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Find the value of $\sin(\frac{1}{4}\arcsin\frac{\sqrt{63}}{8})$


Let $\sin(\frac{1}{4}\arcsin\frac{\sqrt{63}}{8})=x$

$\arcsin\frac{\sqrt{63}}{8}=4\arcsin x$

$\arcsin\frac{\sqrt{63}}{8}=\arcsin(4x\sqrt{1-x^2})(2x^2-1)$

$\frac{\sqrt{63}}{8}=(4x\sqrt{1-x^2})(2x^2-1)$

$\frac{63}{64}=16x^2(1-x^2)(2x^2-1)^2$

Let $x^2=t$

$\frac{63}{64}=16t(1-t)(2t-1)^2$ $64t^4-128t^3+80t^2-16t+\frac{63}{64}=0$

Now solving this fourth degree equation is getting really difficult,even after using rational roots theorem.
Is there a better method possible?The answer given is $x=\frac{\sqrt2}{4}$.Please help me.

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Let

$$x = \arcsin{\frac{\sqrt{63}}{8}} = \arccos{\frac18} $$

Use succession of half-angle formulae:

$$\sin{\frac{x}{4}} = \sqrt{\frac{1-\cos{(x/2)}}{2}} = \sqrt{\frac{1-\sqrt{\frac{1+\cos{(x)}}{2}} }{2}} $$

In this case, $\cos{x} = 1/8$ so that

$$\sin{\left (\frac14 \arcsin{\frac{\sqrt{63}}{8}} \right )} = \sqrt{\frac{1-\sqrt{\frac{9/8}{2}} }{2}} = \frac1{2 \sqrt{2}}$$

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