Given-that-7-cos-2-24-sin-2-R-cos-2-where-R-gt-0-and-0-lt-lt-pi-2-find-the-maximum-value-of-14-cos-2-48-sin-cos-

Question Number 153736 by ZiYangLee last updated on 09/Sep/21

Given that 7 \cos 2 θ + 24 \sin^{2} θ = R \cos (2 θ - α),

where R > 0 and 0 < α < \frac{π}{2}, find the maximum

value of 14 \cos^{2} θ + 48 \sin θ \cos θ .

Answered by mr W last updated on 09/Sep/21

7 \cos 2 θ + 24 \sin^{2} θ = R \cos (2 θ - α)

7 \cos 2 θ - 12 \cos 2 θ + 12 = R \cos (2 θ - α)

- 5 \cos 2 θ + 12 = R \cos 2 θ \cos α - R \sin 2 θ \sin α

12 = (R \cos α + 5) \cos 2 θ - R \sin α \sin 2 θ

\frac{12}{\sqrt{{(R \sin α)}^{2} + {(R \cos α + 5)}^{2}}} = \cos φ \cos 2 θ - \sin φ \sin 2 θ

\frac{12}{\sqrt{R^{2} + 25 + 10 R \cos α}} = \cos (2 θ + φ)

2 θ = \cos^{- 1} \frac{12}{\sqrt{R^{2} + 25 + 10 R \cos α}} - φ

\cos 2 θ = \frac{12 (R \cos α + 5)}{R^{2} + 25 + 10 R \cos α} + \frac{R \sin α \sqrt{R^{2} - 119 + 10 R \cos α}}{R^{2} + 25 + 10 R \cos α}

\sin 2 θ = \frac{(R \cos α + 5) \sqrt{R^{2} - 119 + 10 R \cos α}}{R^{2} + 25 + 10 R \cos α} - \frac{12 R \sin α}{R^{2} + 25 + 10 R \cos α}

14 \cos^{2} θ + 48 \sin θ \cos θ

= 7 \cos 2 θ + 24 \sin 2 θ + 7

= \frac{7 (12 (R \cos α + 5) + R \sin α \sqrt{R^{2} - 119 + 10 R \cos α}) + 24 ((R \cos α + 5) \sqrt{R^{2} - 119 + 10 R \cos α} - 12 R \sin α)}{R^{2} + 25 + 10 R \cos α} + 7

= \frac{84 R \cos α - 288 R \sin α + 420 + (24 R \cos α + 7 R \sin α + 120) \sqrt{R^{2} - 119 + 10 R \cos α}}{R^{2} + 25 + 10 R \cos α} + 7

\dots . .

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