The range of f(x) = cos 2x - sin 2x contains | vedclass

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(B) Given $f(x) = \cos 2x - \sin 2x$.We can write this as $f(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos 2x - \frac{1}{\sqrt{2}} \sin 2x \right)$.Usi

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$[-1, 1]$

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(B) Given $f(x) = \cos 2x - \sin 2x$.We can write this as $f(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos 2x - \frac{1}{\sqrt{2}} \sin 2x ight)$.Using the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$,we have $f(x) = \sqrt{2} \cos\left( 2x + \frac{\pi}{4} ight)$.The range of $\cos	heta$ is $[-1, 1]$,so the range of $f(x)$ is $[-\sqrt{2}, \sqrt{2}]$.Since $\sqrt{2} \approx 1.414$,the interval $[-\sqrt{2}, \sqrt{2}]$ is approximately $[-1.414, 1.414]$.Thus,the set $[-1, 1]$ is a subset of $[-\sqrt{2}, \sqrt{2}]$.Therefore,the correct option is $B$.

The range of $f(x) = \cos 2x - \sin 2x$ contains the set

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