三角函数化简[2cos^3θ+sin^2(π-θ)

三角函数化简[2cos^3θ+sin^2(π-θ)+sin(π/2+θ)-3]/[2+2cos^2(π+θ)+cos(-

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f(θ)=2cos＾3 θ+sin＾(2π-θ)+sin(π/2+θ）-3 / 2+2cos＾(π+θ)+cos(-θ)
求证cos(π-θ)/cosθ[sin(3π/2-θ)-1]+cos(2π-θ)/cos(π+θ)sin(π/2+θ-s
化简：cot(θ+4π)•cos(θ+π)•sin2(θ+3π)tan(π+θ)•cos3(−π−θ)．
把1/2【sin2θ+cos(π/3-2θ)】-sinπ/12*cos(π/12+2θ)化简得?
化简sin(θ-π)cos(θ-3/2π)cot(-θ-π)/tan(θ+3π)sec(-θ-2π)csc(π/2-θ)
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
化简sin(θ−5π)tan(3π−θ)•1tan(θ−3π2)tan(π2−θ)•cos(8π−θ)sin(−θ−4π
化简 sin(θ-5π)•cot( π 2 -θ)•cos(8π-θ) tan(3π-θ)•tan(θ- 3 2 π)•
已知sin（3π+θ）=[1/3]，求cos(π+θ)cosθ[cos(π−θ)−1]+cos(θ−2π)sin(θ−3