1.è¯ætan²Î²-sin²Î²=tan²Î²sin²Î²
è¯æ:âµ(tan²Î²-sin²Î²)/(tan²Î²sinvβ)=1/sin²Î²-1/tan²Î²=csc²Î²-cot²Î²=1
â´tan²Î²-sin²Î²=tan²Î²sin²Î²
2.æ±è¯(1-2sinxcosx)/(cos²x-sin²x)=(1-tanx)/(1+tanx)
è¯æ:左边=(cosx-sinx)²/(cosx+sinx)(cosx-sinx)=(cosx-sinx)/(cosx+sinx)
=(1-sinx/cosx)/(1+sinx/cosx)=(1-tanx)/(1+tanx)=å³è¾¹.
3.æ±è¯ tanxsinx/(tanx-sinx)=(tanx+sinx)/tanxsinx
è¯æ:左边=sin²x/(sinx-sinxcosx)=sinx/(1-cosx)=cot(x/2)
\å³è¾¹=(sinx+sinxcosx)/sin²x=(1+cosx)/sinx=cot(x/2)
æ
左边=å³è¾¹.
4.æ±è¯(1-sin⁶x-cos⁶x)/(1-sin⁴x-cos⁴x)=3/2
è¯æ:左边=[1-(sin²x+cos²x)(sin⁴x-sin²xcos²x+cos⁴x)]/[1-(sin²x+cos²x)²+2(sin²xcos²x)]
=[1-(sin²x+cos²x)²+3sin²xcos²x]/(2sin²xcos²x)=3sin²xcos²x/2sin²xcos²x=3/2=å³è¾¹
5,å·²ç¥1/sinx, 1/cosxæ¯äºæ¬¡æ¹ç¨x²-(â2)ax+a=0çä¸¤ä¸ªæ ¹,æ±tanxåaçå¼.
解:1/sinx+1/cox=(sinx+cosx)/sinxcosx=(â2)a...............(1)
1/sinxcosx=a.........................................................(2)
(2)代å
¥(1)å¼å¾ (sinx+cosx)a=(â2)a
æ
(â2/2)(sinx+cosx)=1
sinxcos(Ï/4)+cosxsin(Ï/4)=sin(x+Ï/4)=1
æ
x+Ï/4=Ï/2+2kÏ, x=Ï/4+2kÏ
â´tanx=tan(Ï/4+2kÏ)=tan(Ï/4)=1
a=1/sin(Ï/4)cos(Ï/4)=1/(2/4)=2.
6.å¨â³ABCä¸,(â2)sinA=â(3cosA),æ±â Aç弧度æ°.
解:å°åå¼ä¸¤è¾¹å¹³æ¹ä¹,å¾ 2sin²A=3cosA
2(1-cos²A)=3cosA, 2cos²A+3cosA-2=(2cosA-1)(cosA+2)=0
â´cosA=1/2, Aæ¯éè§,æ
A=Ï/3.
7,sinx-cosx=(1-â3)/2, xâ(0, Ï),æ±sinx, cosx.
解:(â2)[sinxcos(Ï/4)-cosxsin(Ï/4)]=(1-â3)/2
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(â2)sin(x-Ï/4)=(1-â3)/2=(1/2)-(â3/2)=sin(Ï/6)-sin(Ï/3)=2cos(Ï/2)sin(-Ï/6)=0
â´sin(x-Ï/4)=0, x-Ï/4=0, x=Ï/4
â´ sinx=sin(Ï/4)=(â2)/2
cosx=cos(Ï/4)=(â2)/2
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