Question

Show that over one complete cycle, the average value of a sine function squared is \(\frac{1}{2} .\) [Hint: Use the following trigonometric identities: \(\sin ^{2} a+\cos ^{2} a=1 ; \cos 2 a=\) \(\left.\cos ^{2} a-\sin ^{2} a .\right]\)

Short answer

Answer: The average value of a sine function squared over one complete cycle is \(\frac{1}{2}\).

Step-by-step solution

Express sine squared using given trigonometric identities

Using the given identities, we can express sine squared in a more suitable form for integration. We will use the identity \(\cos 2a = \cos^2a - \sin^2a\). To express \(\sin^2a\), we rearrange the equation: \(\sin^2a = \cos^2a - \cos2a\) Now, we have \(\sin^2a\) in terms of cosines which makes integration easier.

Integrate sine squared over one complete cycle

To find the average value over one complete cycle, we need to integrate \(\sin^2a\) over the interval \([0, 2\pi]\) (since one cycle is \(2\pi\) long). We integrate the expression we found in step 1: \(\int_0^{2\pi} (\cos^2a - \cos2a) \; da\) Since \(\cos^2a\) and \(\cos2a\) are individual functions, we can separate the integrals: \(\int_0^{2\pi} \cos^2a \; da - \int_0^{2\pi} \cos2a \; da\) Now, we can integrate each part: \(\left[\frac{a}{2} + \frac{\sin 2a}{4}\right]_0^{2\pi} - \left[\frac{\sin 2a}{2}\right]_0^{2\pi}\)

Evaluate the integral

Now, we need to evaluate the values of our integrals at our integration limits. Plugging in \(2\pi\) and \(0\): \(\left(\frac{2\pi}{2} + \frac{\sin 4\pi}{4}\right) - \left(\frac{0}{2} + \frac{\sin 0}{4}\right) - \left(\frac{\sin 4\pi}{2}\right) + \left(\frac{\sin 0}{2}\right)\) Simplifying and calculating the values: \(\left(\pi + 0\right) - 0 - 0 + 0\) \(\pi\)

Calculate the average value

To find the average value of sine squared over one complete cycle, we need to divide the integral value we found in step 3 by the length of that cycle (\(2\pi\)): Average value = \(\frac{\pi}{2\pi}\) Average value = \(\boxed{\frac{1}{2}}\) Now, we have shown that the average value of a sine function squared over one complete cycle is \(\frac{1}{2}\).