Question

f（x）=「2cos^3θ+sin^2（2π－θ）+cos（－θ）」／2+2cos^2（2π-θ）求f(π/3)

Answer 1

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.