Confused About Transformation of x: Help Needed!

ElDavidas
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Not sure exactly where to post this, but I don't see where this is coming from:

x = Asin \omega t + Bcos \omega t

to

x = Rsin(\omega t + \phi)

where

R = \sqrt {A^2 + B^2} and cos \phi = A/R and sin \phi = B/R

I'd be grateful if someone could point out what is going is on here.
 
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That's close to what you use for representing a signal in the complex plane:

v(t) = A[ cos(wt) + jsin(wt) ]

where you draw a circle of radius A, and horizontal and vertical axes centered on the circle. The horizontal axis is the real axis, and the real Acos(wt) component is projected onto it. The vertical axis is the imaginary component axis, and has the jAsin(wt) component projected onto it.

But your equations are slightly different -- where do they come from?
 
Use the addition formula of sin(a+b) on your new sine function.

\begin{array}{l}<br /> \sin \left( {\alpha + \beta } \right) = \sin \alpha \cos \beta + \sin \beta \cos \alpha \\ <br /> R\sin \left( {\omega t + \phi } \right) = R\cos \phi \sin \omega t + R\sin \phi \cos \omega t \\ <br /> \end{array}

Then identify the coefficients by comparing this to the initial expresseion, to match with A and B.
Notice the similarity with polar coördinates once you set up the equations, if you've seen those.
 
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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