3sin^2x 2sinxcosx-5cos^2x=0

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sin阝=1/3,sin(a+阝)=1,求sin(2a+3阝)

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

已知3(sinα)^2-2sinα+2(sinβ)^2=0,试求(sinα)^2+(sinβ)^2的取值范围

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=

∫(3 sin t+sin^2t/1) dt

∫(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

已知3SIN^2A+2SIN^2B=5SINA,求SIN^2A+SIN^2B范围

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

三角函数恒等变形证明sin( pi/3 ) + sin( 2*pi/3) + ...+ sin( n * pi/3)=

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)

化简sin^2 1度+sin^2 2度+sin^2 3度+.+sin^2 89度

sin89=sin(90-1)=cos1同理,sin88=cos2,in87=cos3,……,sin45=cos44所以原式=[(sin1)^2+(cos1)^2]+[(sin2)^2+(cos2)^

s = 2*sin(x)-sin(2*x)+2/3*sin(3*x)-1/2*sin(4*x)+2/5*sin(5*x)

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)

sinα=-2cosα,求sin^2α-3sinαcosα+1

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

化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x

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)等于Sin&还是-Sin&?

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&

∫(3 sin t+sin^2 t 分之1) dt .

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

已知sin(2α+ β)=5sinβ,求证:2sin(α+ β)=3sinα

本题题目应是要证:2tan(α+ β)=3tanα,答案见图片:

sin a+sin 2a +sin 3a +...+sin na怎么求和?

/>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

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

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

已知sin平方30度+sin平方90度+sin平方150度=3/2,sin平方5度+sin平方65度+sin平方125度

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

问一道数学题:sin平方+sin平方2度+sin平方3度+.sin平方89度=?

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)+

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

(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x

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