sin^3xdx sinx cosx
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sin(11*pi/3)=sin(4π-π/3)=sin(-π/3)=-sinπ/3=-√3/2
∫(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
我知道这个题是个定积分题,请追问我给出积分限.我按我以前做过的同一题给你做吧,积分限是0→π∫[0→π]√(sin^3x-sin^5x)dx=∫[0→π]√[sin³x(1-sin²
sin89=sin(90-1)=cos1同理,sin88=cos2,in87=cos3,……,sin45=cos44所以原式=[(sin1)^2+(cos1)^2]+[(sin2)^2+(cos2)^
先积化和差sin3xsin5x=0.5(cos2x-cos8x)∫sin3xsin5xdx=∫0.5(cos2x-cos8x)dx=0.25sin2x-0.0625sin8x+c
(cosx+sinx)³=(cosx+sinx)²(cosx+sinx)=(cosx²+sinx²+2cosxsinx)(cosx+sinx)=cosx
令A=arccos(-1/3)B=arcsin(-1/4)那么有cosA=-1/3sinB=-1/4由arccos,arcsin的取值范围可知,A为钝角(90°-180°之间)B为一个负的锐角(-90
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(π3-α)=sin[π2-(π6+α)]=cos(π6+α)=14,∴cos(π3+2α)=cos2(π6+α)=2cos2(π6+α)-1=2×(14)2-1=-78.故答案为:-78
sin(π/2-x)=cosx原式=sin^21°+……+sin^244°+1/2+cos^244°+……+cos^21°=44+1/2=89/2
再答: