4分之根号2SIN(4分之π-π)

f(x)=√2*(2x-π/4)/sin(x+π/2)∵cosa=3/5∴a=arccos3/5f(a)=√2*(2a-π/4)/sin(a+π/2)=√2*(2arccos3/5-π/4)/cosx

cosx=1/7cos(x-y)=13/14siny=sin(x-(x-y))=sinxcos(x-y)-cosxsin(x-y)=4√3/7*13/14-1/7*3√3/14=√3/2所以y=π/3

sina=4√3/7≈0.9897433a=arcsin0.9897433≈81.79°

立方根号10分之1=10的-1/3次方所以-sinα=-1/3sinα=1/3不知你要求什么

sqrt(2)*cos（2x-π/4）=cos(2x)+sin(2x)sin(x+pi/2)=-cos(x)1+cos(2x)=2*cos^2(x)sin(2x)=2*sin(x)*cos(x)所以最

根号5分之2-根号10分之1=1/5√10-1/10√10=1/10√10根号28-根号7分之4=2√7-2/7√7=12/7√7根号5分之4-根号5+根号6分之1=2/5√5-√5+1/6√6=1/

有个公式先告诉你：asinθ+bcosθ=[根号下（a^2+b^2）]*sin(θ+r)  或者 asinθ-bcosθ=[根号下（a^2+b^2）]*sin(θ-r)其

(1)首先利用降幂公式：f(x)=1-cos(π/2+2x)-根号3cos2x再利用诱导公式：f(x)=1+sin2x-√3cos2x最后是辅助角公式:f(x)=2*(1/2*sin2x-√3/2*c

√6/4+√2*√6/2=√6/4+√12/2=√6/4+2√3/2=√6/4+√3

sin(α+π/3)+sinα=-4√3/5sinαcosπ/3+cosαsinπ/3+sinα=-4√3/53/2*sinα+√3/2*cosα=-4√3/53/2*sinα=-√3/2*cosα-

1/2*2*√2-4*1/3(√1/3)+1/4*4√3-1/3*3√2=√2-4/3*(√1/3)+√3-√2=-4/3*1/3*√3+√3=-4/9√3+√3=5/9√3

化简后得（cosa）²-(sina)²／sinacosπ/4-cosasinπ/4=-√2/2（cosa+sina）（cosa-sina）/√2/2（sina-cosa）=-√2/

=2/3×4/3√(6×24)=8/9√144=8/9×12=32/3

1/(√2+1)+1/(√3+√2)+1/(√4+√3)+.+1/(√2007+√2008)=(√2-1)/(√2+1)(√2-1)+(√3-√2)/(√3+√2)(√3-√2)+(√4-√3)/(√

sin(a+π/3)+sina=-4√3/51/2sina+√3/2cosa+sina=-4√3/53/2sina+√3/2cosa=-4√3/5√3/2sina+1/2cosa=-4/5cos(a-

f(x)=[1-√2sin(2x-π/4)]/cosx=[1-√2(sin2xcosπ/4-cos2xsinπ/4)]/cosx=[1-√2(√2/2*sin2x-√2/2*cos2x)]/cosx=

化简后发现右式与左式为相反数,即右式=-√2/4sin(a+b)=sinacosb+cosasinbsin(a-b)=sinacosb-cosasinbcos(a+b)=cosacosb-sinasi

√(4分之9-√2)=√(2-√2+4分之1)=√(√2-2分之1)²=√2-2分之1

根号4分之27等于2分之根号27!

解:f(x)=1+√2sin(2x-π/4).1.函数f(x)的最小正周期为T=2π/2=π.当思念(2x-π/4)=1时,函数f(x)具有最大值,且f(x)min=1+√2.2.函数的增区间:∵si