Please help me understand where this expansion came from

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Homework Statement



(asinz + bsin3z)^3 = (3a/4)(a^2 – ab + 2b^2)sinz – (1/4)(a^3 – 6a^2 – 3b^3)sin3z...

Homework Equations





The Attempt at a Solution



How is this so !? What is being used here (taylor/binomial...?)
 
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Nothing that exotic. It's just algebra and the application of trig identities to get everything in terms of just sin z and sin 3z.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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