求高一数学必修二关于sin(90-x)之类的公式
公式一 设α为任意角,终边相同的角的同一三角函数的值相等:对于x轴正半轴为起点轴而言 弧度制下的角的表示: sin(2kπ+α)=sinα (k∈Z) cos(2kπ+α)=cosα (k∈Z) tan(2kπ+α)=tanα ...绛旓細sinA+sinB=2sin((A+B)/2)cos((A-B)/2 cosA+cosB=2cos((A+B)/2)sin((A-B)/2)tanA+tanB=sin(A+B)/cosAcosB tanA-tanB=sin(A-B)/cosAcosB ctgA+ctgBsin(A+B)/sinAsinB -ctgA+ctgBsin(A+B)/sinAsinB 鏌愪簺鏁板垪鍓峮椤瑰拰 1+2+3+4+5+6+7+8+9+鈥+n=n(n+1)/2 1+...
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绛旓細蹇呬慨浜鎴戣鐩寸嚎涓庢柟绋,鍦嗕笌鏂圭▼鐨勫叕寮忓ぇ鍏,鎴戣繕瑕佸繀淇竴鐨勯泦鍚,鍑芥暟鐨勫叕寮,鏈濂借繕缁欐垜涓浜涘父鑰冮鐨勮В棰樻濊矾缁,灏忓紵浼氭劅婵涓嶅敖鐨!!!... 蹇呬慨浜屾垜瑕佺洿绾夸笌鏂圭▼,鍦嗕笌鏂圭▼鐨勫叕寮忓ぇ鍏,鎴戣繕瑕佸繀淇竴鐨勯泦鍚,鍑芥暟鐨勫叕寮,鏈濂借繕缁欐垜涓浜涘父鑰冮鐨勮В棰樻濊矾缁,灏忓紵浼氭劅婵涓嶅敖鐨!!! 灞曞紑 2涓洖绛...
绛旓細tan8A=-8*tanA*(-1+7*tanA^2-7*tanA^4+tanA^6)/(1-28*tanA^2+70*tanA^4-28*tanA^6+tanA^8) 涔濆嶈鍏紡: sin9A=(sinA*(-3+4*sinA^2)*(64*sinA^6-96*sinA^4+36*sinA^2-3)) cos9A=(cosA*(-3+4*cosA^2)*(64*cosA^6-96*cosA^4+36*cosA^2-3)) tan9A=tanA*(9-84*tanA^...
绛旓細cosA=12\13,,,sinA=5\13 cosB=12\13 ,sinB=5\13 cosC=cos(180-2A)=cos180cos2A+sin180sin2A =-cos2A=-2(cosA)2+1 =-2*144\169+1 =-119\169 sinC=sin(180-2A)=sin180cos2A-sin2Acos180=sin2A=2sinAcosA=120\169 tanC=sinA\cosA=-120\119 ...
绛旓細f(x)=sinx+鈭3cosx+2=2锛sinx/2+鈭3cosx/2锛+2=2sin锛x+蟺/3锛+2 1銆佅=1锛孴=2蟺/蠅锛屾墍浠(x)鐨勬渶灏忔鍛ㄦ湡涓2蟺 2銆佸綋sin锛坸+蟺/3锛=1鏃讹紝鍑芥暟鏈夋渶澶у硷紝姝ゆ椂x+蟺/3=蟺/2+2k蟺锛屾墍浠=蟺/3+2k蟺 3銆佹寮﹀嚱鏁扮殑瀵圭О杞翠负k蟺+蟺/2锛屾墍浠(x)鐨勫绉拌酱锛歺+蟺/3=...
绛旓細鍥炵瓟锛(1)f(伪)=[-sin伪(-cos伪)sin伪]/[-cos伪sin伪(-cos伪)]=tan伪 (2)f(伪)=tan伪=4/3, 涓斘卞湪绗笁璞¢檺,鎵浠in伪=-4/5, cos伪=-3/5 cos伪-sin伪=-3/5+4/5=1/5
绛旓細tan(A-B)=(tanA-tanB)/(1+tanAtanB) ctg(A+B)=(ctgActgB-1)/(ctgB+ctgA) ctg(A-B)=(ctgActgB+1)/(ctgB-ctgA) 鍊嶈鍏紡 tan2A=2tanA/(1-tan2A) ctg2A=(ctg2A-1)/2ctga cos2a=cos2a-sin2a=2cos2a-1=1-2sin2a 鍗婅鍏紡 sin(A/2)=鈭((1-cosA)/2) sin(...
绛旓細sinA+sinB=2sin((A+B)/2)cos((A-B)/2 cosA+cosB=2cos((A+B)/2)sin((A-B)/2)tanA+tanB=sin(A+B)/cosAcosB tanA-tanB=sin(A-B)/cosAcosB ctgA+ctgBsin(A+B)/sinAsinB -ctgA+ctgBsin(A+B)/sinAsinB 鏌愪簺鏁板垪鍓峮椤瑰拰 1+2+3+4+5+6+7+8+9+鈥+n=n(n+1)/2 1+...
绛旓細锛2锛 B=0 X=-C/A (3) A=0, y=-C/B (4) A鈮0锛屼笖B鈮0,2,1銆丆=0锛2銆丅=0锛3銆丄=0锛4銆丄B閮戒笉绛変簬0,2,涓閬鍏充簬楂樹竴鏁板蹇呬慨浜鐩寸嚎涓庢柟绋嬮棶棰橈紙姹傝瑙o級璁剧洿绾縇鐨勬柟绋嬩负Ax+By+C=0,(A,B涓嶅悓鏃朵负0),鏍规嵁涓嬪垪鏉′欢,姹傚嚭A,B,C搴旀弧瓒崇殑鏉′欢锛1.鐩寸嚎L杩囧師鐐, 2.鐩寸嚎L...