高一数学必修4降幂公式是什么
\u9ad8\u4e00\u6570\u5b66\u5fc5\u4fee4\u516c\u5f0f\uff0c\u8981\u5168\uff01\u540c\u89d2\u4e09\u89d2\u51fd\u6570\u95f4\u7684\u57fa\u672c\u5173\u7cfb\u5f0f\uff1a
\u00b7\u5e73\u65b9\u5173\u7cfb\uff1a
sin^2(\u03b1)+cos^2(\u03b1)=1
tan^2(\u03b1)+1=sec^2(\u03b1)
cot^2(\u03b1)+1=csc^2(\u03b1)
\u00b7\u79ef\u7684\u5173\u7cfb\uff1a
sin\u03b1=tan\u03b1*cos\u03b1 cos\u03b1=cot\u03b1*sin\u03b1
tan\u03b1=sin\u03b1*sec\u03b1 cot\u03b1=cos\u03b1*csc\u03b1
sec\u03b1=tan\u03b1*csc\u03b1 csc\u03b1=sec\u03b1*cot\u03b1
\u00b7\u5012\u6570\u5173\u7cfb\uff1a
tan\u03b1\u00b7cot\u03b1=1
sin\u03b1\u00b7csc\u03b1=1
cos\u03b1\u00b7sec\u03b1=1
\u4e09\u89d2\u51fd\u6570\u6052\u7b49\u53d8\u5f62\u516c\u5f0f
\u00b7\u4e24\u89d2\u548c\u4e0e\u5dee\u7684\u4e09\u89d2\u51fd\u6570\uff1a
cos(\u03b1+\u03b2)=cos\u03b1\u00b7cos\u03b2-sin\u03b1\u00b7sin\u03b2
cos(\u03b1-\u03b2)=cos\u03b1\u00b7cos\u03b2+sin\u03b1\u00b7sin\u03b2
sin(\u03b1\u00b1\u03b2)=sin\u03b1\u00b7cos\u03b2\u00b1cos\u03b1\u00b7sin\u03b2
tan(\u03b1+\u03b2)=(tan\u03b1+tan\u03b2)/(1-tan\u03b1\u00b7tan\u03b2)
tan(\u03b1-\u03b2)=(tan\u03b1-tan\u03b2)/(1+tan\u03b1\u00b7tan\u03b2)
\u00b7\u8f85\u52a9\u89d2\u516c\u5f0f\uff1a
Asin\u03b1+Bcos\u03b1=(A^2+B^2)^(1/2)sin(\u03b1+t)\uff0c\u5176\u4e2d
sint=B/(A^2+B^2)^(1/2)
cost=A/(A^2+B^2)^(1/2)
\u00b7\u500d\u89d2\u516c\u5f0f\uff1a
sin(2\u03b1)=2sin\u03b1\u00b7cos\u03b1=2/(tan\u03b1+cot\u03b1)
cos(2\u03b1)=cos^2(\u03b1)-sin^2(\u03b1)=2cos^2(\u03b1)-1=1-2sin^2(\u03b1)
tan(2\u03b1)=2tan\u03b1/[1-tan^2(\u03b1)]
\u00b7\u4e09\u500d\u89d2\u516c\u5f0f\uff1a
sin3\u03b1=3sin\u03b1-4sin^3(\u03b1)
cos3\u03b1=4cos^3(\u03b1)-3cos\u03b1
\u00b7\u534a\u89d2\u516c\u5f0f\uff1a
sin(\u03b1/2)=\u6b63\u8d1f\u221a((1-cos\u03b1)/2)
cos(\u03b1/2)=\u6b63\u8d1f\u221a((1+cos\u03b1)/2)
tan(\u03b1/2)=\u6b63\u8d1f\u221a((1-cos\u03b1)/(1+cos\u03b1))=sin\u03b1/(1+cos\u03b1)=(1-cos\u03b1)/sin\u03b1
\u00b7\u964d\u5e42\u516c\u5f0f
sin^2(\u03b1)=(1-cos(2\u03b1))/2
cos^2(\u03b1)=(1+cos(2\u03b1))/2
tan^2(\u03b1)=(1-cos(2\u03b1))/(1+cos(2\u03b1))
\u00b7\u4e07\u80fd\u516c\u5f0f\uff1a
sin\u03b1=2tan(\u03b1/2)/[1+tan^2(\u03b1/2)]
cos\u03b1=[1-tan^2(\u03b1/2)]/[1+tan^2(\u03b1/2)]
tan\u03b1=2tan(\u03b1/2)/[1-tan^2(\u03b1/2)]
\u00b7\u79ef\u5316\u548c\u5dee\u516c\u5f0f\uff1a
sin\u03b1\u00b7cos\u03b2=(1/2)[sin(\u03b1+\u03b2)+sin(\u03b1-\u03b2)]
cos\u03b1\u00b7sin\u03b2=(1/2)[sin(\u03b1+\u03b2)-sin(\u03b1-\u03b2)]
cos\u03b1\u00b7cos\u03b2=(1/2)[cos(\u03b1+\u03b2)+cos(\u03b1-\u03b2)]
sin\u03b1\u00b7sin\u03b2=-(1/2)[cos(\u03b1+\u03b2)-cos(\u03b1-\u03b2)]
\u00b7\u548c\u5dee\u5316\u79ef\u516c\u5f0f\uff1a
sin\u03b1+sin\u03b2=2sin[(\u03b1+\u03b2)/2]cos[(\u03b1-\u03b2)/2]
sin\u03b1-sin\u03b2=2cos[(\u03b1+\u03b2)/2]sin[(\u03b1-\u03b2)/2]
cos\u03b1+cos\u03b2=2cos[(\u03b1+\u03b2)/2]cos[(\u03b1-\u03b2)/2]
cos\u03b1-cos\u03b2=-2sin[(\u03b1+\u03b2)/2]sin[(\u03b1-\u03b2)/2]
\u00b7\u5176\u4ed6\uff1a
sin\u03b1+sin(\u03b1+2\u03c0/n)+sin(\u03b1+2\u03c0*2/n)+sin(\u03b1+2\u03c0*3/n)+\u2026\u2026+sin[\u03b1+2\u03c0*(n-1)/n]=0
cos\u03b1+cos(\u03b1+2\u03c0/n)+cos(\u03b1+2\u03c0*2/n)+cos(\u03b1+2\u03c0*3/n)+\u2026\u2026+cos[\u03b1+2\u03c0*(n-1)/n]=0 \u4ee5\u53ca
sin^2(\u03b1)+sin^2(\u03b1-2\u03c0/3)+sin^2(\u03b1+2\u03c0/3)=3/2
tanAtanBtan(A+B)+tanA+tanB-tan(A+B)=0
sinxcosx=1/2 sin2x
sin²x=(1-cos2x)/2
cos²x=（1+cos2x）/2
降幂升角才可以 如(cosx)^2=(1+cos2x)/2
绛旓細闄嶅箓鍏紡鏄閲嶈鐨勫叕寮忥紝瑕佽寰楀摝 sinxcosx=1/2 sin2x sin²x=(1-cos2x)/2 cos²x=锛1+cos2x锛/2
绛旓細楂樹竴鏁板蹇呬慨4鍏充簬涓夎鐨闄嶅箓鍏紡 灞曞紑 鎴戞潵绛 1涓洖绛 #鐑# 瀛╁瓙涔嬮棿鎵撴灦 鐖舵瘝瑕佷笉瑕佸共棰?鍖垮悕鐢ㄦ埛 2013-05-24 灞曞紑鍏ㄩ儴 鎺ㄥ鍏紡:(a+b+c)/(sinA+sinB+sinC)=2R(鍏朵腑,R涓哄鎺ュ渾鍗婂緞) 鐢辨寮﹀畾鐞嗘湁 a/sinA=b/sinB=c/sinC=2R 鎵浠 a=2R*sinA b=2R*sinB c=2R*sinC 鍔犺捣鏉+b+c=2...
绛旓細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)鍊嶈鍏紡 tan2a=2t...
绛旓細cot锛埾锛嵨憋級锛濓紞cot伪 鍏紡浜 鍏紡浜旓細鍒╃敤鍏紡涓鍜屽叕寮忎笁鍙互寰楀埌2蟺-伪涓幬辩殑涓夎鍑芥暟鍊间箣闂寸殑鍏崇郴锛
绛旓細鍥涳級鐢ㄤ簩鍊嶈涓殑浣欏鸡鍙帹鍑洪檷骞傚叕寮 (sinA)^2=(1-cos2A)/2 (cosA)^2=(1+cos2A)/2 浜旓級鐢ㄤ互涓婇檷骞傚叕寮忓彲鎺ㄥ嚭浠ヤ笅甯哥敤鐨勫寲绠鍏紡 1-cosA=sin^(A/2)*2 1-sinA=cos^(A/2)*2 + 涓锛変袱瑙掑拰宸叕寮 锛堝啓鐨勯兘瑕佽锛塻in(A+B)=sinAcosB+cosAsinB sin(A-B)=sinAcosB-sinBcosA &#x...
绛旓細鈷嬪崐瑙掔殑姝ｅ鸡銆佷綑寮﹀拰姝ｅ垏鍏紡(闄嶅箓鎵╄鍏紡)1-cos伪sin^2(伪/2)=---21+cos伪cos^2(伪/2)=---21-cos伪tan^2(伪/2)=---1+cos伪涓囪兘鍏紡鈷屼竾鑳藉叕寮2tan(伪/2)sin伪=---1+tan^2(伪/2)1-tan^2(伪/2)cos伪=---1+tan^2(伪/2)2tan(伪/2)tan伪=---1-tan^2(伪/2)涓囪兘鍏紡鎺ㄥ...
绛旓細楂樹竴鏁板蹇呬慨涓鎵鏈夊叕寮忓綊绾虫槸濡備笅锛1銆侀攼瑙掍笁瑙掑嚱鏁板叕寮忥細sin伪=鈭犖辩殑瀵硅竟/鏂滆竟銆2銆佷笁鍊嶈鍏紡锛歴in3伪=4sin伪路sin(蟺/3+伪)sin(蟺/3-伪)銆3銆佽緟鍔╄鍏紡锛欰sin伪+Bcos伪=(A^2+B^2)^(1/2)sin(伪+t)銆4銆闄嶅箓鍏紡锛歴in^2(伪)=(1-cos(2伪))/2=versin(2伪)/2銆5銆佹帹瀵...
绛旓細闄嶅箓鍏紡 sin²伪=[1-cos(2伪)]/2 cos²伪=[1+cos(2伪)]/2 tan²伪=[1-cos(2伪)]/[1+cos(2伪)]涓夎鍜 sin(伪+尾+纬)=sin伪路cos尾路cos纬+cos伪路sin尾路cos纬+cos伪路cos尾路sin纬-sin伪路sin尾路sin纬 cos(伪+尾+纬)=cos伪路cos尾路cos纬-cos伪路...
绛旓細鈷嬪崐瑙掔殑姝ｅ鸡銆佷綑寮﹀拰姝ｅ垏鍏紡锛闄嶅箓鎵╄鍏紡锛1锛峜os伪 sin^2(伪/2)锛濃斺斺2 1锛媍os伪 cos^2(伪/2)锛濃斺斺2 1锛峜os伪 tan^2(伪/2)锛濃斺斺1锛媍os伪 涓囪兘鍏紡 鈷屼竾鑳藉叕寮 2tan(伪/2)sin伪锛濃斺斺1锛媡an^2(伪/2)1锛峵an^2(伪/2)cos伪锛濃斺斺1锛媡an^2(伪/2)2tan(伪...
绛旓細鍊嶈鍏紡锛歵an2A=2tanA/(1-tan2A) ctg2A=(ctg2A-1)/2ctga cos2a=cos2a-sin2a=2cos2a-1=1-2sin2a 鍗婅鍏紡锛歴in(A/2)=鈭((1-cosA)/2) sin(A/2)=-鈭((1-cosA)/2)cos(A/2)=鈭((1+cosA)/2) cos(A/2)=-鈭((1+cosA)/2)闈㈢Н 闀挎柟褰㈢殑闈㈢Н = 闀棵楀 S = ab 姝...
