3sin(2x pai 4)=1

sin²(a+阝)+cos²(a+阝)=1cos²(a+阝)=1-sin²(a+阝)=1-1=0cos(a+阝)=0∴sin2(a+阝)=2sin(a+阝)co

证明：sin(α+β)=sinαcosβ+cosαsinβ=1/2（1）sin(α-β)=sinαcosβ-cosαsinβ=1/3（2）（2）*3-(1)*2得：sinαcosβ-5cosαsinβ

用公式a³+b³=(a+b)(a²-ab+b²)cos^6x+sin^6x=(cos²x)³+(sin²x)³=(cos

用复数w=cos(2π/n)+isin(2π/n)w'=cos(2π/n)-isin(2π/n)z^n=1(z-1)(z^(n-1)+z^(n-2)＋……＋z+1)=0z^(n-1)+z^(n-2)＋

答案是：2/5.令A=(α+β)/2,B=(α-β)/2,则有：2*sinA*cosA=2/3,2*sinB*cosB=3/5,2*sinA*cosB=1/2要求的是：cosA*sinB=(2/3)*

明显的用正弦定理嘛.

已知tan=32sinα-3cosα/sinα-cosα=2sinα-2cosα-cosα/sinα-cosα=(2sinα-2cosα/sinα-cosα)-(cosα/sinα-cosα)=2-(

sin(x+π/6)=1/3sin(5π/6-x)=sin[π-(x+π/6)]=1/3sin^2(π/3-x)=sin^2[π/2-(x+π/6)]=cos^2(x+π/6)=1-sin^2(x+π

(cosα-sinα)/(cosα+sinα)+(cosα+sinα)/(cosα-sinα)＝[(cosα-sinα)^2+(cosa+sinα)^2]/[(cosα)^2-(sinα)^2]＝2[

x=0:0.1:2*pi;s=2*sin(x)-sin(2*x)+2/3*sin(3*x)-1/2*sin(4*x)+2/5*sin(5*x);plot(x,s)

sina=-2cosatana=-2sin²a-3sinacosa+1=(sin²a-3sinacosa+sin²a+cos²a)/(sin²a+co

sin^2x+cos^2y=1/2∴sin^2x=1/2-cos^2y3sin^2x+sin^2y=3（1/2-cos^2y）+sin^2y=1.5-3cos^2y）+sin^2y又有sin^2y+c

f(x)=(√3/2)sin2x-(1/2)[(cosx)^2-(sinx)^2]-1=(√3/2)sin2x-(1/2)cos2x-1=sin(2x-π/6)-1f(x)的最大值是0,最小值是-2,

题目有问题...改:α、β为锐角,且3sin²α+2sin²β=1,3sin2α-2sin2β=0求证：α+2β=π/2.方法多,其一证明:由3sin²α+2sin&su

cos[(α+β)/2]*sin[(α-β)/2]=(1/2)·(sinα-sinβ)(用积化和差公式,或把乘式的每一部分按两角和差的正,余弦展开求出);sin(α+β)=sinαcosβ+sinβc

已知两边同除以余弦得到Tanα=1/3sin²α-2sinαcosα+3cos²α+1=（sin²α-2sinαcosα+3cos²α+sin²α+c

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

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

答：sin^2a+sin^2(a+60)+sin^2（a+120）=3/2.证明：左边=sin^2a+sin^2(a+60)+sin^2（a+120）=sin^2a+(sinacos60+cosasi

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