高一数学作业三角函数的2道题，请写过程，谢谢
2\u9053\u9ad8\u4e00\u6570\u5b66\u4e09\u89d2\u51fd\u6570\u98981.\u539f\u5f0f=(sina/cosa)(cosa-sina)+[sina+(sina/cosa)]/[(cos/sina)+(1/sina)]
=sina-(sin²a/cosa)+[sin²a+(sin²a/cosa)]/(cosa+1)
=sina-(sin²a/cosa)(cosa+1)/(cosa+1)+(sin²a/cosa)/(cosa+1)+sin²a/(cosa+1)
=sina+(sin²a/cosa)[1-(cosa+1)]/(cosa+1)+sin²a/(cosa+1)
=sin-sin²a/(cosa+1)+sin²a/(cosa+1)
=sina
2.\u539f\u5f0f=(1-cos²a)(sina/cosa)+(1-sin²a)(cosa/sina)+2sinacosa
=(sina/cosa)-sinacosa+(cosa/sina)-sinacosa+2sinacosa
=(sina/cosa)+(cosa/sina)
=(sin²a+cos²a)/sinacosa
=1/sinacosa
=2/2sinacosa
=2/sin2a
1)\u5de6\u9762\u5f97\u5316\u7b80\uff0c\u53f3\u9762\u7684\u901a\u5206\u5316\u7b80\uff0c\u6ce8\u610f\u8fd0\u7528\u500d\u89d2\u516c\u5f0f\u548c\u534a\u89d2\u516c\u5f0f\uff0c\u591a\u7b97\u51e0\u6b65\u5c31\u80fd\u8bc1\u660e
2\uff09\u539f\u5f0f=sin2A-\uff082cosA*cosA-1\uff09+1
=sin2A -cos2A+1
=\uff08\u6839\u53f72\uff09*sin\uff082A-\u03c0/4\uff09+1
\u6240\u4ee5\u6700\u5927\u503c\u4e3a1+\u6839\u53f72
\u500d\u89d2\u516c\u5f0f \u500d\u89d2\u516c\u5f0f\u662f\u4e09\u89d2\u51fd\u6570\u4e2d\u975e\u5e38\u5b9e\u7528\u7684\u4e00\u7c7b\u516c\u5f0f.
\u73b0\u5217\u51fa\u516c\u5f0f\u5982\u4e0b:
sin2\u03b1=2sin\u03b1cos\u03b1
tan2\u03b1=2tan\u03b1/(1-tan^2(\u03b1))
cos2\u03b1=cos^2(\u03b1)-sin^2(\u03b1)=2cos^2(\u03b1)-1=1-2sin^2(\u03b1)
\u53ef\u522b\u8f7b\u89c6\u8fd9\u4e9b\u5b57\u7b26,\u5b83\u4eec\u5728\u6570\u5b66\u5b66\u4e60\u4e2d\u4f1a\u8d77\u5230\u91cd\u8981\u4f5c\u7528.
\u53e6\u5916:
tan(\u03b1/2)=sin\u03b1/(1+cos\u03b1)=(1-cos\u03b1)/sin\u03b1 tan(2\u03b1)=2tan\u03b1/[1-tan^2(\u03b1)]
\u00b7\u500d\u89d2\u516c\u5f0f\uff1a
sin(2\u03b1)=2sin\u03b1\u00b7cos\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)]
\u5176\u4ed6\u4e00\u4e9b\u516c\u5f0f
\u00b7\u4e09\u500d\u89d2\u516c\u5f0f\uff1a
sin3\u03b1=3sin\u03b1-4sin^3(\u03b1)
cos3\u03b1=4cos^3(\u03b1)-3cos\u03b1
tan3\u03b1=tan(\u03b1)*(-3+tan(\u03b1)^2)/(-1+3*tan(\u03b1)^2)
\u00b7\u534a\u89d2\u516c\u5f0f\uff1a
sin^2(\u03b1/2)=(1-cos\u03b1)/2
cos^2(\u03b1/2)=(1+cos\u03b1)/2
tan^2(\u03b1/2)=(1-cos\u03b1)/(1+cos\u03b1)
tan(\u03b1/2)=sin\u03b1/(1+cos\u03b1)=(1-cos\u03b1)/sin\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]
\u628a\u8fd9\u4e9b\u516c\u5f0f\u719f\u7ec3\u638c\u63e1\u8fd0\u7528\uff0c\u505a\u8fd92\u9898\u90fd\u5f88\u7b80\u5355
4、将已知sinx+cosx=√10/5两边平方，可得
(sinx+cosx)²=1+2sinxcosx=10/25=2/5 (1)
sinxcosx=(2/5-1)/2=-3/10
∴(sinx-cosx)²=1-2sinxcosx=1+3/5=8/5 (2)
(1),(2)相除可得
(sinx+cosx)²/(sinx-cosx)²=(tanx+1)²/(tanx-1)²=1/4
∵sinxcosx=-3/10<0，∴tanx=sinx/cosx<0，∴1-tanx>0
∴(tanx+1)/(1-tanx)=±1/2
可解得tanx=-1/3或-3
5、∵tanx=-1/2，∴sinx≠-cosx，即sinx+cosx≠0
(1+2sinxcosx)/(sin²x-cos²x)
=(sinx+cosx)²/[(sinx-cosx)(sinx+cosx)]
=(sinx+cosx)/(sinx-cosx)
=(tanx+1)/(tanx-1)
=(-1/2+1)/(-1/2-1)
=-1/3
5.容易点，tanα=sinα/cosα=-1/2, cosα=-2sinα;;
代入式中，原式=（sinα+cosα）∧2/{（sinα+cosα）*（sinα-cosα)]=（sinα+cosα）/（sinα-cosα）
=（sinα-2sinα)/(sinα+2sin α)=-1/3,
4题也可类似解方程，但很麻烦，不知有无简单方法。
绛旓細绗浜岄搴旂敤姝ｅ鸡瀹氱悊鏉ヨВ.鐢变簬a/sinA=c/sinC(鎶夾瑙嗕负閽濊涓夎褰腑鐨勯挐瑙)鍙堝洜涓轰笁鍐呰鎴愮瓑宸暟鍒,鎵浠ヨB鍗充负60搴,A+C=120搴.閭ｄ箞鏈澶ц竟姣旀渶灏忚竟涔嬫瘮鍗充负a/c=sinA/sinC=sinA/sin(120-A)=sinA/(浜屽垎涔嬫牴鍙蜂笁鍊嶇殑COSA锛嬩簩鍒嗕箣涓鍊嶇殑SINA锛夊彇鍊掓暟鍙緱1锛忥紙1锛2锛嬩簩鍒嗕箣鏍瑰彿涓夊嶇殑COTA锛...
绛旓細1\ Sina+cosb=3/4 cosa+sinb=5/4 (SinA)2+(cosB)2+2sinAcosB=9/16 , (cosA)2+(sinB)2+2cosAsinB=25/16 涓寮忕浉鍔 2+2锛坰inAcosB+cosAsinB锛=34/16 Sin(A+B)=1/16 鍙傝冭祫鏂欙細鍘熷垱
绛旓細4銆佸皢宸茬煡sinx+cosx=鈭10/5涓よ竟骞虫柟锛屽彲寰 (sinx+cosx)²=1+2sinxcosx=10/25=2/5 (1)sinxcosx=(2/5-1)/2=-3/10 鈭(sinx-cosx)²=1-2sinxcosx=1+3/5=8/5 (2)(1),(2)鐩搁櫎鍙緱 (sinx+cosx)²/(sinx-cosx)²=(tanx+1)²/(tanx-1)²...
绛旓細1. 宸茬煡鍦ㄢ柍ABC涓紝AB=-鈭2+鈭6锛屸垹C=30掳 璁锯垹A>鈭燘锛岃繃A鐐逛綔AD鈯C锛屼氦BC浜嶥鐐广傚湪鐩磋鈻矨CD涓 鈭燙=30掳锛孉D=AC/2锛CD=AC*cos30掳=锛堚垰3/2锛*AC 鍦ㄧ洿瑙掆柍ABD涓 BD^2=AB^2-AD^2 =锛-鈭2+鈭6锛塣2-锛圓C/2锛塣2 =8-4鈭3-AC^2/4 BD=鈭氾紙8-4鈭3-AC^2/4锛塀...
绛旓細sin(pi/3)+cos(pi/3)=0.5*(1+sqtr(3))锛宼an(pi/3)=sqrt(3)銆俿inx+cosx<tanx,鍙互鐪嬪埌锛孋閫夐」鐨勫尯闂涓涓寖鍥达紝涓涓弧瓒硈inx+cosx>tanx锛屼竴涓弧瓒硈inx+cosx<tanx锛屾墍浠ユ弧瓒硈inx+cosx=tanx鐨剎搴旇鍦–閫夐」鐨勫尯闂翠腑銆傜浜岄鐢ㄥ叕寮忓寲绠涓涓嬶紝cos锛坅+b锛=cosa*cosb-sina*sinb锛岃宑osa=1/...
绛旓細2sin2xcos2x=sin4x A 鍛ㄦ湡T=2蟺/w y=2sin(蟺/3-x)-cos(蟺/6+x)=2sin[蟺/2-(蟺/6+x)]-cos(蟺/6+x)=2cos(蟺/6+x)-cos(蟺/6+x)=cos(蟺/6+x) 鏈灏忓-1 ,C
绛旓細sin(a-b)=1/2 {cos[(a+b)-(a-b)]-cos[(a+b)+(a-b)]}=1/2[cos2b-cos2a]=1/2[(2cos^2 b-1)-(2cos^2 a-1)]=cos^2 b-cos^2 a 5.cosA=12/13,sinA=5/13 cosB=8/17,sinB=15/17 sinC=sin(A+B)=sinAcosB+cosAsinB=5/13*8/17+12/13*15/17=220/221 ...
绛旓細sin2C/b=sinC/c 2sinCcosC=sinC/c cosC=1/(2c)1/c=2cosC cosB=cos(2C)=2cosC鐨勫钩鏂-1=(a^2+c^2-1)/(2ac)鍖栫畝寰 c^3+ca^2+2ac^2-c-a=0 c^2(c+a)+ac(a+c)-(c+a)=0 (c+a)(c^2+ac-1)=0 c+a涓嶇瓑浜0 鎵浠^2+ac=1 鍗砤+c=1/c=2cosC 灞炰簬(...
绛旓細瑙ｇ瓟濡備笅锛歵an锛2伪-尾锛=tan銆愶紙伪-尾锛+伪銆=銆恡an锛埼-尾锛+tan伪銆/銆1-tan锛埼-尾锛塼an伪銆=锛1/2+tan伪锛/锛1-1/2*tan伪锛夎宼an锛埼-尾锛=锛坱an伪-tan尾锛/锛1-tan伪tan尾锛=锛坱an伪+1/7锛/锛1+1/7*tan伪锛=1/2 鈭2tan伪+2/7=1+1/7*tan伪 13/7*tan伪=5...
绛旓細1.瑙:(1)f(x)=asinx+bcosx鐨勫浘鍍忚繃鐐癸紙蟺/3锛0锛夊拰锛埾/2锛1),鍒欐湁:asin(蟺/3)+bcos(蟺/3)=0,鍗:(鈭3/2)a+(1/2)b=0,asin(蟺/2)+bcos(蟺/2)=0,鍗:a=1,鈭磋В寰:a=1,b=-鈭3.(2)鐢(1)寰:f(x)=sinx-鈭3cosx =2sin(x-蟺/3).鈭村綋x-蟺/3=蟺/2+2k蟺,鍗:...
