设tan(2π- θ )=2,-π/2＜ θ ＜π

设tan(2π- θ )=2,-π/2＜ θ ＜π/2那么sin^2θ+cos( θ-2π）=

1个回答

设f（θ）=2cos(2 π-θ)sin(π2+θ)1tan(π-θ)•cos(3π2-θ)．
求值已知tanθ=2（1）sin(π2+θ)−cos(π−θ)sin(π2−θ)−sin(π−θ)；（2）2cos2θ2
化简sin(θ−5π)tan(3π−θ)•1tan(θ−3π2)tan(π2−θ)•cos(8π−θ)sin(−θ−4π
化简sin(θ-π)cos(θ-3/2π)cot(-θ-π)/tan(θ+3π)sec(-θ-2π)csc(π/2-θ)
已知：tan(−5π−θ)•cos(θ−2π)•sin(−3π−θ)tan(7π2+θ)•sin(−4π+θ)•cot(
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(-
化简 sin(θ-5π)•cot( π 2 -θ)•cos(8π-θ) tan(3π-θ)•tan(θ- 3 2 π)•
已知tanθ=2，求f（x）=sin(θ−3π2)+2sin(π−θ)+4sin(7π2−θ)cos(π+θ)+2cos
已知tan（π/4+θ）+tan（π/4-θ)=4,且-π＜θ＜-π/2,求sin²θ-2sinθcosθ-c