 | 余り自信がありません。しかも途中までです。
(1)
積和公式から、sin(x-t)sin(2t-a) = (1/2){cos((x-t)-(2t-a))-cos((x-t)+(2t-a))} = (1/2){cos(x-3t+a))-cos(x+t-a)}
f(x) = ∫[0,π](1/2){cos(x-3t+a))-cos(x+t-a)}dt
= (1/2)[sin(x-3t+a)/(-3)-sin(x+t-a)]_[0,π]
= (1/2){(-sin(x-3π+a)/3-sin(x+π-a))-(-sin(x+a)/3-sin(x-a))
= (1/2){(sin(x+a)/3+sin(x-a))-(-sin(x+a)/3-sin(x-a))
= sin(x+a)/3+sin(x-a)
(2)
f(x) = {sin(x)cos(a)+cos(x)sin(a)}/3+{sin(x)cos(a)-cos(x)sin(a)}
= (4/3)sin(x)cos(a)-(2/3)cos(x)sin(a)
u = (4/3)cos(a), v = (2/3)sin(a)とおくと、
f(x) = u*sin(x)-v*cos(x)
= (√(u^2+v^2)){sin(x)*u/(√(u^2+v^2))-cos(x)*v/(√(u^2+v^2))}
u/(√(u^2+v^2)) = cos(b), v/(√(u^2+v^2)) = sin(b)となる実数b(0 ≦ b < 2π)が存在し、
f(x) = (√(u^2+v^2)){sin(x)cos(b)-cos(x)sin(b)} = (√(u^2+v^2))sin(x-b)となります。
よって、最大値はM(a) = √(u^2+v^2), 最小値はm(a) = -√(u^2+v^2)となります。
u^2+v^2 = {(4/3)cos(a)}^2+{(2/3)sin(a)}^2
= (16/9)cos(a)^2+(4/9)sin(a)^2
= (12/9)cos(a)^2+(4/9)
= (4/9)(3cos(a)^2+1)
以上から、M(a) = (2/3)√(3cos(a)^2+1), m(a) = -(2/3)√(3cos(a)^2+1)
(3)
Ｉ = ∫[0,π/2]{M(a)sin(m(2a))}daの計算ですよね?
分かったら書き込みます。
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