3sin^2x 2sinxcosx-5cos^2x=0
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/29 21:50:49
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
可设a=sinα,b=sinβ,-1
sina2+sinb2=sina-sina2/2只需求该区间就可以了令x=sina可得f(x)=x-x2/2-1=
∫(3sint+sin^2t)dt第一项直接积出,第二项利用二倍角降次,然后再积分
3(sinA)^2+2(sinB)^2=5sinA(sinA)^2+(sinB)^2=5sinA/2-(sinA)^2/25sinA/2-(sinA)^2/2=-(1/2)(sinA-5/2)^2+2
应用数学归纳法.1.当n=1时,左边=sin(pi/3),右边=sin(pi/3).则命题成立2.假设当n=k时,命题成立.即sin(pi/3)+sin(2*pi/3)+...+sin(k*pi/3)
sin89=sin(90-1)=cos1同理,sin88=cos2,in87=cos3,……,sin45=cos44所以原式=[(sin1)^2+(cos1)^2]+[(sin2)^2+(cos2)^
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^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^2x+cos^2x)^2-3sin^2xcos^2x
Cos(&-3派/2)=Cos(3派/2-&)余弦在第三象限,为负值所以Cos(&-3派/2)=Cos(3派/2-&)=-sin&
题目是∫[1/(3sint+sin²t)]dt还是∫[3sint+sin²(1/t)]dt请说明一下,不然没法帮你.再问:求不定积分:∫(3sint+(1/sint^2t))dt求
本题题目应是要证:2tan(α+ β)=3tanα,答案见图片:
/>利用积化和差公式,达到裂项的效果.2sinka*sin(a/2)=-cos[(k+1/2)a]+[cos(k-1/2)a]∴2sin(a/2)*(sina+sin2a+sin3a+...+sinn
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的平方1度+sin的平方2度+sin的平方3度+.+sin的平方89度=sin^2(90-89)+sin^2(90-88)+sin^2(90-87)+.+sin^2(89)=cos^2(89)+
sin(π/2-x)=cosx原式=sin^21°+……+sin^244°+1/2+cos^244°+……+cos^21°=44+1/2=89/2
sin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^2x+cos^2x)^2-3sin^2xcos^2x