Solve 2sin(2x): Table of Values & Graph (-π to π)

DethRose
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I have a question that says: prepare a table of values and plot the graph for the domain (-pie,pie) for the function f(x)=2sin(2x)

I understand how to do that for y=sinx, but for this function the wave compresses horizontally and stretches vertically so isn't the domain now (-pie/2,pie/2) and then therefore how can you make a table of values for (-pie,pie).

thanks for any help
andrew
 
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No, the domain is still (-pi,pi). Just use a calculator.
 
why is that because doesn't the 2x compress it horizontally?
 
Yes, the period will be cut in half, but you can still put x=pi, f(x)=2*sin(2*pi) in your table.
 
The period is now [-\pi /2, \pi /2]. The "domain" is still all real numbers just as it is for sin(x).
 
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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