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I have problem with calculating average of sine. In my book, there is claim, that $$\langle\sin^2\omega t\rangle=\frac12$$ Now I have a question how to get this result mathematically? I know you must integrate by cycle, but I can not get to this result

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    $\begingroup$ Hint: there is an identity for writing $\sin^2{x}$ in terms of a constant plus $\cos(2x)$. $\endgroup$
    – The Photon
    May 21, 2014 at 20:56

7 Answers 7

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There is a nice trick. You know that:

$$\sin^2 \omega t+\cos^2 \omega t = 1$$

Calculate the average of this equalty, since the average over a cycle is the same for the sine and the cosine and $\langle 1 \rangle = 1$:

$$\langle \sin^2 \omega t \rangle= \frac 1 2$$

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The mean of a function over an interval $T$ is given by,

$$\langle f(t) \rangle = \frac{1}{T}\int_0^T f(t) \, \mathrm{d}t$$

which is roughly the continuous analogue of the arithmetical mean. For $\sin^2 \omega t$, we obtain,

$$\langle \sin^2 \omega t \rangle = \frac{1}{T} \int_0^T \sin^2 \omega t \, \mathrm{d} t = \frac{1}{T} \left( \frac{T}{2} - \frac{\sin(2\omega T)}{4\omega}\right)$$

If we choose $T=2\pi \omega^{-1}$, the result simplifies to,

$$\langle \sin^2 \omega t \rangle = \frac{1}{2}.$$

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    $\begingroup$ Instead of choosing $T = \pi$, usually $\omega = \frac{2\pi}{T}$, and the second term vanishes. (I.e., $T$ is a cycle/period) $\endgroup$
    – Minethlos
    Apr 15, 2015 at 22:25
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\begin{align} &\frac{1}{T}\int_{-T/2}^{+T/2}\sin^2\omega t\,dt\\ &\overset{1-2\sin^2\omega t=\cos2\omega t}{=} \frac{1}{T}\int_{-T/2}^{+T/2}\left(\frac{1}{2}-\frac{1}{2}\cos2\omega t\right)\,dt\\ &=\frac{1}{T}\left(\frac{1}{2}t\big|_{-T/2}^{T/2}-\frac{1}{4\omega}\sin2\omega t\big|_{-T/2}^{+T/2}\right)\\ &\overset{T\to+\infty,\,\omega=\frac{2\pi}{T}}{=}\frac{1}2-\frac{1}{T}\cdot\frac{T}{8\pi}\sin\left(\frac{4\pi}{T}\cdot\frac{T}{2}\right)\cdot2\\ &=\frac{1}{2} \end{align}

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Another way to see this is by using the fact that $$\sin^2 \omega t=\frac{1-\cos(2\omega t)}{2}$$

and the average value of $\cos(2\omega t)$ is zero.

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The average of discrete quantities is:

$$\langle x \rangle = \frac{1}{N}\sum_i x_i$$

The extension to a continuous quantity should seem fairly natural if you think of an integral as a continuous sum:

$$ \langle f(x) \rangle = \frac{1}{b-a} \int_a^b f(x) dx$$

where $[a,b]$ is the interval you want to average over. Then you just need to evaluate this integral. The bounds $(-\infty,\infty)$ make it seem a bit difficult, but for a periodic function the average over one cycle is the same as the average over many cycles.

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It can be easily answered graphically by realizing $\displaystyle \sin^2(x) - \frac{1}{2} $ is symmetric about the x-axis. Therefore $$\left\langle \sin^2(x) - \frac{1}{2} \right\rangle = 0$$

Adding $\displaystyle \frac{1}{2} $ to both sides of the equation yields the desired result.

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The valve of sine function is between 0 and 1 . So we know the average formula

Average = (Minimum+Maximum)÷2 So, Average = (0+1)÷2 = 1/2

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  • $\begingroup$ Good Nice well done ✅ $\endgroup$
    – Chuwa FF
    Apr 9 at 7:35

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