f(x)=「2cos^3θ+sin^2(2π-θ)+cos(-θ)」/2+2cos^2(2π-θ)求f(π/3)
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![f(x)=「2cos^3θ+sin^2(2π-θ)+cos(-θ)」/2+2cos^2(2π-θ)求f(π/3)](/uploads/image/z/14573952-0-2.jpg?t=f%EF%BC%88x%EF%BC%89%3D%E3%80%8C2cos%5E3%CE%B8%2Bsin%5E2%EF%BC%882%CF%80%EF%BC%8D%CE%B8%EF%BC%89%2Bcos%EF%BC%88%EF%BC%8D%CE%B8%EF%BC%89%E3%80%8D%EF%BC%8F2%2B2cos%5E2%EF%BC%882%CF%80-%CE%B8%EF%BC%89%E6%B1%82f%28%CF%80%2F3%29)
f(x)=「2cos^3θ+sin^2(2π-θ)+cos(-θ)」/2+2cos^2(2π-θ)求f(π/3)
f(x)=「2cos^3θ+sin^2(2π-θ)+cos(-θ)」/2+2cos^2(2π-θ)求f(π/3)
f(x)=「2cos^3θ+sin^2(2π-θ)+cos(-θ)」/2+2cos^2(2π-θ)求f(π/3)
f(x)=[2cos³θ+sin²(2π-θ)+cos(-θ)]/[2+2cos²(2π-θ)],求f(π/3)
f(x)=(2cos³θ+sin²θ+cosθ)/(2+2cos²θ)
f(π/3)=[2×(1/2)³+(√3/2)²+(1/2)]/[2+2×(1/2)²]=[(1/4)+(3/4)+(1/2)]/[2+(1/2)]=(3/2)/(5/2)=3/5.
好简单