Question

高一数学题，求详细解答，谢谢

已知【f(θ)=sin²(2π-θ)+sin（0.5π+θ）-3】/【2+2cos²（π+θ）+cos（-θ）】，求f（π/3）的值

Answer 1

f(θ)=【sin²(2π-π/3)+sin（0.5π+π/3）-3】/【2+2cos²（π+π/3）+cos（-π/3）】=【sin²(-π/3)+sin（5π/6）-3】/【2+2cos²（-2π/3）+cos（-π/3）】=【3/4+1/2-3】/【2+1/2+1/2】=-7/12 应该没有错吧，哦也！

Answer 2

解：f(θ)=[sin^2(2π-θ)+sin(π/2+θ)-3]/[2+2cos^2(π+θ)+cos(-θ)] ---题目该是这样吧。
f(θ)=(sin^2θ+cosθ-3)/(2cos^2θ+cosθ+2)
f(π/3)=[sin^2(π/3)+cos(π/3)-3]/[2cos^2(π/3)+cos(π/3)+2].
=(3/4)+1/2-3)/(1/2+1/2+2).
=-7/12.

Answer 3

-7/12