f(θ)=2cos＾3 θ+sin＾(2π-θ)+sin(π/2+θ）-3 / 2+2cos＾(π+θ)+cos(-θ),
f(π/6)=3√3/4+1/4+√3/2-3/2+3/2+√3/2
=(5√3+1)/4
f(θ)=2cos＾3 θ+sin＾(2π-θ)+sin(π/2+θ）-3 / 2+2cos＾(π+θ)+cos(-θ)
1个回答
相关问题
-
求证cos(π-θ)/cosθ[sin(3π/2-θ)-1]+cos(2π-θ)/cos(π+θ)sin(π/2+θ-s
-
急.设f(θ)=[2cos2θ+sin2(2π-θ)+sin(π/2+θ)-3]/[2+2cos2(π+θ)+cos(-
-
设f( θ)=2cos^3θ + sin^2 (2π-θ)+cos(-θ)-3 / 2+2cos^2(π+θ) + co
-
设f（θ）=2cos3θ−sin2(θ+π)−2cos(−θ−π)+12+2cos2(7π+θ)+cos(−θ)，求f(
-
设f（θ）=2cos(2 π-θ)sin(π2+θ)1tan(π-θ)•cos(3π2-θ)．
-
已知sin（3π+θ）=[1/3]，求cos(π+θ)cosθ[cos(π−θ)−1]+cos(θ−2π)sin(θ−3
-
已知sin（3π+θ）=[1/3]，求cos(π+θ)cosθ[cos(π−θ)−1]+cos(θ−2π)sin(θ−3
-
三角函数化简[2cos^3θ+sin^2(π-θ)+sin(π/2+θ)-3]/[2+2cos^2(π+θ)+cos(-
-
f（x）=「2cos^3θ+sin^2（2π－θ）+cos（－θ）」／2+2cos^2（2π-θ）求f(π/3)
-
已知tanθ=2，求f（x）=sin(θ−3π2)+2sin(π−θ)+4sin(7π2−θ)cos(π+θ)+2cos