三角形公式全部，求证三角形的公式，sina. cosa. tana公式，求大神解答！ sinA，cosA，tanA之间存在的一般关系有哪些
\u8c01\u80fd\u628a\u9ad8\u4e00\u6709\u5173\u4e09\u89d2\u5f62\u6052\u7b49\u53d8\u5f62\u6709\u5173sina\uff0ccosa\uff0ctana\u7684\u6240\u6709\u5173\u7cfb\u5f0f\u5217\u51fa\u6765\uff0c\u9ad8\u60ac\u8d4f\u540c\u89d2\u4e09\u89d2\u51fd\u6570\u7684\u57fa\u672c\u5173\u7cfb
\u5012\u6570\u5173\u7cfb: tan\u03b1 \u00b7cot\u03b1\uff1d1 sin\u03b1 \u00b7csc\u03b1\uff1d1 cos\u03b1 \u00b7sec\u03b1\uff1d1 \u5546\u7684\u5173\u7cfb\uff1a sin\u03b1/cos\u03b1\uff1dtan\u03b1\uff1dsec\u03b1/csc\u03b1 cos\u03b1/sin\u03b1\uff1dcot\u03b1\uff1dcsc\u03b1/sec\u03b1 \u5e73\u65b9\u5173\u7cfb\uff1a sin^2(\u03b1)\uff0bcos^2(\u03b1)\uff1d1 1\uff0btan^2(\u03b1)\uff1dsec^2(\u03b1) 1\uff0bcot^2(\u03b1)\uff1dcsc^2(\u03b1)
\u5e73\u5e38\u9488\u5bf9\u4e0d\u540c\u6761\u4ef6\u7684\u5e38\u7528\u7684\u4e24\u4e2a\u516c\u5f0f
sin² \u03b1+cos² \u03b1=1 tan \u03b1 *cot \u03b1=1
\u4e00\u4e2a\u7279\u6b8a\u516c\u5f0f
\uff08sina+sin\u03b8\uff09*\uff08sina+sin\u03b8\uff09=sin\uff08a+\u03b8\uff09*sin\uff08a-\u03b8\uff09 \u8bc1\u660e\uff1a\uff08sina+sin\u03b8\uff09*\uff08sina+sin\u03b8\uff09=2 sin[(\u03b8+a)/2] cos[(a-\u03b8)/2] *2 cos[(\u03b8+a)/2] sin[(a-\u03b8)/2] =sin\uff08a+\u03b8\uff09*sin\uff08a-\u03b8\uff09
\u9510\u89d2\u4e09\u89d2\u51fd\u6570\u516c\u5f0f
\u6b63\u5f26\uff1a sin \u03b1=\u2220\u03b1\u7684\u5bf9\u8fb9/\u2220\u03b1 \u7684\u659c\u8fb9 \u4f59\u5f26\uff1acos \u03b1=\u2220\u03b1\u7684\u90bb\u8fb9/\u2220\u03b1\u7684\u659c\u8fb9 \u6b63\u5207\uff1atan \u03b1=\u2220\u03b1\u7684\u5bf9\u8fb9/\u2220\u03b1\u7684\u90bb\u8fb9 \u4f59\u5207\uff1acot \u03b1=\u2220\u03b1\u7684\u90bb\u8fb9/\u2220\u03b1\u7684\u5bf9\u8fb9
\u4e8c\u500d\u89d2\u516c\u5f0f
\u6b63\u5f26 sin2A=2sinA\u00b7cosA \u4f59\u5f26 1.Cos2a=Cos^2(a)-Sin^2(a) =2Cos^2(a)-1 =1-2Sin^2(a) 2.Cos2a=1-2Sin^2(a) 3.Cos2a=2Cos^2(a)-1 \u6b63\u5207 tan2A=\uff082tanA\uff09/\uff081-tan^2(A)\uff09
\u4e09\u500d\u89d2\u516c\u5f0f
sin3\u03b1=4sin\u03b1\u00b7sin(\u03c0/3+\u03b1)sin(\u03c0/3-\u03b1) cos3\u03b1=4cos\u03b1\u00b7cos(\u03c0/3+\u03b1)cos(\u03c0/3-\u03b1) tan3a = tan a \u00b7 tan(\u03c0/3+a)\u00b7 tan(\u03c0/3-a) \u4e09\u500d\u89d2\u516c\u5f0f\u63a8\u5bfc sin(3a) =sin(a+2a) =sin2acosa+cos2asina =2sina(1-sin²a)+(1-2sin²a)sina =3sina-4sin^3a cos3a =cos(2a+a) =cos2acosa-sin2asina =(2cos²a-1)cosa-2(1-cos^a)cosa =4cos^3a-3cosa sin3a=3sina-4sin^3a =4sina(3/4-sin²a) =4sina[(\u221a3/2)²-sin²a] =4sina(sin²60\u00b0-sin²a) =4sina(sin60\u00b0+sina)(sin60\u00b0-sina) =4sina*2sin[(60+a)/2]cos[(60\u00b0-a)/2]*2sin[(60\u00b0-a)/2]cos[(60\u00b0-a)/2] =4sinasin(60\u00b0+a)sin(60\u00b0-a) cos3a=4cos^3a-3cosa =4cosa(cos²a-3/4) =4cosa[cos²a-(\u221a3/2)^2] =4cosa(cos²a-cos²30\u00b0) =4cosa(cosa+cos30\u00b0)(cosa-cos30\u00b0) =4cosa*2cos[(a+30\u00b0)/2]cos[(a-30\u00b0)/2]*{-2sin[(a+30\u00b0)/2]sin[(a-30\u00b0)/2]} =-4cosasin(a+30\u00b0)sin(a-30\u00b0) =-4cosasin[90\u00b0-(60\u00b0-a)]sin[-90\u00b0+(60\u00b0+a)] =-4cosacos(60\u00b0-a)[-cos(60\u00b0+a)] =4cosacos(60\u00b0-a)cos(60\u00b0+a) \u4e0a\u8ff0\u4e24\u5f0f\u76f8\u6bd4\u53ef\u5f97 tan3a=tanatan(60\u00b0-a)tan(60\u00b0+a)
n\u500d\u89d2\u516c\u5f0f
sin\uff08n a\uff09=Rsina sin\uff08a+\u03c0/n\uff09\u2026\u2026sin\uff08a+\uff08n-1\uff09\u03c0/n\uff09\u3002 \u5176\u4e2dR=2^\uff08n-1\uff09 \u8bc1\u660e\uff1a\u5f53sin\uff08na\uff09=0\u65f6\uff0csina=sin\uff08\u03c0/n\uff09\u6216=sin\uff082\u03c0/n\uff09\u6216=sin\uff083\u03c0/n\uff09\u6216=\u2026\u2026\u6216=sin\u3010\uff08n-1\uff09\u03c0/n\u3011 \u8fd9\u8bf4\u660esin\uff08na\uff09=0\u4e0e\uff5bsina-sin\uff08\u03c0/n\uff09\uff5d*\uff5bsina-sin\uff082\u03c0/n\uff09\uff5d*\uff5bsina-sin\uff083\u03c0/n\uff09\uff5d*\u2026\u2026*\uff5bsina- sin\u3010\uff08n-1\uff09\u03c0/n\u3011=0\u662f\u540c\u89e3\u65b9\u7a0b\u3002 \u6240\u4ee5sin\uff08na\uff09\u4e0e\uff5bsina-sin\uff08\u03c0/n\uff09\uff5d*\uff5bsina-sin\uff082\u03c0/n\uff09\uff5d*\uff5bsina-sin\uff083\u03c0/n\uff09\uff5d*\u2026\u2026*\uff5bsina- sin\u3010\uff08n-1\uff09\u03c0/n\u3011\u6210\u6b63\u6bd4\u3002 \u800c\uff08sina+sin\u03b8\uff09*\uff08sina+sin\u03b8\uff09=sin\uff08a+\u03b8\uff09*sin\uff08a-\u03b8\uff09\uff0c\u6240\u4ee5 \uff5bsina-sin\uff08\u03c0/n\uff09\uff5d*\uff5bsina-sin\uff082\u03c0/n\uff09\uff5d*\uff5bsina-sin\uff083\u03c0/n\uff09\uff5d*\u2026\u2026*\uff5bsina- sin\u3010\uff08n-1\u03c0/n\u3011 \u4e0esina sin\uff08a+\u03c0/n\uff09\u2026\u2026sin\uff08a+\uff08n-1\uff09\u03c0/n\uff09\u6210\u6b63\u6bd4\uff08\u7cfb\u6570\u4e0en\u6709\u5173 \uff0c\u4f46\u4e0ea\u65e0\u5173\uff0c\u8bb0\u4e3aRn\uff09\u3002 \u7136\u540e\u8003\u8651sin\uff082n a\uff09\u7684\u7cfb\u6570\u4e3aR2n=R2*(Rn)^2=Rn*(R2)^n.\u6613\u8bc1R2=2\uff0c\u6240\u4ee5Rn= 2^\uff08n-1\uff09
\u534a\u89d2\u516c\u5f0f
tan(A/2)=(1-cosA)/sinA=sinA/(1+cosA); cot(A/2)=sinA/(1-cosA)=(1+cosA)/sinA. sin^2(a/2)=(1-cos(a))/2 cos^2(a/2)=(1+cos(a))/2 tan(a/2)=(1-cos(a))/sin(a)=sin(a)/(1+cos(a))
\u548c\u5dee\u5316\u79ef
sin\u03b8+sin\u03c6 = 2 sin[(\u03b8+\u03c6)/2] cos[(\u03b8-\u03c6)/2]
sin\u03b8-sin\u03c6 = 2 cos[(\u03b8+\u03c6)/2] sin[(\u03b8-\u03c6)/2] cos\u03b8+cos\u03c6 = 2 cos[(\u03b8+\u03c6)/2] cos[(\u03b8-\u03c6)/2] cos\u03b8-cos\u03c6 = -2 sin[(\u03b8+\u03c6)/2] sin[(\u03b8-\u03c6)/2] tanA+tanB=sin(A+B)/cosAcosB=tan(A+B)(1-tanAtanB) tanA-tanB=sin(A-B)/cosAcosB=tan(A-B)(1+tanAtanB)
\u4e24\u89d2\u548c\u516c\u5f0f
cos(\u03b1+\u03b2)=cos\u03b1cos\u03b2-sin\u03b1sin\u03b2cos(\u03b1-\u03b2)=cos\u03b1cos\u03b2+sin\u03b1sin\u03b2sin(\u03b1+\u03b2)=sin\u03b1cos\u03b2+cos\u03b1sin\u03b2sin(\u03b1-\u03b2)=sin\u03b1cos\u03b2 -cos\u03b1sin\u03b2
\u79ef\u5316\u548c\u5dee
sin\u03b1sin\u03b2 = [cos(\u03b1-\u03b2)-cos(\u03b1+\u03b2)] /2 cos\u03b1cos\u03b2 = [cos(\u03b1+\u03b2)+cos(\u03b1-\u03b2)]/2 sin\u03b1cos\u03b2 = [sin(\u03b1+\u03b2)+sin(\u03b1-\u03b2)]/2 cos\u03b1sin\u03b2 = [sin(\u03b1+\u03b2)-sin(\u03b1-\u03b2)]/2
\u53cc\u66f2\u51fd\u6570
sinh(a) = [e^a-e^(-a)]/2 cosh(a) = [e^a+e^(-a)]/2 tanh(a) = sin h(a)/cos h(a) \u516c\u5f0f\u4e00\uff1a \u8bbe\u03b1\u4e3a\u4efb\u610f\u89d2\uff0c\u7ec8\u8fb9\u76f8\u540c\u7684\u89d2\u7684\u540c\u4e00\u4e09\u89d2\u51fd\u6570\u7684\u503c\u76f8\u7b49\uff1a sin\uff082k\u03c0\uff0b\u03b1\uff09= sin\u03b1 cos\uff082k\u03c0\uff0b\u03b1\uff09= cos\u03b1 tan\uff082k\u03c0\uff0b\u03b1\uff09= tan\u03b1 cot\uff082k\u03c0\uff0b\u03b1\uff09= cot\u03b1 \u516c\u5f0f\u4e8c\uff1a \u8bbe\u03b1\u4e3a\u4efb\u610f\u89d2\uff0c\u03c0+\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e0e\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a sin\uff08\u03c0\uff0b\u03b1\uff09= -sin\u03b1 cos\uff08\u03c0\uff0b\u03b1\uff09= -cos\u03b1 tan\uff08\u03c0\uff0b\u03b1\uff09= tan\u03b1 cot\uff08\u03c0\uff0b\u03b1\uff09= cot\u03b1 \u516c\u5f0f\u4e09\uff1a \u4efb\u610f\u89d2\u03b1\u4e0e -\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a sin\uff08-\u03b1\uff09= -sin\u03b1 cos\uff08-\u03b1\uff09= cos\u03b1 tan\uff08-\u03b1\uff09= -tan\u03b1 cot\uff08-\u03b1\uff09= -cot\u03b1 \u516c\u5f0f\u56db\uff1a \u5229\u7528\u516c\u5f0f\u4e8c\u548c\u516c\u5f0f\u4e09\u53ef\u4ee5\u5f97\u5230\u03c0-\u03b1\u4e0e\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a sin\uff08\u03c0-\u03b1\uff09= sin\u03b1 cos\uff08\u03c0-\u03b1\uff09= -cos\u03b1 tan\uff08\u03c0-\u03b1\uff09= -tan\u03b1 cot\uff08\u03c0-\u03b1\uff09= -cot\u03b1 \u516c\u5f0f\u4e94\uff1a \u5229\u7528\u516c\u5f0f-\u548c\u516c\u5f0f\u4e09\u53ef\u4ee5\u5f97\u52302\u03c0-\u03b1\u4e0e\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a sin\uff082\u03c0-\u03b1\uff09= -sin\u03b1 cos\uff082\u03c0-\u03b1\uff09= cos\u03b1 tan\uff082\u03c0-\u03b1\uff09= -tan\u03b1 cot\uff082\u03c0-\u03b1\uff09= -cot\u03b1 \u516c\u5f0f\u516d\uff1a \u03c0/2\u00b1\u03b1\u53ca3\u03c0/2\u00b1\u03b1\u4e0e\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a sin\uff08\u03c0/2+\u03b1\uff09= cos\u03b1 cos\uff08\u03c0/2+\u03b1\uff09= -sin\u03b1 tan\uff08\u03c0/2+\u03b1\uff09= -cot\u03b1 cot\uff08\u03c0/2+\u03b1\uff09= -tan\u03b1 sin\uff08\u03c0/2-\u03b1\uff09= cos\u03b1 cos\uff08\u03c0/2-\u03b1\uff09= sin\u03b1 tan\uff08\u03c0/2-\u03b1\uff09= cot\u03b1 cot\uff08\u03c0/2-\u03b1\uff09= tan\u03b1 sin\uff083\u03c0/2+\u03b1\uff09= -cos\u03b1 cos\uff083\u03c0/2+\u03b1\uff09= sin\u03b1 tan\uff083\u03c0/2+\u03b1\uff09= -cot\u03b1 cot\uff083\u03c0/2+\u03b1\uff09= -tan\u03b1 sin\uff083\u03c0/2-\u03b1\uff09= -cos\u03b1 cos\uff083\u03c0/2-\u03b1\uff09= -sin\u03b1 tan\uff083\u03c0/2-\u03b1\uff09= cot\u03b1 cot\uff083\u03c0/2-\u03b1\uff09= tan\u03b1 (\u4ee5\u4e0ak\u2208Z) A\u00b7sin(\u03c9t+\u03b8)+ B\u00b7sin(\u03c9t+\u03c6) = \u221a{(A² +B² +2ABcos(\u03b8-\u03c6)} \u00b7 sin{ \u03c9t + arcsin[ (A\u00b7sin\u03b8+B\u00b7sin\u03c6) / \u221a{A^2 +B^2; +2ABcos(\u03b8-\u03c6)} } \u221a\u8868\u793a\u6839\u53f7,\u5305\u62ec{\u2026\u2026}\u4e2d\u7684\u5185\u5bb9
\u8bf1\u5bfc\u516c\u5f0f
sin(-\u03b1) = -sin\u03b1 cos(-\u03b1) = cos\u03b1 tan (-\u03b1)=-tan\u03b1 sin(\u03c0/2-\u03b1) = cos\u03b1 cos(\u03c0/2-\u03b1) = sin\u03b1 sin(\u03c0/2+\u03b1) = cos\u03b1 cos(\u03c0/2+\u03b1) = -sin\u03b1 sin(\u03c0-\u03b1) = sin\u03b1 cos(\u03c0-\u03b1) = -cos\u03b1 sin(\u03c0+\u03b1) = -sin\u03b1 cos(\u03c0+\u03b1) = -cos\u03b1 tanA= sinA/cosA tan\uff08\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dcot\u03b1 tan\uff08\u03c0/2\uff0d\u03b1\uff09\uff1dcot\u03b1 tan\uff08\u03c0\uff0d\u03b1\uff09\uff1d\uff0dtan\u03b1 tan\uff08\u03c0\uff0b\u03b1\uff09\uff1dtan\u03b1 \u8bf1\u5bfc\u516c\u5f0f\u8bb0\u80cc\u8bc0\u7a8d\uff1a\u5947\u53d8\u5076\u4e0d\u53d8\uff0c\u7b26\u53f7\u770b\u8c61\u9650
\u4e07\u80fd\u516c\u5f0f
sin\u03b1=2tan(\u03b1/2)/[1+(tan(\u03b1/2))²] cos\u03b1=[1-(tan(\u03b1/2))²]/[1+(tan(\u03b1/2))²] tan\u03b1=2tan(\u03b1/2)/[1-(tan(\u03b1/2))²]
\u5176\u5b83\u516c\u5f0f
(1) (sin\u03b1)²+(cos\u03b1)²=1 (2)1+(tan\u03b1)²=(sec\u03b1)² (3)1+(cot\u03b1)²=(csc\u03b1)² \u8bc1\u660e\u4e0b\u9762\u4e24\u5f0f,\u53ea\u9700\u5c06\u4e00\u5f0f,\u5de6\u53f3\u540c\u9664(sin\u03b1)²\uff0c\u7b2c\u4e8c\u4e2a\u9664(cos\u03b1)²\u5373\u53ef (4)\u5bf9\u4e8e\u4efb\u610f\u975e\u76f4\u89d2\u4e09\u89d2\u5f62,\u603b\u6709 tanA+tanB+tanC=tanAtanBtanC \u8bc1: A+B=\u03c0-C tan(A+B)=tan(\u03c0-C) (tanA+tanB)/(1-tanAtanB)=(tan\u03c0-tanC)/(1+tan\u03c0tanC) \u6574\u7406\u53ef\u5f97 tanA+tanB+tanC=tanAtanBtanC \u5f97\u8bc1 \u540c\u6837\u53ef\u4ee5\u5f97\u8bc1,\u5f53x+y+z=n\u03c0(n\u2208Z)\u65f6,\u8be5\u5173\u7cfb\u5f0f\u4e5f\u6210\u7acb \u7531tanA+tanB+tanC=tanAtanBtanC\u53ef\u5f97\u51fa\u4ee5\u4e0b\u7ed3\u8bba (5)cotAcotB+cotAcotC+cotBcotC=1 (6)cot(A/2)+cot(B/2)+cot(C/2)=cot(A/2)cot(B/2)cot(C/2) (7)(cosA\uff09²+(cosB\uff09²+(cosC\uff09²=1-2cosAcosBcosC (8)\uff08sinA\uff09²+\uff08sinB\uff09²+\uff08sinC\uff09²=2+2cosAcosBcosC \u5176\u4ed6\u975e\u91cd\u70b9\u4e09\u89d2\u51fd\u6570 csc(a) = 1/sin(a) sec(a) = 1/cos(a)
\u7f16\u8f91\u672c\u6bb5\u5185\u5bb9\u89c4\u5f8b
\u4e09\u89d2\u51fd\u6570\u770b\u4f3c\u5f88\u591a\uff0c\u5f88\u590d\u6742\uff0c\u4f46\u53ea\u8981\u638c\u63e1\u4e86\u4e09\u89d2\u51fd\u6570\u7684\u672c\u8d28\u53ca\u5185\u90e8\u89c4\u5f8b\u5c31\u4f1a\u53d1\u73b0\u4e09\u89d2\u51fd\u6570\u5404\u4e2a\u516c\u5f0f\u4e4b\u95f4\u6709\u5f3a\u5927\u7684\u8054\u7cfb\u3002\u800c\u638c\u63e1\u4e09\u89d2\u51fd\u6570\u7684\u5185\u90e8\u89c4\u5f8b\u53ca\u672c\u8d28\u4e5f\u662f\u5b66\u597d\u4e09\u89d2\u51fd\u6570\u7684\u5173\u952e\u6240\u5728. 1\u3001\u4e09\u89d2\u51fd\u6570\u672c\u8d28\uff1a
[1] \u6839\u636e\u53f3\u56fe\uff0c\u6709 sin\u03b8=y/ r; cos\u03b8=x/r; tan\u03b8=y/x; cot\u03b8=x/y\u3002 \u6df1\u523b\u7406\u89e3\u4e86\u8fd9\u4e00\u70b9\uff0c\u4e0b\u9762\u6240\u6709\u7684\u4e09\u89d2\u516c\u5f0f\u90fd\u53ef\u4ee5\u4ece\u8fd9\u91cc\u51fa\u53d1\u63a8\u5bfc\u51fa\u6765\uff0c\u6bd4\u5982\u4ee5\u63a8\u5bfc sin(A+B) = sinAcosB+cosAsinB \u4e3a\u4f8b\uff1a \u63a8\u5bfc\uff1a \u9996\u5148\u753b\u5355\u4f4d\u5706\u4ea4X\u8f74\u4e8eC\uff0cD\uff0c\u5728\u5355\u4f4d\u5706\u4e0a\u6709\u4efb\u610fA\uff0cB\u70b9\u3002\u89d2AOD\u4e3a\u03b1\uff0cBOD\u4e3a\u03b2\uff0c\u65cb\u8f6cAOB\u4f7fOB\u4e0eOD\u91cd\u5408\uff0c\u5f62\u6210\u65b0A'OD\u3002 A(cos\u03b1,sin\u03b1),B(cos\u03b2,sin\u03b2),A'(cos(\u03b1-\u03b2),sin(\u03b1-\u03b2)) OA'=OA=OB=OD=1,D(1,0) \u2234[cos(\u03b1-\u03b2)-1]^2+[sin(\u03b1-\u03b2)]^2=(cos\u03b1-cos\u03b2)^2+(sin\u03b1-sin\u03b2)^2 \u548c\u5dee\u5316\u79ef\u53ca\u79ef\u5316\u548c\u5dee\u7528\u8fd8\u539f\u6cd5\u7ed3\u5408\u4e0a\u9762\u516c\u5f0f\u53ef\u63a8\u51fa\uff08\u6362(a+b)/2\u4e0e(a-b)/2\uff09 \u5355\u4f4d\u5706\u5b9a\u4e49 \u5355\u4f4d\u5706 \u516d\u4e2a\u4e09\u89d2\u51fd\u6570\u4e5f\u53ef\u4ee5\u4f9d\u636e\u534a\u5f84\u4e3a\u4e00\u4e2d\u5fc3\u4e3a\u539f\u70b9\u7684\u5355\u4f4d\u5706\u6765\u5b9a\u4e49\u3002\u5355\u4f4d\u5706\u5b9a\u4e49\u5728\u5b9e\u9645\u8ba1\u7b97\u4e0a\u6ca1\u6709\u5927\u7684\u4ef7\u503c\uff1b\u5b9e\u9645\u4e0a\u5bf9\u591a\u6570\u89d2\u5b83\u90fd\u4f9d\u8d56\u4e8e\u76f4\u89d2\u4e09\u89d2\u5f62\u3002\u4f46\u662f\u5355\u4f4d\u5706\u5b9a\u4e49\u7684\u786e\u5141\u8bb8\u4e09\u89d2\u51fd\u6570\u5bf9\u6240\u6709\u6b63\u6570\u548c\u8d1f\u6570\u8f90\u89d2\u90fd\u6709\u5b9a\u4e49\uff0c\u800c\u4e0d\u53ea\u662f\u5bf9\u4e8e\u5728 0 \u548c \u03c0/2 \u5f27\u5ea6\u4e4b\u95f4\u7684\u89d2\u3002\u5b83\u4e5f\u63d0\u4f9b\u4e86\u4e00\u4e2a\u56fe\u8c61\uff0c\u628a\u6240\u6709\u91cd\u8981\u7684\u4e09\u89d2\u51fd\u6570\u90fd\u5305\u542b\u4e86\u3002\u6839\u636e\u52fe\u80a1\u5b9a\u7406\uff0c\u5355\u4f4d\u5706\u7684\u7b49\u5f0f\u662f\uff1a \u56fe\u8c61\u4e2d\u7ed9\u51fa\u4e86\u7528\u5f27\u5ea6\u5ea6\u91cf\u7684\u4e00\u4e9b\u5e38\u89c1\u7684\u89d2\u3002\u9006\u65f6\u9488\u65b9\u5411\u7684\u5ea6\u91cf\u662f\u6b63\u89d2\uff0c\u800c\u987a\u65f6\u9488\u7684\u5ea6\u91cf\u662f\u8d1f\u89d2\u3002\u8bbe\u4e00\u4e2a\u8fc7\u539f\u70b9\u7684\u7ebf\uff0c\u540c x \u8f74\u6b63\u534a\u90e8\u5206\u5f97\u5230\u4e00\u4e2a\u89d2 \u03b8\uff0c\u5e76\u4e0e\u5355\u4f4d\u5706\u76f8\u4ea4\u3002\u8fd9\u4e2a\u4ea4\u70b9\u7684 x \u548c y \u5750\u6807\u5206\u522b\u7b49\u4e8e cos \u03b8 \u548c sin \u03b8\u3002\u56fe\u8c61\u4e2d\u7684\u4e09\u89d2\u5f62\u786e\u4fdd\u4e86\u8fd9\u4e2a\u516c\u5f0f\uff1b\u534a\u5f84\u7b49\u4e8e\u659c\u8fb9\u4e14\u957f\u5ea6\u4e3a1\uff0c\u6240\u4ee5\u6709 sin \u03b8 = y/1 \u548c cos \u03b8 = x/1\u3002\u5355\u4f4d\u5706\u53ef\u4ee5\u88ab\u89c6\u4e3a\u662f\u901a\u8fc7\u6539\u53d8\u90bb\u8fb9\u548c\u5bf9\u8fb9\u7684\u957f\u5ea6\uff0c\u4f46\u4fdd\u6301\u659c\u8fb9\u7b49\u4e8e 1\u7684\u4e00\u79cd\u67e5\u770b\u65e0\u9650\u4e2a\u4e09\u89d2\u5f62\u7684\u65b9\u5f0f\u3002 \u4e24\u89d2\u548c\u516c\u5f0f
sin(A+B) = sinAcosB+cosAsinB sin(A-B) = sinAcosB-cosAsinB cos(A+B) = cosAcosB-sinAsinB cos(A-B) = cosAcosB+sinAsinB tan(A+B) = (tanA+tanB)/(1-tanAtanB) tan(A-B) = (tanA-tanB)/(1+tanAtanB) cot(A+B) = (cotAcotB-1)/(cotB+cotA) cot(A-B) = (cotAcotB+1)/(cotB-cotA)
\u5b58\u5728\u7684\u4e00\u822c\u5173\u7cfb\u6709\uff1a
\uff081\uff09 sin2A+cos2A=1\uff1b
\uff082\uff09tanA=\uff1b
\uff081\uff09 \u8bc1\u660e\uff1a\u2235 sinA=,cosA=,
a2+b2=c2,\u2234 sin2A+cos2A=\uff1b
\uff082\uff09 \u8bc1\u660e\uff1a\u2235 sinA=,cosA=,\u2234 tanA=.
\u4e09\u89d2\u51fd\u6570\u516c\u5f0f\uff1a
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倒数关系： tanα ·cotα=1 sinα ·cscα=1 cosα ·secα=1
商的关系： sinα/cosα=tanα=secα/cscα cosα/sinα=cotα=cscα/secα
平方关系： sin^2(α)+cos^2(α)=1 1+tan^2(α)=sec^2(α) 1+cot^2(α)=csc^2(α)
平常针对不同条件的常用的两个公式
sin^2(α)+cos^2(α)=1 tan α *cot α=1
一个特殊公式
（sina+sinθ）*（sina-sinθ）=sin（a+θ）*sin（a-θ）
证明：（sina+sinθ）*（sina-sinθ）=2 sin[(θ+a)/2] cos[(a-θ)/2] *2 cos[(θ+a)/2] sin[(a-θ)/2] =sin（a+θ）*sin（a-θ）
二倍角公式
正弦
sin2A=2sinA·cosA
余弦
1.Cos2a=Cos^2(a)-Sin^2(a)
2.Cos2a=1-2Sin^2(a)
3.Cos2a=2Cos^2(a)-1
即Cos2a=Cos^2(a)-Sin^2(a)=2Cos^2(a)-1=1-2Sin^2(a)
正切 tan2A=（2tanA）/（1-tan^2(A)）
和差化积
sinθ+sinφ = 2 sin[(θ+φ)/2] cos[(θ-φ)/2]
sinθ-sinφ = 2 cos[(θ+φ)/2] sin[(θ-φ)/2]
cosθ+cosφ = 2 cos[(θ+φ)/2] cos[(θ-φ)/2]
cosθ-cosφ = -2 sin[(θ+φ)/2] sin[(θ-φ)/2]
tanA+tanB=sin(A+B)/cosAcosB=tan(A+B)(1-tanAtanB)
tanA-tanB=sin(A-B)/cosAcosB=tan(A-B)(1+tanAtanB)
积化和差
sinαsinβ =-[cos(α+β)-cos(α-β)] /2
cosαcosβ = [cos(α+β)+cos(α-β)]/2
sinαcosβ = [sin(α+β)+sin(α-β)]/2
cosαsinβ = [sin(α+β)-sin(α-β)]/2
三角函数的诱导公式（六公式）
公式一 sin(-α) = -sinα cos(-α) = cosα tan (-α)=-tanα
公式二sin(π/2-α) = cosα cos(π/2-α) = sinα
公式三 sin(π/2+α) = cosα cos(π/2+α) = -sinα
公式四sin(π-α) = sinα cos(π-α) = -cosα
公式五sin(π+α) = -sinα cos(π+α) = -cosα
公式六tanA= sinA/cosA tan（π/2+α）=－cotα tan（π/2－α）=cotα tan（π－α）=－tanα tan（π+α）=tanα
诱导公式记背诀窍：奇变偶不变，符号看象限
1.面积公式S=(1/2)a×ha
S=(1/2)ab×sinC
S=rs
S=abc/(4R)
S=2R²×sinAsinBsinC
S=s(s-a)×tan(A/2)
S=√[(s-a)(s-b)(s-c)s] (海伦公式)
S=s²×tan(A/2)tan(B/2)tan(C/2)
S=(a²-b²)sinAsinB/[2sin(A-B)]
2.中线.a边中线长Ma=(1/2)×√(2b²+2c²-a²)
=(1/2)×√(b²+c²+2bc×cosA)
3.高.a边高长ha=c×sinB=b×sinC
ha=a×sinBsinC/sinA
ha=√[b²-(a²+b²-c²)²/(2a)² ]
4.角平分线.a边角平分线长la=2bc×cos(A/2)/(b+c)
la=√{bc[(b+c)²-a²]}/(b+c)
5.内切圆,外接圆半径:
r=S/s=4R×sin(A/2)sin(B/2)sin(C/2)
r=s×tan(A/2)tan(B/2)tan(C/2)
R=a/(2sinA)=abc/(4s)=abc/[2r(a+b+c)]
6.同角三角函数间的关系:
sinα×cscα=1
cosα×secα=1
tanα×cotα=1
tanα=sinα/cosα,cotα=cosα/sinα
(sinα)²+(cosα)²=1
1+(tanα)²=(secα)²
1+(cotα)²=(cscα)²
7.正弦定理：
a/sinA=b/sinB=c/sinC=2R
8.余弦定理：
a²=b²+c²-2bc cosA
b²=a²+c²-2ac cosB
c²=a²+b²-2ab cosC
9.倍角公式:
sin(2α)=2sinαcosα
cos(2α)=(cosα)²-1=1-2(sinα)²
tan(2α)=2tanα/[1-(tanα)²]
sin(3α)=3sinα-4(sinα)^3
cos(3α)=4(cosα)^3-3cosα
单位圆
没必要
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绛旓細1.娴蜂鸡鍏紡 鈻矨BC涓 涓夎竟涓篴,b,c銆 p=(a+b+c)/2.S(abc)=鈭歔p(p-a)(p-b)(p-c)]鍗冲凡鐭ヤ笁瑙掑舰涓夎竟姹傞潰绉殑娴蜂鸡鍏紡銆2.宸茬煡涓夎褰㈠簳a锛岄珮h锛屽垯S锛漚h/2 3.宸茬煡涓夎褰袱杈筧,b,杩欎袱杈瑰す瑙扖锛屽垯S锛漚bsinC/2 4.璁句笁瑙掑舰涓夎竟鍒嗗埆涓篴銆乥銆乧锛屽唴鍒囧渾鍗婂緞涓簉 鍒欎笁瑙掑舰闈㈢Н=(a...
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绛旓細涓夎褰㈣绠楀叕寮忔槸Sina=1锛屼笁瑙掑舰鏄敱鍚屼竴骞抽潰鍐呬笉鍦ㄥ悓涓鐩寸嚎涓婄殑涓夋潯绾挎棣栧熬椤烘杩炴帴鎵缁勬垚鐨勫皝闂浘褰紝鍦ㄦ暟瀛︺佸缓绛戝鏈夊簲鐢ㄣ傜敱涓夋潯绾挎棣栧熬椤烘鐩歌繛锛屽緱鍒扮殑灏侀棴鍑犱綍鍥惧舰鍙綔涓夎褰備笁瑙掑舰鏄嚑浣曞浘妗堢殑鍩烘湰鍥惧舰銆傚父瑙佺殑涓夎褰㈡寜杈瑰垎鏈夋櫘閫氫笁瑙掑舰锛堜笁鏉¤竟閮戒笉鐩哥瓑锛夛紝绛夎叞涓夎锛堣叞涓庡簳涓嶇瓑鐨勭瓑鑵...
绛旓細鈶ｇ洿瑙掍笁瑙掑舰涓30搴﹁鎵瀵圭殑鐩磋杈圭瓑浜庢枩杈圭殑涓鍗娿傜洿瑙涓夎褰㈢殑鍒ゅ畾锛氣憼鏈変袱涓浜掍綑鐨勪笁瑙掑舰鏄洿瑙掍笁瑙掑舰銆傗憽濡傛灉涓夎褰㈢殑涓夎竟闀縜銆乥 銆乧鏈変笅闈㈠叧绯籥^2+b^2=c^2锛岄偅涔堣繖涓笁瑙掑舰鏄洿瑙掍笁瑙掑舰锛堝嬀鑲″畾鐞嗙殑閫嗗畾鐞嗭級銆備互涓婂鏁板鐩磋涓夎褰㈠畾鐞鍏紡鐨勫唴瀹硅瑙ｅ涔狅紝鍚屽浠兘鑳藉緢濂界殑鎺屾彙浜嗗惂锛...
绛旓細娴蜂鸡鍏紡:涓夎竟闈㈢Н娴蜂鸡鍏紡 鈻矨BC涓璸=(a+b+c)/2:S(ABC)=鈭歔p(p-a)(p-b)(p-c)]鍗冲凡鐭涓夎褰涓夎竟姹傞潰绉殑娴蜂鸡鍏紡銆傚凡鐭ヤ笁瑙掑舰搴昦锛岄珮h锛屽垯S锛漚h/2 闈㈢Н鍏紡 宸茬煡涓夎褰笁杈筧,b,c,鍗婂懆闀縫,鍒橲锛 鈭歔p(p - a)(p - b)(p - c)] 锛堟捣浼﹀叕寮忥級锛坧=(a+b+c)/2锛夊凡鐭...
绛旓細6.鍚岃涓夎鍑芥暟闂寸殑鍏崇郴:sin伪脳csc伪=1 cos伪脳sec伪=1 tan伪脳cot伪=1 tan伪=sin伪/cos伪,cot伪=cos伪/sin伪 (sin伪)2+(cos伪)2=1 1+(tan伪)2=(sec伪)2 1+(cot伪)2=(csc伪)2 7.姝ｅ鸡瀹氱悊锛歛/sinA=b/sinB=c/sinC=2R 8.浣欏鸡瀹氱悊锛歛2=b2+c2-2bc cosA b2=a2+c2-2ac ...
绛旓細涓夎褰㈢殑闈㈢Н鍏紡 (1)S鈻=1/2ah 锛坅鏄笁瑙掑舰鐨勫簳锛宧鏄簳鎵瀵瑰簲鐨勯珮锛(2)S鈻=1/2acsinB锛1/2bcsinA锛1/2absinC 锛堜笁涓涓衡垹A鈭燘鈭燙锛屽杈瑰垎鍒负a,b,c锛屽弬瑙佷笁瑙掑嚱鏁帮級(3)S鈻=鈭氥攑(p-a)(p-b)(p-c)銆 銆攑=1/2(a+b+c)銆曪紙娴蜂鸡鈥旂Е涔濋煻鍏紡锛(4)S鈻=abc/(4R) (...
绛旓細涓鑸涓夎褰 鍥犲紡鍒嗛潰绉細S锛漚h/2 (2).宸茬煡涓夎褰笁杈筧,b,c锛屽垯锛堟捣浼鍏紡锛夛紙p=(a+b+c)/2锛塖=鈭歔p(p-a)(p-b)(p-c)]=锛1/4锛夆垰[(a+b+c)(a+b-c)(a+c-b)(b+c-a)](3).宸茬煡涓夎褰袱杈筧,b,杩欎袱杈瑰す瑙扖锛屽垯S锛1/2 absinC (4).璁句笁瑙掑舰涓夎竟鍒嗗埆涓篴銆乥銆乧锛屽唴...
