 | f(θ) = 6sin(θ)cos(θ)-8((sin(θ))^3)cos(θ)+2((cos(θ))^2)-1
= 3sin(2θ)-4((sin(θ))^2)sin(2θ)+cos(2θ)
= 3sin(2θ)-2(1-cos(2θ))sin(2θ)+cos(2θ)
= 2cos(2θ)sin(2θ)+sin(2θ)+cos(2θ)
(1)
t = sin(2θ)+cos(2θ), t^2 = 1+2sin(2θ)cos(2θ)ですから、
f(θ) = (t^2-1)+t = t^2+t-1
(2)
t = (√2)(sin(2θ)*(1/√2)+cos(2θ)*(1/√2))
= (√2)(sin(2θ)*cos(π/4)+cos(2θ)*sin(π/4))
= (√2)sin(2θ+π/4)
よって-√2 ≦ t ≦ √2です。
また(d/dt)f(θ) = 2t+1なので、t = -1/2で(d/dt)f(θ) = 0です。
t = -√2: f(θ) = 2-√2-1 = 1-√2
-√2 < t < -1/2: (d/dt)f(θ) < 0, f(θ)は減少
t = -1/2: f(θ) = 1/4+1/2-1 = -1/4は極小
-1/2 < t < √2: (d/dt)f(θ) > 0, f(θ)は増加
t = √2: f(θ) = 2+√2-1 = 1+√2
よってf(θ)の最大値は1+√2
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