2sin²225°-cos330°×tan405°

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计算[3/(sin^2)20°-[1/(cos^2)20°]+64(sin^2)20°

原式=[3(cos20)^2-(sin20)^2]/(sin20)^2(cos20)^2+64(sin20)^2=[(√3cos20+sin20)(√3cos20-sin20)]/(sin20cos2

2sin@+cos@等于?

(2sina+cosa)=-5(sina-3cosa)7sina=14cosasina=2cosa

求证sin3°=(sin^2°-sin^1°)/sin1°

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

在Rt△ABC中,角C=90°,求证sin^2A+sin^2B=1,并利用上式求sin^21°+sin^22°+sin^

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

求2sin(-1110°)-sin960°+根号2cos(-225°)-cos(-210°)

2sin(-1110°)-sin960°+根号2cos(-225°)-cos(-210°)=2sin(-30-3*360)-sin(360*3-120)+根号2cos(135)-cos150=2*(-

求sin^2(44°)+sin^2(46°)+tan53°+tan37°的值

2再问:要过程再答:∵44º+46º=90º∴sin44º=cos46º∴sin²44º+sin²46º=co

证明sin(a+b)sin(a-b)=sin^2 a-sin^2 b,

左边=(sinacosb+cosasinb)(sinacosb-cosasinb)=sin²acos²b-cos²asin²b=sin²a(1-sin

已知2sin

2sin2α+2sinαcosα1+tanα=2sinα(sinα+cosα)1+sinαcosα=2sinαcosα(sinα+cosα)sinα+cosα=2sinαcosα=k.当0<α<π4时

sin²1°+sin²2°+sin²89°+sin²88°=

原式=sin²1+sin²2+cos²(90-89)+cos²(90-88)=(sin²1+cos²1)+(sin²2+cos&#

sin²1°+sin²2°+……sin²88°+sin²89°

sin²1°+sin²2°+……sin²88°+sin²89°=(sin²1°+sin²89°)+(sin²2°+sin²

求证:sin(2α+β)sinα

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

化简sin(x+60°)+2sin(x-60°)-3

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是什么,sin^260°等于什么,还有cos^43兀/2等于什么

sin^2是sinx的平方sin260等于sin80最后一个没太看明白……

求证:sin^2/(sin-cos) - (sin+cos)/(tan^2 -1) =sin+cos

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

sin^2 (0°)+sin^2( 1°)+.+sin^2(90du) 求和

°省略原式=sin²0+sin²1+sin²2+……+sin²44+sin²45+cos²(90-46)+……+cos²(90-8

【证明】Sin A+sin B=2Sin 22

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(

sin²40°+sin²50°+2-tan135°怎么算=

原式=sin²40+cos²(90-50)+2-tan(180-45)=(sin²40+cos²40)+2-(-tan45)=1+2-(-1)=4再问:为什么-

计算:sin²1°+sin²2°+sin²3°...+sin²45°+sin&#

sin²1°+sin²2°+sin²3°...+sin²45°+sin²46°...+sin²89°=sin^2(90-89)+sin^2(

数列求和 sin²1°+sin²2°+sin²3°+.+sin²88°+sin&

sin(π/2-x)=cosx原式=sin^21°+……+sin^244°+1/2+cos^244°+……+cos^21°=44+1/2=89/2