Convergence of Sin(x) Series: Understanding the Pattern and Proving Convergence

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Homework Statement



Why does sin(x) - 1/2sin^2(x) + 1/4sin^3(x) - 1/8 sin^4(x) + ... = 2sin(x)/2+sin(x)

How do you know for certain the series converges for all real values of x?

Homework Equations





The Attempt at a Solution



Have no clue where to even start...

Thanks for any help...
 
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You should use the standard series identity

\sum_{n=0}^{\infty} y^n = \frac{1}{1-y}

The series converges for |y|<1. With an appropriate substitution you can make this series appear in your problem.
 
You might want to consider, separately, what happens when x= \pi/2or x= -\pi/2.
 
To make dhris's hint more obvious, just show that

\sum_{n=0}^{\infty} (\frac{- \sin x}{2})^n = \frac{2}{2+ \sin x}
then multiply through out by 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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