已知cos(π/4 -x)=1/7,sin(3π/4 +y)=11/14,π/4
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/09/25 23:11:47
已知cos(π/4 -x)=1/7,sin(3π/4 +y)=11/14,π/4
(3π/4 +y)-(π/4 -x)
=π/2+(x+y)
所以
sin(x+y)=-cos[π/2+(x+y)]
=-cos[(3π/4 +y)-(π/4 -x)]
令A=(3π/4 +y),B=(π/4 -x)
π/2
再问: 方法没错,但似乎算错了 sin^2B=1-cos^2B=48/49, sinB=-(4/7)根号2 似乎错了
再答: 我去,打错了,是根号3.。。 sinB=-(4/7)根号3 sin(x+y) =-[cosAcosB+sinAsinB] =-[-(5/14)根号3*1/7+11/14*-(4/7)根号3] =[5根号3+44根号3]/98 =(49/98)根号3 =(1/2)根号3
=π/2+(x+y)
所以
sin(x+y)=-cos[π/2+(x+y)]
=-cos[(3π/4 +y)-(π/4 -x)]
令A=(3π/4 +y),B=(π/4 -x)
π/2
再问: 方法没错,但似乎算错了 sin^2B=1-cos^2B=48/49, sinB=-(4/7)根号2 似乎错了
再答: 我去,打错了,是根号3.。。 sinB=-(4/7)根号3 sin(x+y) =-[cosAcosB+sinAsinB] =-[-(5/14)根号3*1/7+11/14*-(4/7)根号3] =[5根号3+44根号3]/98 =(49/98)根号3 =(1/2)根号3
已知cos(π/4 -x)=1/7,sin(3π/4 +y)=11/14,π/4
y=sin(-3x) =-sin3x y=cos(3x+π\4) =cos(π/2+3x-π/4) =-sin(3x-π
已知sin(x-3π/4)cos(x-π/4)=-1/4,则cosx等于
1.y=cos^4x+sin^4x 求周期 2.y=(sin2x+sin(2x+π/3))/( cos2x+cos(2x
三角函数已知P(-4,3) 求(cos((π/2)+x)sin(-π-x))/(cos((11π/2)-x)sin((9
已知函数f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)
已知函数f(x)=(1+1/tanx)sin(x)^2 -2sin(x+π/4)*cos(x+π/4)
已知函数f(x)=sin^2*x-根号3*sinπ/4*x*cosπ/4*x
函数y=【cos(3x+π/4)-sin(3x+π/4)】/[cos(3x+π/4)+sin(3x+π/4)]的定义域,
已知f(x)=cos^2x/1+sin^2x求f'(π/4)
已知函数f(x)=sin²(π/4+x)+cos²x+1/2求最值
已知函数f(x0=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)