Cant understand from where the numbers come from

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suppose that the nonlinear resistor R has a characteristics specified by the equation:
<br /> v=20i+i^2+0.5i^3<br />
express v as a sum of sinusoids for
<br /> i(t)=cos \omega _1t +2cos \omega _2t \\<br />
the solution in the book is
http://i43.tinypic.com/eve807.gif

i get a differnt expression why??
<br /> v=20cos \omega _1t +40cos \omega _2t + cos^2 \omega _1t +4cos (\omega _1t) cos (\omega _2t)+4cos^2 \omega _2t+0.5cos^3 \omega _1t+2cos^2 \omega _1tcos \omega _2t+0.5cos^3 \omega _1t+2cos \omega _1tcos^2 \omega _2t+4cos^3 \omega _2t<br />
 
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You haven't completely reduced. For example, you have powers and products of cosines. Those can be reduced to first power by trig identitities: cos^2(\omega_1t)= (1/2)(1+ cos(2\omega_1t)) and cos(\omega_1t)cos(\omega_2t)= (1/2)(cos((\omega_1+\omega_2)t)+ cos((\omega_1- \omega_2)t)).
cos^3(\omega_1t) can be written as cos(\omega_1t)cos^2(\omega_1t)= cos(\omega_1t)(1/2)(1+ cos(2\omega_1t)) which can be reduced, and so on.
 
still i can't see i get numbers here
like they did
as far is i know only sin^2 a +cos^2 a =1

and my expression and not your has anything near that
 
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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