I have a function $$f(x)=\sqrt{1-\cos(x)}$$ with the fundamental period $2\pi$. But I can also write this as $$\sqrt{2} \sin(x/2)$$ whose fundamental period is $4\pi$. Why has the fundamental period changed.
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The period for the second one is actually $4\pi$. Regardless, there is still a mismatch because you made a simplification mistake. What is $\sqrt{x^2}$ ? – Ninad Munshi May 18 '22 at 10:36
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First of all, period of $\sqrt{2}\sin(x/2)$ is $4\pi$ not $\pi$. (Just check by putting some values, like is $\sqrt2\sin(0/2)=\sqrt2\sin(\pi/2)$?)
The actual value of $\sqrt{1-\cos(x)}$ is $\sqrt2|\sin(x/2)|$ which indeed has a period of $2π$.
Hope this helps. :)
ultralegend5385
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The two functions are not identical.
Take, for example, $x=-\pi$.
Then, $f(x) = \sqrt{1-\cos (-\pi)} = \sqrt{1 - (-1)} = \sqrt{2}$, while
$$\sqrt 2 \sin\left(-\frac \pi2\right) = -\sqrt{2}\cdot\sin\frac\pi2 = -\sqrt 2$$
5xum
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