3

Why is the average of both $\sin^2 = \frac{1}{2}$ and $\cos^2 = \frac{1}{2}$

I was revising Simple Harmonic motion notes and in the average of Kinetic energy derivation

$$KE = \frac12 k A^2 \cos^2(\omega t)$$

And the solution is given as $\frac{1}{4} kA^2$

I haven't gone that deep in integration help

TShiong
  • 1,270
PsyScar
  • 31
  • 6
    What do you know about the sum $\sin^2(x)+\cos^2(x)$? – Lee Mosher Sep 29 '23 at 13:07
  • 1
    Maybe try to go deeper in integration? Or try to use some trigonometric formulae? – julio_es_sui_glace Sep 29 '23 at 13:07
  • Also, use $\cos^2x-\sin^2x=\cos 2x$ – Andrei Sep 29 '23 at 13:10
  • 4
    The average of either one is the average of the other, since $\cos^2 x=\sin^2(x+\pi/2).$ But the average of $\sin^2+\cos^2$ is obviously $1.$ – Thomas Andrews Sep 29 '23 at 13:12
  • 2
    BTW, your title is confusing - you seem to be talking about the average value of $\sin^2 x,$ but "the average of $\sin^2$ and $\cos^2$ could mean like "the average of $1$ and $3$ is $2.$" – Thomas Andrews Sep 29 '23 at 13:15
  • A look at the graph of $y = \cos^2 x$ makes it intuitively obvious, and also could be used as the inspiration for a proof based on moving areas around. – aschepler Sep 29 '23 at 13:21
  • Average value of the integrable function $~f(x),~$ in the interval $~[a,b] ~: ~a,b \in \Bbb{R}, ~a < b~$ is defined to be $$\frac{\int_a^b f(x)dx}{b - a}.$$ Then, by symmetry, you know that in the interval $~[0,\pi/2]~$ the average value of $~\cos^2(x)~$ will equal the average value of $~\sin^x(x).~$ Now consider $$\frac{\int_0^{\pi/2} \left[ ~\sin^2(x) + \cos^2(x) ~\right] ~dx}{\pi/2 - 0} = \frac{\int_0^{\pi/2} 1~dx}{\pi/2 - 0} = 2 \times \text{Average value}.$$ – user2661923 Sep 29 '23 at 14:03

1 Answers1

2

Set $\text{avg}(\cos^2(x))=a$ and $ \text{avg}(\sin^2(x))=b$. The two functions are horisontal translations of one another, so the averages are the same. $a=b$

It is known that $\forall x:\sin^2(x)+\cos^2(x)=1$, so $\text{avg}\big(\sin^2(x)\big)+\text{avg}\big(\cos^2(x)\big)=1$, which means that $a+b=1$ so $2a=1$ so $a=\frac12$

$$\text{avg}(\cos^2(x))=\text{avg}(\sin^2(x))=\frac12$$

Q.E.D

Michael Morad
  • 172