos \uff0c\u5219 = .
\u4e09\u3001\u89e3\u7b54\u9898\uff08\u672c\u5927\u9898\u51716\u5c0f\u9898\uff0c\u517170\u5206\uff09
15\u3001\uff0812\u5206\uff09\u5df2\u77e5 \uff0c\u4e14 \u3001 \u662f\u65b9\u7a0b \u7684\u4e24\u4e2a\u6839\uff0c\u6c42COS\uff08 \uff09\u7684\u503c
16\uff0814\u5206\uff09 \u25b3ABC\u7684\u4e09\u4e2a\u5185\u89d2A\u3001B\u3001C\uff0c\u6c42\u5f53A\u4e3a\u4f55\u503c\u65f6\uff0c \u53d6\u5f97\u6700\u5927\u503c\uff0c\u5e76\u6c42\u51fa\u8fd9\u4e2a\u6700\u5927\u503c
17\uff0e\uff0814\u5206\uff09\u5df2\u77e5\u51fd\u6570 \uff08\u2160\uff09\u6c42 \u7684\u6700\u5c0f\u6b63\u5468\u671f\uff1b
\uff08\u2161\uff09\u6c42 \u7684\u6700\u5927\u503c\u548c\u6700\u5c0f\u503c\uff1b\uff08\u2162\uff09\u82e5 \uff0c\u6c42 \u7684\u503c.
18\uff0e\uff0812\u5206\uff09\u5982\u56fe\uff0c\u5f53\u7532\u8239\u4f4d\u4e8eA\u5904\u65f6\u83b7\u6089\uff0c\u5728\u5176\u6b63\u4e1c\u65b9\u5411\u76f8\u8ddd20\u6d77\u91cc\u7684B\u5904\u6709\u4e00\u8258
\u6e14\u8239\u9047\u9669\u7b49\u5f85\u8425\u6551\uff0e\u7532\u8239\u7acb\u5373\u524d\u5f80\u6551\u63f4\uff0c\u540c\u65f6\u628a\u6d88\u606f\u544a\u77e5\u5728\u7532\u8239\u7684\u5357\u504f\u897f30 \uff0c\u76f8\u8ddd10
\u6d77\u91ccC\u5904\u7684\u4e59\u8239\uff0c\u8bd5\u95ee\u4e59\u8239\u5e94\u671d\u5317\u504f\u4e1c\u591a\u5c11\u5ea6\u7684\u65b9\u5411\u6cbf\u76f4\u7ebf\u524d\u5f80B\u5904\u6551\u63f4\uff08\u89d2\u5ea6\u7cbe\u786e\u5230 \uff09\uff1f
19.\uff0814\u5206\uff09\uff0e\u5df2\u77e5 \u662f\u4e09\u89d2\u5f62 \u4e09\u5185\u89d2\uff0c\u5411\u91cf \uff0c\u4e14
\uff08\u2160\uff09\u6c42\u89d2 \uff1b\uff08\u2161\uff09\u82e5 \uff0c\u6c42
20\u3001\uff08 14\u5206\uff09\u5df2\u77e5b\u3001c\u662f\u5b9e\u6570\uff0c\u51fd\u6570f(x)= \u5bf9\u4efb\u610f\u03b1\u3001\u03b2 R\u6709\uff1a \u4e14
\uff081\uff09\u6c42f\uff081\uff09\u7684\u503c\uff1b\uff082\uff09\u8bc1\u660e\uff1aC \uff1b(3\uff09\u8bbe \u7684\u6700\u5927\u503c\u4e3a10\uff0c\u6c42f\uff08x\uff09\u3002
\u4e09\u89d2\u51fd\u6570\u6d4b\u8bd5\u5377\u53c2\u8003\u7b54\u6848
\u4e00\u3001\u9009\u62e9\u9898
DBBDB CCDDA
\u4e8c\u3001\u586b\u7a7a\u9898
11\u3001 12\u3001 13\u3001 14\u3001\u20144
\u4e09\u3001\u89e3\u7b54\u9898
15\u3001
16\u3001\u89e3: \u7531A+B+C=\u03c0, \u5f97B+C2 = \u03c02 \uff0dA2 , \u6240\u4ee5\u6709cosB+C2 =sinA2 .
cosA+2cosB+C2 =cosA+2sinA2 =1\uff0d2sin2A2 + 2sinA2
=\uff0d2(sinA2 \uff0d 12)2+ 32
\u5f53sinA2 = 12 , \u5373A=\u03c03 \u65f6, cosA+2cosB+C2\u53d6\u5f97\u6700\u5927\u503c\u4e3a32
17\uff0e\u89e3\uff1a
\uff08\u2160\uff09 \u7684\u6700\u5c0f\u6b63\u5468\u671f\u4e3a ;
\uff08\u2161\uff09 \u7684\u6700\u5927\u503c\u4e3a \u548c\u6700\u5c0f\u503c \uff1b
\uff08\u2162\uff09\u56e0\u4e3a \u5373 ,\u5373
18\u3001\u89e3 \u8fde\u63a5BC,\u7531\u4f59\u5f26\u5b9a\u7406\u5f97
BC2=202+102\uff0d2\u00d720\u00d710COS120\u00b0=700.
\u4e8e\u662f,BC=10 .
\u2235 , \u2234sin\u2220ACB= ,
\u2235\u2220ACB<90\u00b0 \u2234\u2220ACB=41\u00b0
\u2234\u4e59\u8239\u5e94\u671d\u5317\u504f\u4e1c71\u00b0\u65b9\u5411\u6cbf\u76f4\u7ebf\u524d\u5f80B\u5904\u6551\u63f4.
19\u3001\u89e3\uff1a\uff08\u2160\uff09\u2235 \u2234
\u5373
,
\u2235 \u2234 \u2234
\uff08\u2161\uff09\u7531\u9898\u77e5 \uff0c
\u6574\u7406\u5f97
\u2234 \u2234
\u2234 \u6216
\u800c \u4f7f \uff0c\u820d\u53bb \u2234
\u2234
20\u3001\u89e3\uff1a\uff081\uff09\u4ee4\u03b1= \uff0c\u5f97 \u4ee4\u03b2= \uff0c\u5f97 \u56e0\u6b64 \uff1b
\uff082\uff09\u8bc1\u660e\uff1a\u7531\u5df2\u77e5\uff0c\u5f53 \u65f6\uff0c
\u5f53 \u65f6\uff0c \u901a\u8fc7\u6570\u5f62\u7ed3\u5408\u7684\u65b9\u6cd5\u53ef\u5f97\uff1a \u5316\u7b80\u5f97c \uff1b
\uff083\uff09\u7531\u4e0a\u8ff0\u53ef\u77e5\uff0c[-1\uff0c1]\u662f \u7684\u51cf\u533a\u95f4\uff0c\u90a3\u4e48 \u53c8 \u8054\u7acb\u65b9\u7a0b\u7ec4\u53ef\u5f97 ,
\u6240\u4ee5
方法：将sinα、α、tanα转化为单位圆中的线段或弧线。
如图，P是角α的终边与单位圆的交点，PM⊥x轴，AT是切线。
则有sinα=MP/OP=MP，cosα=OM/OP=OM，tanα=AT/OA=AT (注：单位圆的半径为1 ）
α=弧AP/r=弧AP (注：弧长与半径的比等于圆心角的弧度数的绝对值。r=1为半径）
（1）易知 MP<弦AP<弧AP ，所以 sinα<α
(2) 三角形两边之和大于么三边，|MP|+|OM|>|OP|，即
sinα+cosα>1
（3）S扇形OAP= 弧AP•r/2=弧AP/2=α/2，
S⊿OAT=OA•AT/2=AT/2=(tanα)/2
因为 S扇形OAP < S⊿OAT ，所以α/2<(tanα)/2，
即α<tanα
从而，当α为锐角时，有 sinα<α<tanα
如图，在单位圆O中，角OAB=a，则弧AB=a，BN=sin
a，AM=tan
a,比较这几条段可知有：sin
a
<
a
<
tan
a，又ON=cos
a，根据三角形两边之和大于第三边可知，ON+NB>OB=1，故：sin
a
+
cos
a
>1
α为单位圆（以原点O为中心的）中一个锐角，A为圆上一点，B为圆到x轴上的点，C为X轴与圆的交点，D为延长OA与C点处垂线的交点。
则，AB=sin
α,OB=cos
α,OC=1,CD=tan
α,则AB+OB>OA,即sin
α+cos
α>1
又
三角形OAC的面积<扇形OAC的面积<三角形ODC的面积,则1/2sin
α<
1/2α<1/2tan
α，即sin
α<
α<tan
α
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绛旓細13锛夎В鏋愶細鈭礷(x)=coswx(w>0)锛屽叾鍥惧儚鍏充簬鐐筂(3蟺/4锛0)瀵圭О锛屽湪鍖洪棿[0, 蟺/2]涓婂崟璋 f(x)=coswx=sin(wx+蟺/2)f(3蟺/4)=sin(w3蟺/4+蟺/2)=0 w3蟺/4+蟺/2=蟺==>w=2/3 鈭磃(x)=cos2/3x=sin(2/3x+蟺/2)14锛夎В鏋愶細鈭-蟺/6<=尾<蟺/4锛3(sin伪)^2-2(sin尾...
绛旓細涓銆佸畾涔夋硶 鐩存帴鍒╃敤鍛ㄦ湡鍑芥暟鐨勫畾涔夋眰鍑哄懆鏈熴備簩銆佸叕寮忔硶 鍒╃敤涓嬪垪鍏紡姹傝В涓夎鍑芥暟鐨勬渶灏忔鍛ㄦ湡銆備笁銆佽浆鍖栨硶 瀵硅緝澶嶆潅鐨勪笁瑙掑嚱鏁板彲閫氳繃鎭掔瓑鍙樺舰杞寲涓 绛夌被鍨嬶紝鍐嶇敤鍏紡娉曟眰瑙 鍥涖佹渶灏忓叕鍊嶆暟娉 鐢变笁瑙掑嚱鏁扮殑浠ｆ暟鍜岀粍鎴愮殑涓夎鍑芥暟寮忥紝鍙厛鎵惧嚭鍚勪釜鍔犲嚱鏁扮殑鏈灏忔鍛ㄦ湡锛岀劧鍚庢壘鍑烘墍鏈夊懆鏈熺殑鏈灏忓叕鍊...
