minimum-value-of-function-f-x-3sin-x-4cos-x-10-3sin-x-4cos-x-10-

Question Number 159642 by tounghoungko last updated on 19/Nov/21

m i n i m u m v a l u e o f f u n c t i o n

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

Commented by bobhans last updated on 19/Nov/21

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

g (x) = {(3 \sin x - 10)}^{2} - 16 \cos^{2} x

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

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

g (x) = {(5 \sin x - 6)}^{2} + 48

- 5 ⩽ 5 \sin x ⩽ 5 \Rightarrow - 11 ⩽ 5 \sin x - 6 ⩽ - 1

\Rightarrow 0 < {(5 \sin x - 6)}^{2} ⩽ 121

g {(x)}_{\min} = {(5.1 - 6)}^{2} + 48 = 49

f {(x)}_{\min} = \sqrt{49} = 7

Answered by mr W last updated on 19/Nov/21

f (x) = \sqrt{{(3 \sin x - 10)}^{2} - {(4 \cos x)}^{2}}

f (x) = \sqrt{9 \sin^{2} x + 100 - 60 \sin x - 16 \cos^{2} x}

f (x) = \sqrt{25 \sin^{2} x - 60 \sin x + 84}

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

f {(x)}_{m i n} = \sqrt{25 {(1 - \frac{6}{5})}^{2} + 48} = 7

f {(x)}_{m a x} = \sqrt{25 {(- 1 - \frac{6}{5})}^{2} + 48} = 13

Commented by mnjuly1970 last updated on 19/Nov/21

b r a v o s i r W

Answered by FongXD last updated on 19/Nov/21

f (x) = \sqrt{(3 sinx - 10 - 4 cosx) (3 sinx - 10 + 4 cosx)}

f (x) = \sqrt{{(3 sinx - 10)}^{2} - {(4 cosx)}^{2}}

f (x) = \sqrt{{9 \sin}^{2} x - 60 sinx + 100 - 16 (1 - \sin^{2} x)}

f (x) = \sqrt{{25 \sin}^{2} x - 60 sinx + 84}

f (x) = \sqrt{{(5 sinx - 6)}^{2} + 48}

Min of f (x) occurs when \min of ∣ 5 sinx - 6 ∣ occurs

we have : - 1 ⩽ sinx ⩽ 1, \forall x \in R

\Leftrightarrow - 11 ⩽ 5 sinx - 6 ⩽ - 1

\Leftrightarrow 1 ⩽∣ 5 sinx - 6 ∣⩽ 11, \Rightarrow Min of ∣ 5 sinx - 6 ∣= 1

therefore, Min of f (x) = \sqrt{1^{2} + 48} = 7

Commented by mnjuly1970 last updated on 19/Nov/21

Commented by mnjuly1970 last updated on 19/Nov/21

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