1sin3xex在什么条件等于1x吗

蒲狭树5084三角函数的诱导公式我们书本上有一条sin三分之11派,求它的值,结果是负二分之根号三,还有sin负三分之十六派的值是多少啊? -
支褚步13167814746 ______[答案] sinx周期是2π sin11π/3=sin(2*2π-π/3)=sin(-π/3)=-sinπ/3=-√3/2 sin16π/3=sin(3*2π-2π/3)=sin(-2π/3)=-sin2π/3=-√3/2

蒲狭树5084f(x)=sin三分之πx,则f(1)+f(2)+f(3)...+f(2013)=? -
支褚步13167814746 ______ 解:f(x)=sin(πx/3) 最小正周期是:2π/(π/3)=6 即:f(x)=f(x+6) f(1)+f(2)+f(3)+f(4)+f(5)+f(6)=sin(π/3)+sin(2π/3)+sin(π)+sin(4π/3)+sin(5π/3)+sin(2π)=(√3)/2+(√3)/2+0-(√3)/2-(√3)/2+0=0 因为:2013÷6=335余3 所以:f(1)+f(2)+f(3)+……+f(2013)=0*355+f(2011)+f(2012)+f(2013)=f(1)+f(2)+f(3)=sin(π/3)+sin(2π/3)+sin(π)=(√3)/2+(√3)/2+0=√3

蒲狭树5084已知向量a=(cos3/2x,sin3/2x),b=(cos1/2x, - sin1/2x),x属于[0,π/2] -
支褚步13167814746 ______ 已知向量a=(cos(3/2)x,sin(3/2)x),b=(cos(1/2)x,-sin(1/2)x),x属于[0,π/2],若f(x)=a•b-2t|a+b|的最小值为g(t),求g(t) 解:f(x)=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)-2t√{[cos(3x/2)+cos(x/2)]²+[sin(3x/2)-sin(x/2)]²}=cos2x-2t√{2+2[cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)]}=cos2x-2t√(2+2cos2x) x∈[0,π/2],当2x=π/2,即x=π/4时得g(t)=-2(√2)t

蒲狭树5084在三角函数中,如sin(x)= - 12/13,其中的x为第三象限角,这里的“x为第三象线角”一条件有什么作用? -
支褚步13167814746 ______ sin(x)=-12/13,可以使得x 的值确定下来啊,x即可为第三象限角也可是第四象限角啊

蒲狭树5084比较下列各组数的大小(1)sin194°与cos160° (2)sin(sin3π/8),sin(cos3π/8) -
支褚步13167814746 ______ sin194°>cos160° sin(sin3π/8)>sin(cos3π/8)

蒲狭树5084已知sin(3π+a)=1/3, -
支褚步13167814746 ______ sin(3π+a)=1/3 则sina=-1/3 则cosa=±2√2/3 cos(π+a)/cosa[cos(π-a)-1]+cos(a-2π)/sin(a-3π/2)cos(a-π)-sin(3π/2+a)=(-cosa)/cosa(-cosa)+cosa/cosa(-cosa)+cosa=cosa=±2√2/3

蒲狭树50841+sinθ/cosθ=根号3 -
支褚步13167814746 ______ 1.解方程 A:(1+sinθ)= √3cosθ(1+sinθ)^2 = 3 cos^2θ=3(1-sin^2 θ) sin^2θ + 2sinθ + 1=3-3sin^2 θ4sin^2θ + 2sinθ - 2 = 02sin^2θ + sinθ - 1 = 0 sinθ = x2x^2 + x - 1 = 0 x1 = (-1 + √9)/4 = 0.5 x2 = (- 1 - 3)/4 = -1 θ1 = π/6 θ2 = 3π/22.化简A式: A:(1+sinθ)= √3cosθ 与 B:(sinθ - 1/2)(sinθ + 1) = 0 等价.

蒲狭树5084已知sin(3π+θ)=1/3,求cos(π+θ)/cosθ【cos(π - θ) - 1】+cos(θ - 2π)/sin(θ - 3π/2) -
支褚步13167814746 ______ sin(3π+θ)=1/3 sinθ=-1/3 cosθ=±根号(1-sin²θ)=±2√2/3 【cos(π+θ)】/【cosθ[cos(π-θ)-1]】+【cos(θ-2π)】/【sin(θ-3π/2)cos(θ-π)-sin(3π/2+θ)】=-cosθ/[cosθ(-cosθ-1)]+cosθ/(cosθ+cosθ)=1/(cosθ+1)+1/2=3(3±2√2)+1/2=(19±12√2)/2

蒲狭树5084化简二次根号下1+2sin(π - 3)cos(π+3) -
支褚步13167814746 ______ sin(π-3)=sin3 cos(π+3)=-cos31+2sin(π-3)cos(π+3)=1-2sin3cos3=(sin3)^2-2sin3cos3+(cos3)^2=(sin3-cos3)^2 因为3弧度的角在第二象限 所以sin3>0,cos3sin3>cos3 所以原式=sin3-cos3

蒲狭树5084已知sin(3π+θ)=1/3 -
支褚步13167814746 ______ [cos(π+θ)]/[cosθ(cos(π-θ)-1)]+[cos(θ-2π)]/[sin(θ-3π/2)cos(θ-π)-sin(3π/2+θ)]=(-cosθ)/[cosθ(-cosθ-1)]+(cosθ)/[cosθ(-cosθ)-(-cosθ)]=1/(cosθ+1)+(cosθ)/(cosθ-cos²θ)=1/(1+cosθ)+1/(1-cosθ)=2/(1-cos²θ)=2/(sin²θ)=2/(1/9)=18