Trigonometry-What-is-the-minimum-value-of-3sin-x-4cos-x-10-3sin-x-4cos-x-10-

Question Number 132016 by bramlexs22 last updated on 10/Feb/21

Trigonometry

What is the minimum value of

\sqrt{(3 \sin x - 4 \cos x - 10) (3 \sin x + 4 \cos x - 10)} .

Answered by EDWIN88 last updated on 11/Feb/21

consider ((3 \sin x - 10) - 4 \cos x) ((3 \sin x - 10) + 4 \cos x) =

{(3 \sin x - 10)}^{2} - 16 \cos^{2} x = 9 \sin^{2} x - 60 \sin x + 100 - 16 (1 - \sin^{2} x)

= 25 \sin^{2} x - 60 \sin x + 84

= 25 (\sin^{2} x - \frac{12}{5} \sin x + \frac{84}{25})

= 25 [{(\sin x - \frac{6}{5})}^{2} + \frac{48}{25}]

let J = \sqrt{(3 \sin x - 4 \cos x - 10) (3 \sin x + 4 \cos x - 10)}

J = \sqrt{25 [{(\sin x - \frac{6}{5})}^{2} + \frac{48}{25}]}

J = 5 \sqrt{{(\sin x - \frac{6}{5})}^{2} + \frac{48}{25}}

J will be minimum if g (x) = {(\sin x - \frac{6}{5})}^{2} + \frac{48}{25}

minimum \Rightarrow take g^{'} (x) = 2 \cos x (\sin x - \frac{6}{5}) = 0

we get \cos x = 0 since \sin x = \frac{6}{5} is rejected

then from \cos x = 0 \Rightarrow \sin x = 1

J_{\min} = 5 \sqrt{{(1 - \frac{6}{5})}^{2} + \frac{48}{25}} = 7

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