What are the absolute extrema of $f (x) = | sin (x) - cos (x) |$ $f(x)= |sin(x) - cos(x)|$ on the interval [-pi,pi]?

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What are the absolute extrema of $f (x) = | sin (x) - cos (x) |$ $f(x)= |sin(x) - cos(x)|$ on the interval [-pi,pi]?

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Answer 1 · 2016-04-03T03:22:23

0 and $\sqrt{2}$ $sqrt2$ .

Explanation:

$0 \leq | sin θ | \leq 1$ $0<=|sin theta|<=1$
$sin x - cos x = sin x - sin (\frac{π}{2} - x) = 2 cos (\frac{x + \frac{π}{2} - x}{2}) sin (\frac{x - (\frac{π}{2} - x)}{2}) = - 2 cos (\frac{π}{4}) sin (x - \frac{π}{4}) = - \sqrt{2} sin (x - \frac{π}{4})$ $sin x - cos x = sin x -sin (pi/2-x) =2 cos ((x+pi/2-x)/2) sin ((x-(pi/2-x))/2) =- 2 cos (pi/4) sin (x-pi/4) = -sqrt2 sin (x-pi/4)$
so, $| sin x - cos x | = ∣ ∣ - \sqrt{2} sin (x - \frac{π}{4}) ∣ ∣ = \sqrt{2} ∣ ∣ sin (x - \frac{π}{4}) ∣ ∣ \leq \sqrt{2}$ $|sin x - cos x|=| -sqrt2 sin (x-pi/4)| =sqrt2|sin (x-pi/4)| <=sqrt2$ .

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What are the absolute extrema of $f (x) = | sin (x) - cos (x) |$ $f(x)= |sin(x) - cos(x)|$ on the interval [-pi,pi]?

1 Answer

Explanation:

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Science

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Humanities

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What are the absolute extrema of f(x)=|sin(x)−cos(x)| f(x)= |sin(x) - cos(x)| on the interval [-pi,pi]?

1 Answer

Explanation:

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What are the absolute extrema of $f (x) = | sin (x) - cos (x) |$ $f(x)= |sin(x) - cos(x)|$ on the interval [-pi,pi]?