$f(x)=\sqrt{1-\sin^2x} = \sqrt{\cos^2x} = \vert \cos x\vert$, period is $\pi$. Is this a correct way to find the period of this function? Can I just state that the period of $\vert\cos x\vert$ is $\pi$, or should I prove it somehow?
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you may look at the graph of $f(x)=|\cos x|$ , just reflect negavite portion of $\cos x$ above the x-axis then it will lie entirely above the x-axis with period $\pi$.
Mathronaut
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You can prove it is a period by doing the following: $$|\cos (x+ \pi)| = |\cos x \cos \pi - \sin x \sin \pi| = | - \cos x + 0| = |\cos x|$$ because recall that the definition of a period of some function is that $f(x+p) = f(x)$ for some non zero period $p$.
Sheheryar Zaidi
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