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谁来帮我解一道竞赛题？急　急f(

f(x) =asinθ+bcosθ =根号(a^2+b^2)*sin(θ+t1) sint1 =b/根号(a^2+b^2), cost1 =a/根号(a^2+b^2) g(x) =csinθ+dcosθ =根号(c^2+d^2)*sin(θ+t2) sint2 =d/根号(c^2+d^2), cost2 =c/根号(c^2+d^2) f(x)+g(x) =根号[(a+c)^2+(b+d)^2]*sin(θ+t2) 根号(a^2+b^2) =3, a^2+b^2 =9 根号(c^2+d^2) =5, c^2+d^2 =25 根号[(a+c)^2+(b+d)^2] =6, (a+c)^2+(b...全部

  f(x) =asinθ+bcosθ =根号(a^2+b^2)*sin(θ+t1) sint1 =b/根号(a^2+b^2), cost1 =a/根号(a^2+b^2) g(x) =csinθ+dcosθ =根号(c^2+d^2)*sin(θ+t2) sint2 =d/根号(c^2+d^2), cost2 =c/根号(c^2+d^2) f(x)+g(x) =根号[(a+c)^2+(b+d)^2]*sin(θ+t2) 根号(a^2+b^2) =3, a^2+b^2 =9 根号(c^2+d^2) =5, c^2+d^2 =25 根号[(a+c)^2+(b+d)^2] =6, (a+c)^2+(b+d)^2 =36 ==> ac+bd =1 f(x)*g(x) = [根号(a^2+b^2)*sin(θ+t1)][根号(c^2+d^2)*sin(θ+t2)] = [3*sin(θ+t1)][5*sin(θ+t2)] = (15/2)[cos(t1-t2) -cos(2θ+t1+t2)] <= (15/2)[cos(t1-t2) +1] = 15/2 +(15/2)(cost1cost2+sint1sint2) = 15/2 +(15/2)(ac+bd)/15 = 8 f(x)*g(x)的最大值 = 8。
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