sin^4θ+sin^4(π/3-θ)+sin^4(π/3+θ)化简求值
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/15 05:59:33
xO
@0j$/ ݠmN3%Үqtf2ڨ|p,ܯJ(νJ<eWp?YVGn@865";,\6CI h4l*q
A-tq;ReRJ~-@It};bOPQ>wm=~ Ui0g0EA٤!PQx~haOî+wejlڿ m(
sin^4θ+sin^4(π/3-θ)+sin^4(π/3+θ)化简求值
sin^4θ+sin^4(π/3-θ)+sin^4(π/3+θ)化简求值
sin^4θ+sin^4(π/3-θ)+sin^4(π/3+θ)化简求值
sin^4(π/3-θ)+sin^4(π/3+θ)=[sin²(π/3-θ)+sin²(π/3+θ)]²-2sin²(π/3-θ)sin²(π/3+θ)
=[1/2-cos(2π/3-2θ)/2+1/2-cos(2π/3+2θ)/2]²-(1/2)[2sin(π/3-θ)sin(π/3+θ)]²
=[1-cos(2π/3)cos(2θ)]²-(1/2)[cos(2π/3)+cos(2θ)]²
=[1+cos(2θ)/2]²-(1/2)[-1/2+cos(2θ)]²=1+cos(2θ)+(1/4)cos²(2θ)-(1/8)+(1/2)cos(2θ)-(1/2)cos²(2θ)
=7/8+(3/2)cos(2θ)-(1/4)cos²(2θ)=7/8+(3/2)cos(2θ)-(1/8)[1+cos(4θ)]
=3/4+(3/2)cos(2θ)-(1/8)cos(4θ)
sin^4θ=[1/2-cos(2θ)/2]²=1/4-(1/2)cos(2θ)+(1/4)cos²(2θ)=1/4-(1/2)cos(2θ)+(1/8)+(1/8)cos(4θ)
所以sin^4θ+sin^4(π/3-θ)+sin^4(π/3+θ)
=9/8+cos(2θ)
化简求值sin^4θ+sin^4(π/3-θ)+sin^4(π/3+θ)
sin^4θ+sin^4(π/3-θ)+sin^4(π/3+θ)化简求值
提问sin(π/4-θ)=1/3,则sin(π/4+θ)=
2sinθ+cosθ/sinθ-3cosθ=-5,求cos2θ+4sinθ
已知sinθ+sin^2θ=1,求3cos^2θ+cos^4θ-2sinθ+1
已知sinθ=4/5,sinθcosθ
3cosθ+4sinθ=
若sin(θ/2)=4/5,且sinθ
求值:(tan10°-√3)×cos10°/sin50° 求证:[sin(2α+β)]/sinα- 2cos(α+β)=sinβ/cosα 化简cosθ+cos(θ+2π/3)+cos(θ+4π/3) 证明:sin(α+β)sin(α-β)=sin^2α-sin^2β
sin(θ-5π)cos[(-π/2)-θ]cos(8π-θ)}/sin[θ-(3π/2)]sin(-θ-4π)我晕,正的还是负的
已知sinθ=4/5,π/2<θ<π,求sin∧2+2sinθcosθ/3sin^2θ+cos^2θ的值
已知sin(π+θ)
已知tanθ=(sin α-cos α)/(sin α+cos α) a,θ(0,π/2) 求证cos(3/2兀+ α) -sin(5π/2-α)=根号2sin(θ-4π)
已知 sin(θ+kπ)=-2cos (θ+kπ) 求 ⑴4sinθ-2cosθ/5cosθ+3sinθ; ⑵(1/4)sin平方θ+(2/5)cos平方θ已知sin(θ+kπ)=-2cos (θ+kπ)求⑴ 4sinθ-2cosθ/5cosθ+3sinθ⑵(1/4)sin平方θ+(2/5)cos平方θ
sin(π/4+θ)*sin(π/4-θ)=2/5,则cos2θ=
2(sinθ+cosθ)= 2√2sin(θ+π/4)为什么?; (-π/2
求值: 2(sin^6 θ+cos^6 θ)-3(sin^4 θ+cos^4 θ)
ρ(cosθ-sinθ)+1=0 ρ=4sin(θ-π/3)怎么化成直角坐标方程