求函数f(a)=(sina-1) (cosa-2)大值和小值

f(x)=cos^2x+sinxcosx=1/2(1+cos2x)+1/2sin2x=1/2sin2x+1/2cos2x=√2/2(√2/2sin2x+√2/2cos2x)=√2/2sin(2x+π/

(sinA-cosA)^2=sinA^2-2sinAcosA+cosA^2=1/252sinAcosA=24/25（sinA+cosA)^2=1+2sinAcosA=49/25已知A为锐角所以sinA

f(x)=cos^2x/2-sinx/2cosx/2-1/2=1/2(cosx+1-sinx)-1/2=(cosx-sinx)/2=√2/2cos(x+π/4)f(a)=√2/2cos(x+π/4)=

f(a)=(2sina*cosa+1+3/2)/(sina+cosa)=((sina+cosa)^2+3/2)/(sina+cosa)=(sina+cosa)+(3/2)/(sina+cosa)(si

上下除以cosxf(x)=(1+tanx)/(1-tanx)=tan(π/4+x)所以T=π

(1)sinA+2cosA=0==>sinA=-2cosA==>tanA=-2(2)f(x)=cos²x-2sinx=1-sin²x-2sinx=－sin²x－2sinx

1/(4,sina)//(3,cosa)4cosa=3sina,即tana=4/3f(x)=1+根号2cos(2x-π/4)=1+cos2x+sin2x=1+(1-(tana)^2)/(1+(tana

很简单...若,令sina=m,则f(m)+f(-m)=5/2,又由于f(m)=√[(1+m)/(1-m)],代入,解出m可得sina=3/5或-3/5

由定义域-1/2

sina+f(a)=2/3sina+sin(a+π/2)=2/3sina+sin(-a+π/2)=2/3sina+cosa=2/31+2sinacosa=4/92sinacosa=-5/9(√2sin

(1)用引入辅助角的方法,-(3+2^(1/2))=1吧?

已知函数f(x)=2sinxcosx+cos2x=sin2x+cos2x=sin（2x+╥/2）/2再答：当f=√2/2时，x=a/2=45度，a=90度

f(x)=sin2x+cos2x=√2sin(2x+π/4)f(a/2)=√2sin(a+π/4)=√2/2sin(a+π/4)=1/2a+π/4∈(π/4,5π/4),a+π/4=5π/6a=7π/

令x=sina+cosax²=sin²a+cos²a+2sinacosa=1+2sinacosasinacosa=(x²-1)/2f(x)=(x²-1

1/sinB+1/sinA=(sinA+sinB)/(sinA*sinB)=2(sinA+sinB)/[(sinA+sinB)^2-1]设X=sinA+sinB∈（1,根2]则上式=2/(x-1/x)

a为锐角cosa为正cosa=根号下1---sina方=3/5sin2a=2sinacosacos2a=1-2sina方tan（a-4/5π)=(tana+tan4/5π)/(1-tanatan4/5

观察f(x),把cosx^2转化下,具体二倍角公式忘了

(1±√7)/4f(a/2)=-√3sin^2(a/2)+sin(a/2)cos(a/2)=-√3(1-cosa)/2+sina/2=-√3/2+cosa/2+sin/2=1/4-√3/2得cosa=

1、定义域x≠kπ+2/π的所以实数2、因角a在第一象限,且cosa=3/5,所以sinα=4/5,f(a)=1+4/3=7/3

2+cos(2a)+sin(2a)就是三角函数变·倍角公式：sin(2α)=2sinα·cosα=2/(tanα+cotα)cos(2α)=(cosα)^2-(sinα)^2=2(cosα)^2-1=