Proving <sin^2(k*r-wt)>= 1/2 using Time Average Integral Method

aakeso1
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Homework Statement



Show that <sin^2(k*r-wt)>= 1/2

Homework Equations


<f(t)>= 1/T integral ( f(t')dt' ) from [t, t+T]

The Attempt at a Solution

 
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Trigonometric identities: You know that \cos(2x)=\cos^{2}(x)-\sin^{2}(x)=1-2\sin^{2}(x)?
 
Svein said:
Trigonometric identities: You know that \cos(2x)=\cos^{2}(x)-\sin^{2}(x)=1-2\sin^{2}(x)?
Right, so I use the appropriate identity ( 1-2sin^2(x) ), and carry out the integration?
 
aakeso1 said:
Right, so I use the appropriate identity ( 1-2sin^2(x) ), and carry out the integration?

Try it and see!
 
aakeso1 said:
Right, so I use the appropriate identity ( 1-2sin^2(x) ), and carry out the integration?
Well, I do not know exactly what you mean, but solve the identity for sin2(x) and integrate (for example from 0 to 2π).
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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