sin阿尔法+sin贝塔=？？？？？ sin(α-β)=？
sin\u963f\u5c14\u6cd5+sin\u8d1d\u5854=\uff1f\u5177\u4f53\u70b9sina+sinb
=sin[(a+b)/2+(a-b)/2]+sin[(a+b)/2-(a-b)/2]
=sin(a+b)/2cos(a-b)/2+cos(a+b)/2sin(a-b)/2+sin(a+b)/2cos(a-b)/2-cos(a+b)/2sin(a-b)/2
=2sin(a+b)/2cos(a-b)/2
sin(\u03b1-\u03b2)=sin\u03b1cos\u03b2-cos\u03b1sin\u03b2
sin(\u03b1+\u03b2)=sin\u03b1cos\u03b2+cos\u03b1sin\u03b2
\u6269\u5c55\u8d44\u6599\uff1a
\u4e24\u89d2\u548c\u7684\u516c\u5f0f
sin(A+B)=sinAcosB+cosAsinB sin(A-B)=sinAcosB-sinBcosA
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)
\u500d\u89d2\u7684\u516c\u5f0f
tan2A=2tanA/(1-tan2A) cot2A=(cot2A-1)/2cota
cos2a=cos2a-sin2a=2cos2a-1=1-2sin2a
sin\u03b1+sin(\u03b1+2\u03c0/n)+sin(\u03b1+2\u03c02/n)+sin(\u03b1+2\u03c03/n)+\u2026\u2026+sin[\u03b1+2\u03c0*(n-1)/n]=0
cos\u03b1+cos(\u03b1+2\u03c0/n)+cos(\u03b1+2\u03c02/n)+cos(\u03b1+2\u03c03/n)+\u2026\u2026+cos[\u03b1+2\u03c0*(n-1)/n]=0
sin2(\u03b1)+sin2(\u03b1-2\u03c0/3)+sin^2(\u03b1+2\u03c0/3)=3/2
tanAtanBtan(A+B)+tanA+tanB-tan(A+B)=0
\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1-\u4e09\u89d2\u51fd\u6570\u516c\u5f0f
绛旓細=sin[(a+b)/2+(a-b)/2]+sin[(a+b)/2-(a-b)/2]=sin(a+b)/2cos(a-b)/2+cos(a+b)/2sin(a-b)/2+sin(a+b)/2cos(a-b)/2-cos(a+b)/2sin(a-b)/2 =2sin(a+b)/2cos(a-b)/2
绛旓細-1=<sin闃垮皵娉<=1,-1=<sin璐濆<=1 sin闃垮皵娉=1,sin璐濆=1,闃垮皵娉=璐濆=蟺/2 cos(闃垮皵娉+璐濆)=-1 鎴杝in=-1sin闃垮皵娉=-1,cos璐濆=-1,闃垮皵娉=璐濆=3蟺/2 cos(闃垮皵娉+璐濆)=-1 缁间笂cos(闃垮皵娉+璐濆)=-1
绛旓細sin(2伪)=2sin伪路cos伪=2/(tan伪+cot伪)cos(2伪)=cos²伪-sin²伪=2cos²伪-1=1-2sin²伪 tan(2伪)=2tan伪/[1-(tan伪)²]cot(2伪)=(cot²伪-1)/(2cot伪)sec(2伪)=sec²伪/(1-tan²伪)csc(2伪)=1/2sec伪路csc伪 ...
绛旓細sin伪+sin尾=2sin[(伪+尾)/2]cos[(伪-尾)/2]
绛旓細cos伪路sin尾=(1/2)[sin(伪+尾)-sin(伪-尾)]cos伪路cos尾=(1/2)[cos(伪+尾)+cos(伪-尾)]sin伪路sin尾=-(1/2)[cos(伪+尾)-cos(伪-尾)]鐩稿叧鍏紡锛氬钩鏂瑰拰鍏崇郴锛氾紙sin伪锛塣2 +锛坈os伪锛塣2=1 绉殑鍏崇郴锛歴in伪 = tan伪 脳 cos伪锛堝嵆sin伪 / cos伪 = tan伪 锛塩os伪 = ...
绛旓細sin伪銆sin尾銆乻in纬鎴愮瓑姣旀暟鍒楋紝鍒檚in²尾=sin伪路sin纬 (1/2)[1-cos(2尾)]=sin伪路sin纬 1-cos(2尾)=sin伪路sin纬 2尾=伪+纬浠ｅ叆 1-cos(伪+纬)=2sin伪路sin纬 1-cos伪cos纬+sin伪sin纬=2sin伪sin纬 cos伪cos纬+sin伪sin纬=1 cos(伪-纬)=1 伪銆佄冲潎涓洪攼瑙掞紝0<伪...
绛旓細sin(伪-尾)=sin伪cos尾-cos伪sin尾锛岀被浼肩殑鍏紡杩樻湁锛1銆乻in(伪+尾)=sin伪cos尾+cos伪sin尾 2銆乧os(伪+尾)=cos伪cos尾-sin伪sin尾 3銆乧os(伪-尾)=cos伪cos尾+sin伪sin尾
绛旓細鐢变簬闃垮皵娉 璐濆灞炰簬(0,娲/2)锛屾墍浠ユ棰樹腑鎵鏈変笁瑙掑嚱鏁板潎涓烘鍊
绛旓細sin(伪锛2)=卤鈭氾紙(1-cos伪锛/2)銆俢os(伪锛2)=卤鈭氾紙(1+cos伪锛/2)銆倀an(伪锛2)=卤鈭氾紙(1-cos伪锛/(1+cos伪锛)=sin伪锛(1+cos伪锛=(1-cos伪锛/sin伪銆傝鍦ㄥ嚑浣曞涓紝鏄敱涓ゆ潯鏈夊叕鍏辩鐐圭殑灏勭嚎缁勬垚鐨勫嚑浣曞璞°傝繖涓ゆ潯灏勭嚎鍙仛瑙掔殑杈癸紝瀹冧滑鐨勫叕鍏辩鐐瑰彨鍋氳鐨勯《鐐广備竴鑸殑瑙掍細...
绛旓細伪,尾 閮芥槸閿愯 伪-尾鈭(-蟺/2锛屜/2)sin(伪-尾)=1/3 鈭碿os(伪-尾)=2鈭2/3 伪+尾鈭(0,蟺)cos(伪+尾)=1/4 鈭磗in(伪+尾)=鈭15/4 sin2伪 =sin[(伪-尾)+(伪+尾)]=sin(伪-尾)cos(伪+尾)+cos(伪-尾)sin(伪+尾)=1/3*1/4+2鈭2/3*鈭15/4 =(2鈭30+1)/12...
