sin(π／6－θ）=﹣3／5,π／6＜θ＜2π／3,
则cos(π／6－θ）=4／5,
sinθ=sin[π／6 -(π／6－θ)]
= sinπ／6cos(π／6－θ)- cosπ／6 sin(π／6－θ）
=(4+3√3)/10.
cos2θ=1-2 sin²θ=(7-24√3)/50.
已知sin(π／6－θ）=﹣3／5,π／6＜θ＜2π／3,求sinθ与cos2θ的值!具体过程,限今晚.谢谢!
1个回答
相关问题
-
已知θ∈（0，π），且sin(π3+θ)−sin(π3−θ)＝35，求sin2θ-cos2θ的值．
-
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
-
已知cos(π/6-θ）=a （|a|≦1)求cos(5π/6+θ)和sin(2π/3-θ)的值
-
已知cos(θ+π/6)=-5/13,θ∈(π/2,π),则sin(θ+π/6)的值
-
已知sinθ+cosθ=根号2/3,θ∈（π/2,π）,求sin^3θ+cos^3θ
-
已知根号3sin-sin(π/2-2θ)/cos(π+θ)*cosθ=1,θ属于（0,π）,求θ的值
-
求值已知tanθ=2（1）sin(π2+θ)−cos(π−θ)sin(π2−θ)−sin(π−θ)；（2）2cos2θ2
-
急求已知根号3sinθ-sin(π/2-2θ)/cos(π+θ)×cosθ=1 θ ∈(0,π) 求θ的值
-
sin(θ+π/3)=3/5 θ∈（π/6,2π/3）则cosθ=