高一数学必修5解三角形公式 高中数学必修五解三角形
\u9ad8\u4e2d\u6570\u5b66\uff0c\u5fc5\u4fee5\uff0c\u89e3\u4e09\u89d2\u5f62\u89e3\uff1a
\u7531\u4f59\u5f26\u5b9a\u7406\u5f97\uff1aa²=b²+c²-2bccosA
A=75\u00b0\uff0cb=\u221a2\uff0cc=\u221a3\u4ee3\u5165\uff0c\u5f97\uff1a
a²=(\u221a2)²+(\u221a3)²-2\u00b7\u221a2\u00b7\u221a3\u00b7cos75\u00b0
=2+3-2\u221a6\u00b7(\u221a6-\u221a2)/4
=2+\u221a3
a=\u221a(2+\u221a3)
=\u221a[(4+2\u221a3)/2]
=\u221a[(\u221a3+1)²/2]
=(\u221a6+\u221a2)/2
a>c>b\uff0cA>C>B\uff0c\u53c8A=75\u00b0\uff0c\u56e0\u6b64A\u3001B\u3001C\u5747\u4e3a\u9510\u89d2
\u7531\u6b63\u5f26\u5b9a\u7406\u5f97\uff1aa/sinA=b/sinB
sinB=bsinA/a
=\u221a2\u00b7sin75\u00b0/[(\u221a6+\u221a2)/2]
=\u221a2\u00b7[(\u221a6+\u221a2)/4]/[(\u221a6+\u221a2)/2]
=\u221a2/2
B\u4e3a\u9510\u89d2\uff0cB=45\u00b0
C=180\u00b0-A-B=180\u00b0-75\u00b0-45\u00b0=60\u00b0
\u7efc\u4e0a\uff0c\u5f97\uff1aa=(\u221a6+\u221a2)/2\uff0cB=45\u00b0\uff0cC=60\u00b0
1\u3001(sinA+cosA)^2=2
2sinAcosA=1
sin2A=1
2A=90\u00b0
A=45\u00b0
2\u3001\u8fc7B\u4f5cBD\u5782\u76f4\u4e8eAC\u4ea4AC\u4e8e\u70b9D
\u53ef\u77e5\u25b3ABD\u4e3a\u7b49\u8170\u76f4\u89d2\u4e09\u89d2\u5f62\uff0c\u8bbeBD=x
S_ABC=3x/2=3
x=2
AB=2\u6839\u53f72
CD=1
BC=\u6839\u53f75
\u9510\u89d2\u4e09\u89d2\u5f62\u6709\u4e2a\u63a8\u8bba\u516c\u5f0f\uff1aa/b=sinA/sinB
sinB=(3\u6839\u53f710)/10
B=arcsin(3\u6839\u53f710)/10
正弦定理a/sinA=b/sinB=c/sinC=2R
余弦定理及推论
a^2
=
b^2+
c^2
-
2·b·c·cosA
b^2
=
a^2
+
c^2
-
2·a·c·cosB
c^2
=
a^2
+
b^2
-
2·a·b·cosC
cosC
=
(a^2
+
b^2
-
c^2)
/
(2·a·b)
cosB
=
(a^2
+
c^2
-b^2)
/
(2·a·c)
cosA
=
(c^2
+
b^2
-
a^2)
/
(2·b·c)
三角函数公式
两角和公式
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)
ctg(A+B)=(ctgActgB-1)/(ctgB+ctgA)
ctg(A-B)=(ctgActgB+1)/(ctgB-ctgA)
倍角公式
tan2A=2tanA/(1-tan2A)
ctg2A=(ctg2A-1)/2ctga
cos2a=cos2a-sin2a=2cos2a-1=1-2sin2a
半角公式
sin(A/2)=√((1-cosA)/2)
sin(A/2)=-√((1-cosA)/2)
cos(A/2)=√((1+cosA)/2)
cos(A/2)=-√((1+cosA)/2)
tan(A/2)=√((1-cosA)/((1+cosA))
tan(A/2)=-√((1-cosA)/((1+cosA))
ctg(A/2)=√((1+cosA)/((1-cosA))
ctg(A/2)=-√((1+cosA)/((1-cosA))
和差化积
2sinAcosB=sin(A+B)+sin(A-B)
2cosAsinB=sin(A+B)-sin(A-B)
2cosAcosB=cos(A+B)-sin(A-B)
-2sinAsinB=cos(A+B)-cos(A-B)
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
正弦定理
a/sinA=b/sinB=c/sinC=2R
注：
其中
R
表示三角形的外接圆半径
余弦定理
b2=a2+c2-2accosB
注：角B是边a和边c的夹角
绛旓細tgh(a) = sinh(a)/cosh(a)鍏紡涓锛氳伪涓轰换鎰忚锛岀粓杈圭浉鍚岀殑瑙掔殑鍚屼竴涓夎鍑芥暟鐨勫肩浉绛夛細sin锛2k蟺锛嬑憋級= sin伪 cos锛2k蟺锛嬑憋級= cos伪 tan锛2k蟺锛嬑憋級= tan伪 cot锛2k蟺锛嬑憋級= cot伪
绛旓細瑙ｄ笁瑙掑舰鍏紡涓昏鏄銆佷綑寮﹀畾鐞 鐭ラ亾A銆丅銆乧,姹侰銆乤銆乥 C=180掳-A-B a/sinA=c/sinC,a=sinA*c/sinC,b=sinB*c/sinB 鐭ラ亾a銆乥銆乧,姹侫銆丆銆丅 鐢ㄤ綑寮﹀畾鐞 cosA=(b^2+c^2-a^2)/2bc cosB=(a^2+c^2-b^2)/2ac C=180-A-B ...
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)ctg(A+B)=(ctgActgB-1)/(ctgB+ctgA)ctg(A-B)=(ctgActgB+1)/(ctgB-ctgA)鍊嶈鍏紡 tan2A=2tanA/(1-tan2A)...
绛旓細1銆(sinA+cosA)^2=2 2sinAcosA=1 sin2A=1 2A=90掳 A=45掳 2銆佽繃B浣淏D鍨傜洿浜嶢C浜C浜庣偣D 鍙煡鈻矨BD涓虹瓑鑵扮洿瑙涓夎褰锛岃BD=x S_ABC=3x/2=3 x=2 AB=2鏍瑰彿2 CD=1 BC=鏍瑰彿5 閿愯涓夎褰㈡湁涓帹璁鍏紡锛歛/b=sinA/sinB sinB=(3鏍瑰彿10)/10 B=arcsin(3鏍瑰彿10)/10 ...
绛旓細(涓)瑙ｆ枩涓夎褰 1銆佽В鏂滀笁瑙掑舰鐨勪富瑕佸畾鐞嗭細姝ｅ鸡瀹氱悊鍜屼綑寮﹀畾鐞嗗拰浣欏鸡鐨勫皠褰卞叕寮忓拰鍚勭褰㈠紡鐨勯潰绉殑鍏紡銆2銆佽兘瑙ｅ喅鐨勫洓绫诲瀷鐨勯棶棰橈細(1)宸茬煡涓よ鍜屼竴鏉¤竟(2)宸茬煡涓よ竟鍜屽す瑙(3)宸茬煡涓夎竟(4)宸茬煡涓よ竟鍜屽叾涓竴杈圭殑瀵硅銆(浜)瑙ｇ洿瑙掍笁瑙掑舰 1銆佽В鐩磋涓夎褰㈢殑涓昏瀹氱悊锛氬湪鐩磋涓夎褰BC涓紝鐩磋涓鸿C...
绛旓細鐢鍏紡x^2+(p+q)x+pq=(x+p)(x+q)浠^2-(鈭3+3)a+(2+鈭3)=0 (a-1)(a-2-鈭3)=0 a=1 a=2+鈭3 tanA=1 tanC=2+鈭3 鍗矨=45掳 C=180-60-45=75掳 杩嘋浣淎B楂楥D銆傚垯BCD涓30銆60銆90鐩磋涓夎褰 ACD涓45銆45銆90鐩磋涓夎褰 鍒橞D锛2鈭3锛孋D锛6锛滱D AC锛6鈭2 S=...
绛旓細涓夎褰ACD涓瑿osC=(AC^2+CD^2-AD^2)/(2AC*CD)=锛7^2+3^2-5^2)/(2*3*7)=11/14锛孲inC=5*鏍瑰彿3/14锛汚B/SinC=AC/SinB锛孉B=AC*SinC/SinB=7*5鏍瑰彿3/[14*锛1/鏍瑰彿2)]=2.5鏍瑰彿6銆
绛旓細-2500=锛堟牴鍙2锛墄y BC²+y²-x²=2BCycos鈭燩BC鍗2500+y²-x²=50y 瑙ｅ緱x=25鍊嶆牴鍙6锛寉=25+25鍊嶆牴鍙3 鍦ㄢ柍ABP涓紝BC²+y²-AP²=2BCycosCBP 鍗2500+2500+1250鏍瑰彿3+1250+1250鏍瑰彿3=AP²锛屽緱AP=25鏍瑰彿[10+4鏍瑰彿3]鈮102.8m ...
绛旓細寰3cosC/5+4sinC/5=鈭3/2 鈶紝4cosC/5-3sinC/5=-1/2 鈶 鐢扁憿*4-鈶*3锛岃В寰梥inC=(4鈭3+3)/10 2) 鐢辨寮﹀畾鐞哸/sinA=b/sinB锛屽緱a=bsinA/sinB=鈭3*3/5/sin(蟺/3)=6/5 鎵浠鈻矨BC=absinC/2=6/5*鈭3*(4鈭3+3)/10/2=(36+9鈭3)/50 18.1) sin(A+C)=sin(蟺-...
绛旓細鍥犱负asinA=bsinB锛屾墍浠/b=sinB/sinA,鎵浠inA/sinB=sinB/sinA,鎵浠inA*sinA=sinB*sinB,鎵浠inA=sinB鎴杝inA=-sinB锛屽洜涓篈銆丅鍧囦负涓夎褰鍐呰锛屾墍浠inA=sinB锛屾墍浠=B锛岀瓑鑵颁笁瑙掑舰銆
