2sin²225°-cos330°tan405°
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/29 15:25:25
原式=[3(cos20)^2-(sin20)^2]/(sin20)^2(cos20)^2+64(sin20)^2=[(√3cos20+sin20)(√3cos20-sin20)]/(sin20cos2
(2sina+cosa)=-5(sina-3cosa)7sina=14cosasina=2cosa
三倍角公式:sin3°=3sin1-4sin1^3化简:(sin^2°-sin^1°)/sin1°=(4sin^1cos^1-sin^1)/sin1=sin^1(4cos^1-1)/sin1=sin1
因为sin^2A+cos^2A=1cos2A=cos^2A-sin^2A所以cos^2A=(cos2A+1)/2sin^2A=(1-cos2A)/2因为A+B=90°所以2A+2B=180°所以cos
2sin(-1110°)-sin960°+根号2cos(-225°)-cos(-210°)=2sin(-30-3*360)-sin(360*3-120)+根号2cos(135)-cos150=2*(-
2再问:要过程再答:∵44º+46º=90º∴sin44º=cos46º∴sin²44º+sin²46º=co
左边=(sinacosb+cosasinb)(sinacosb-cosasinb)=sin²acos²b-cos²asin²b=sin²a(1-sin
2sin2α+2sinαcosα1+tanα=2sinα(sinα+cosα)1+sinαcosα=2sinαcosα(sinα+cosα)sinα+cosα=2sinαcosα=k.当0<α<π4时
原式=sin²1+sin²2+cos²(90-89)+cos²(90-88)=(sin²1+cos²1)+(sin²2+cos
sin²1°+sin²2°+……sin²88°+sin²89°=(sin²1°+sin²89°)+(sin²2°+sin²
证明:∵sin(2α+β)-2cos(α+β)sinα=sin[(α+β)+α]-2cos(α+β)sinα=sin(α+β)cosα+cos(α+β)sinα-2cos(α+β)sinα=sin(α
原式=sin(x+60°)-3cos[180°-(x+60°)]+2sin(x-60°)=sin(x+60°)+3cos(x+60°)+2sin(x-60°)=2sin(x+60°+60°)+2sin
sin^2是sinx的平方sin260等于sin80最后一个没太看明白……
sin^2/(sin-cos)-(sin+cos)/(tan^2-1)=sin^2/(sin-cos)-(sin+cos)/[(sin^2/cos^2)-1]=sin^2/(sin-cos)-(sin
°省略原式=sin²0+sin²1+sin²2+……+sin²44+sin²45+cos²(90-46)+……+cos²(90-8
应该是sinA+sinB=2sin[(A+B)/2]cos[(A-B)/2]A=(A+B)/2+(A-B)/2.B=(A+B)/2-(A-B)/2所以sin(A+B)/2cos(A-B)/2+cos(
原式=sin²40+cos²(90-50)+2-tan(180-45)=(sin²40+cos²40)+2-(-tan45)=1+2-(-1)=4再问:为什么-
sin²1°+sin²2°+sin²3°...+sin²45°+sin²46°...+sin²89°=sin^2(90-89)+sin^2(
sin(π/2-x)=cosx原式=sin^21°+……+sin^244°+1/2+cos^244°+……+cos^21°=44+1/2=89/2