I'm learning trigonometry, and I was just looking at the $y = \sin^2(x)$ graph. To me it looks the same as a $y = -\cos(x)$ shifted up. More specifically, it looks like $y = -0.5\cos(2x) + 0.5$ . Are these two functions the same? And this is my first post here so I'm not sure how good of a question this is. Thanks.
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They are the same, that expression is a common identity – Rob Jun 05 '16 at 02:27
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Thanks! That's pretty cool. – Archie Gertsman Jun 05 '16 at 02:27
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Yes, $\cos(2x) = \cos^2(x) - \sin^2(x) = 1 - 2\sin^2(x)$ thus:
$$ \sin^2(x) = \frac{1 - \cos(2x)}{2} $$
Likewise, $\cos(2x) = 2\cos^2(x) - 1$ and thus:
$$ \cos^2(x) = \frac{\cos(2x) + 1}{2} $$
Jared
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I would like to underline that relationship $\sin(x)^2 =0.5 -0.5\cos(2x)$ can as well be considered as the Fourier series development of $\sin(x)^2$, which - quite an exception - is "exact" with only two terms.
One can in particular recognize that the presence of $\cos(2x)$ means that there is a single harmonic with a frequency doubling (see figure : $y=\sin(x)$ in red, $y=\sin^2(x)$ in blue ; note that, by squaring, simple roots become double roots).
A similar remark can be made for the twin formula $\cos(x)^2 =0.5 +0.5\cos(2x)$.
Jean Marie
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