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I am having trouble trying to show that: $$F\left(\frac{1}{2}+\frac{1}{2}a,\frac{1}{2}-\frac{1}{2}a;\frac{3}{2};\sin^2z\right) = \sin{az}.$$

I have found that the coefficients in the series are given by: $$\frac{\left(\frac{1}{2}+\frac{1}{2}a\right)_n \left(\frac{1}{2}-\frac{1}{2}a\right)_n}{\left(\frac{3}{2}\right)_n\, n!}=(-1)^n \frac{(a^2-1^2)(a^2-3^2)\ldots(a^2-(2n-1)^2)}{(2n+1)!},$$

but have no idea what to do with the $\sin^2z$ term?

Thanks!

user2387654
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1 Answers1

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By comparing the Chebyshev differential equation with the hypergeometric differential equation you may check that:

$$ \phantom{}_{2}F_1\left(\frac{1+a}{2},\frac{1-a}{2};\frac{3}{2};z^2\right)=\frac{\sin(a\arcsin z)}{a z} \tag{1}$$ from which it follows that: $$ \phantom{}_{2}F_1\left(\frac{1+a}{2},\frac{1-a}{2};\frac{3}{2};\sin^2 z\right)=\frac{\sin(a z)}{a\sin z}=\frac{1}{a}\,U_{a-1}(\cos z). \tag{2}$$

Jack D'Aurizio
  • 353,855
  • The presence of $\dfrac{1\pm a}2$ made me think of $\dfrac{1\pm\cos2t}2$ , but a is not necessarily smaller than $1$ in absolute value. – Lucian Oct 28 '15 at 05:25