高中数学三角函数 高中数学三角函数是课本必修几
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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)
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tan2A = 2tanA/(1-tan^2 A)
Sin2A=2SinA?CosA
Cos2A = Cos^2 A--Sin^2 A
=2Cos^2 A\u20141
=1\u20142sin^2 A
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sin3A = 3sinA-4(sinA)^3;
cos3A = 4(cosA)^3 -3cosA
tan3a = tan a ? tan(\u03c0/3+a)? tan(\u03c0/3-a)
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sin(A/2) = \u221a{(1--cosA)/2}
cos(A/2) = \u221a{(1+cosA)/2}
tan(A/2) = \u221a{(1--cosA)/(1+cosA)}
cot(A/2) = \u221a{(1+cosA)/(1-cosA)}
tan(A/2) = (1--cosA)/sinA=sinA/(1+cosA)
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sin(a)+sin(b) = 2sin[(a+b)/2]cos[(a-b)/2]
sin(a)-sin(b) = 2cos[(a+b)/2]sin[(a-b)/2]
cos(a)+cos(b) = 2cos[(a+b)/2]cos[(a-b)/2]
cos(a)-cos(b) = -2sin[(a+b)/2]sin[(a-b)/2]
tanA+tanB=sin(A+B)/cosAcosB
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sin(a)sin(b) = -1/2*[cos(a+b)-cos(a-b)]
cos(a)cos(b) = 1/2*[cos(a+b)+cos(a-b)]
sin(a)cos(b) = 1/2*[sin(a+b)+sin(a-b)]
cos(a)sin(b) = 1/2*[sin(a+b)-sin(a-b)]
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sin(-a) = -sin(a)
cos(-a) = cos(a)
sin(\u03c0/2-a) = cos(a)
cos(\u03c0/2-a) = sin(a)
sin(\u03c0/2+a) = cos(a)
cos(\u03c0/2+a) = -sin(a)
sin(\u03c0-a) = sin(a)
cos(\u03c0-a) = -cos(a)
sin(\u03c0+a) = -sin(a)
cos(\u03c0+a) = -cos(a)
tgA=tanA = sinA/cosA
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sin(a) = [2tan(a/2)] / {1+[tan(a/2)]^2}
cos(a) = {1-[tan(a/2)]^2} / {1+[tan(a/2)]^2}
tan(a) = [2tan(a/2)]/{1-[tan(a/2)]^2}
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a?sin(a)+b?cos(a) = [\u221a(a^2+b^2)]*sin(a+c) [\u5176\u4e2d\uff0ctan(c)=b/a]
a?sin(a)-b?cos(a) = [\u221a(a^2+b^2)]*cos(a-c) [\u5176\u4e2d\uff0ctan(c)=a/b]
1+sin(a) = [sin(a/2)+cos(a/2)]^2;
1-sin(a) = [sin(a/2)-cos(a/2)]^2;;
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csc(a) = 1/sin(a)
sec(a) = 1/cos(a)
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sinh(a) = [e^a-e^(-a)]/2
cosh(a) = [e^a+e^(-a)]/2
tg h(a) = sin h(a)/cos h(a)
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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
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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
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sin\uff08-\u03b1\uff09= -sin\u03b1
cos\uff08-\u03b1\uff09= cos\u03b1
tan\uff08-\u03b1\uff09= -tan\u03b1
cot\uff08-\u03b1\uff09= -cot\u03b1
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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
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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
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sin\uff08\u03c0/2+\u03b1\uff09= cos\u03b1
cos\uff08\u03c0/2+\u03b1\uff09= -sin\u03b1
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sina=2tan(a/2)/[1+tan^2(a/2)],令,tan(a/2)=t,(方便后面运算),则有
4/5=2t/(1+t^2),
2t^2-5t+2=0,t1=1/2,t2=2(不合,舍去,α，β都是锐角).
t=1/2,
cosa=(1-t^2)/(1+t^2)=3/5.
cos(α+β)=5/13,
cosa*cosβ+sina*cosβ=5/13,
cosa+sina*tanβ=5/(13cosβ),
令,tan(β/2)=n,
3/5+4/5*(2n)/(1-n^2)=5(1+n^2)/[13(1-n^2)]
化简得32n^2-52n-7=0,
(4n-7)(8n+1)=0,
n1=7/4(不合,舍去),n2=-1/8(不合,舍去).
tan(β/2)=7/4>1,大于1就等于大于45度啊,β>90.
方程无解啊?
2.α,β都是锐角,tan α =1/7,sinβ=(√10)/10
cosβ=√(1-sin^2β)=3√10/10.
tanβ=sinβ/tanβ=1/3,
tan2β=2tanβ/[1-tan^2(β)]=3/4.
tan(α+2β)=[tana+tan2β]/[1-tana*tan2β]
=(1/7+3/4)/(1-1/7*3/4)
=1.
3.化简：
（1）1/sin10°- (√3)/cos10°
=[1/2cos10-(√2/2)*cos10]/(2sin10*cos10)
=(sin30*cos10-cos30*sin10)/sin20
=sin20/sin20
=1.
（2）sin40°(tan10°-√3)
=2sin40(1/2*sin10-√3/2*cos10)/cos10
=2sin40(sin10*cos60-cos10*sin60)/sin(90-10)
=-2sin40*sin50/(2sin40*cos40)
=-sin50/cos40
=-1.
（3）tan70°cos10°[(√3)tan20°-1]
=2tan70*cos10[(√3/2)sin20-1/2*cos20]/cos20
=2tan70*cos10*[sin20*cos30-cos20*sin30]/cos20
=-2sin10*cos10*tan70/cos20
=-tan20*tan(90-20)
=-tan20*ctn20
=-1.
（4）sin50°[1+(√3)tan10°]
=2sin50*[1/2*cos10+√3/2*sin10]/cos10
=2sin50*[sin30*cos10+cos30*sin10]/cos10
=2sin50*sin(90-50)/cos10
=2sin50*cos50/cos10
=sin100/cos10
=sin80/cos10
=sin(90-10)/cos10
=1.
1.知道sinα . 就知道了cosα .
在利用这个公式:cos(α+β)＝cosαcosβ-sinαsinβ
cosαcosβ-sinαsinβ里面就剩下cosβ和sinβ是未知的了.
接下来我们要求的是sinα// 我们在把cosβ用sinβ表示出来. 即:cosβ=(1-sinβ^2)的开根号.(取正值,因为是锐角). 这样cosαcosβ-sinαsinβ就剩下sinα是未知数了..来个换元法.把sinα看成一个X.最后解方程就行.注意取正值
1.cos(α+β)＝cosαcosβ-sinαsinβ
下面没学，不会
绛旓細涓夎鍑芥暟楂樹腑鎵鏈夊叕寮忓涓嬶細1銆佷袱瑙掑拰鍏紡锛歴in(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)ctg(A+B)=(ctgActgB-1)/(ctgB+ctgA) ctg...
绛旓細楂樹腑闃舵鐨涓夎鍑芥暟鍏紡鏄鏁板瀛︿範涓潪甯搁噸瑕佺殑閮ㄥ垎锛屽叾涓富瑕佸寘鎷悓瑙掍笁瑙掑嚱鏁扮殑鍩烘湰鍏崇郴銆佷袱瑙掑拰涓庡樊鐨勪笁瑙掑嚱鏁板叕寮忋佽緟鍔╄鍏紡銆佸嶈鍏紡銆佷笁鍊嶈鍏紡銆佸崐瑙掑叕寮忎互鍙婂拰宸寲绉叕寮忕瓑銆傜浉鍏崇殑鍐呭濡備笅锛1銆佸悓瑙掍笁瑙掑嚱鏁扮殑鍩烘湰鍏崇郴寮忥細sin^2锛坸锛+cos^2锛坸锛=1銆傝繖涓叕寮忚〃绀哄湪浠讳綍涓涓搴涓嬶紝姝ｅ鸡...
绛旓細鍗婅鍏紡锛歴in(A/2)=鈭歔(1-cosA)/2]锛宑os(A/2)=鈭歔(1+cosA)/2]锛宼an(A/2)=鈭歔(1-cosA)/(1+cosA)]銆4銆佹寮﹀畾鐞嗭細鍦ㄤ换鎰涓夎褰腑锛屽悇杈归暱涓庡搴旇鐨勬寮﹀肩殑姣旂浉绛夛紝鍗砤/sinA=b/sinB=c/sinC銆5銆佷綑寮﹀畾鐞嗭細鍦ㄤ换鎰忎笁瑙掑舰涓紝浠绘剰涓杈圭殑骞虫柟绛変簬鍏朵粬涓よ竟鐨勫钩鏂瑰拰鍑忓幓杩欎袱杈逛笌鍏跺す6...
绛旓細鍦楂樹腑鏁板涓紝甯哥敤鐨涓夎鍑芥暟鏄寮﹀嚱鏁帮紙sin锛夛紝浣欏鸡鍑芥暟锛坈os锛夛紝姝ｅ垏鍑芥暟锛坱an锛夛紝鍓插嚱鏁帮紙sec锛夛紝浣欏壊鍑芥暟锛坈sc锛夛紝浠ュ強瀹冧滑鐨勫掓暟鍑芥暟銆備笁瑙掑嚱鏁板艰〃閫氬父鍖呭惈浠ヤ笅鍐呭锛1. 瑙掑害鍊硷細甯哥敤鐨勮搴﹀煎寘鎷 0掳銆30掳銆45掳銆60掳 鍜 90掳锛屼互鍙婂畠浠殑鏁存暟鍊嶅拰鐩稿叧琛ヨ銆傝繖浜涜搴﹀兼槸甯哥敤鐨勭壒娈婅锛屽...
绛旓細涓夎鍑芥暟鐨勫熀鏈叕寮忥細1銆佸叕寮忎竴锛氫换鎰忚伪涓-伪鐨勪笁瑙掑嚱鏁板间箣闂寸殑鍏崇郴锛歴in锛堬紞伪锛=锛峴in伪cos锛堬紞伪锛=cos伪tan锛堬紞伪锛=锛峵an伪cot锛堬紞伪锛=锛峜ot伪 2銆佸叕寮忎簩锛歴in锛埾+伪锛=鈥攕in伪cos锛埾+伪锛=锛峜os伪tan锛埾+伪锛=tan伪cot锛埾+伪锛=cot伪 3銆佸叕寮忎笁锛氬埄鐢ㄥ叕寮忎簩鍜屽叕寮忎笁鍙互...
绛旓細\)锛孿( \cos a - \cos b = -2\sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right) \)銆傚湪楂樹腑鏁板涓夎鍑芥暟鐨勫涔犱腑锛岃繖浜涘叕寮忔槸鍩虹涓旈噸瑕佺殑锛屽畠浠笉浠呭嚭鐜板湪璇惧爞涓婏紝涔熸槸瑙ｅ喅涓夎鍑芥暟闂鐨勫叧閿傚湪鍋氱浉鍏抽鐩椂锛屽簲鐔熺粌鎺屾彙杩欎簺鍏紡鐨勮繍鐢ㄥ拰鍙樺舰銆
绛旓細楂樹腑鏁板 涓夎鍑芥暟鍏紡澶у叏 涓夎鍑芥暟杩欑珷鍏紡寰堝锛屽挨鍏舵槸璇卞鍏紡灏辨湁浜屽崄澶氫釜锛屽叏閮ㄨ蹇嗘槸姣旇緝鍚冨姏銆傚熀纭寮辩殑鍚屽鏇村簲璇ュソ濂借璁拌繖浜涘叕寮忥紝鐔熺粌杩欎簺鍏紡涔熷氨鎶撲綇浜嗚繖绔犵殑閲嶇偣浜嗐傚涔犺捣鏉ヤ簨鍗婂姛鍊嶃備袱瑙掑拰 sin(A+B)=sinAcosB+cosAsinB sin(A-B)=sinAcosB-sinBcosA cos(A+B)=cosAcosB-sinAsinB ...
绛旓細楂樹腑鏁板涓夎鍑芥暟鍏紡锛氫袱瑙掑拰鍏紡锛歴in(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+...
绛旓細1. 鍩烘湰涓夎姣 姝ｅ鸡瀹氱悊: a/sin A = b/sin B = c/sin C浣欏鸡瀹氱悊: c² = a² + b² - 2ab cos C姝ｅ垏瀹氱悊: tan(伪) = (opposite/adjacent) = sin(伪)/cos(伪)2. 涓夎鍑芥暟鐨勫懆鏈熸у拰鍛ㄦ湡鍏紡 sin(胃 + 2蟺n) = sin 胃 鍜 cos(胃 + 2蟺n) = cos 胃锛...
绛旓細楂樹腑甯哥敤鐨涓夎鍑芥暟鍊艰〃閫氬父鍖呮嫭姝ｅ鸡鍑芥暟锛坰in锛夈佷綑寮﹀嚱鏁帮紙cos锛夈佹鍒囧嚱鏁帮紙tan锛夈佷綑鍒囧嚱鏁帮紙cot锛夈佹鍓插嚱鏁帮紙sec锛夊拰浣欏壊鍑芥暟锛坈sc锛夊湪鐗瑰畾瑙掑害涓婄殑鏁板笺傝繖浜涘嚱鏁板艰〃鎻愪緵浜嗙壒瀹氳搴︾殑涓夎鍑芥暟鍙栧硷紝浣垮緱瀛︾敓浠彲浠ュ湪瑙ｉ鎴栬绠楄繃绋嬩腑蹇熸煡鎵惧弬鑰冦傚父瑙佺殑涓夎鍑芥暟鍊艰〃涓鑸粰鍑轰簡0搴﹀埌360搴︼紙鎴0鍒...
