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in Differential equations by (50.3k points)

Find the minimum value of f(θ) = (3 sin(θ) – 4 cos(θ) – 10) (3 sin(θ) + 4 cos(θ) – 10)

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We have, f(θ) = (3sin(θ) – 4cos(θ) – 10)

(3sin(θ) + 4cos(θ) – 10) = (9sin2(θ) – 16cos2(θ)) – 10(3sinθ + 4cosθ) – 10(3sinθ – 4cosθ)

= (9sin2(θ) – 16cos2(θ))– 10(3sinθ + 4cosθ + 3sinθ – 4cosθ)

= (9sin2(θ) – 16cos2(θ)) – 60sin(θ)

= 25sin2θ – 60sin(θ) – 16

= (5 sinθ – 6)2 – 36 – 16

= (5 sinθ – 6)2 – 52

Hence, the minimum value of f(θ) = 121 – 52 = 69

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