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According to WA$$\sin48^\circ=\frac{1}{4}\sqrt{7-\sqrt5+\sqrt{6(5-\sqrt5)}}$$

What would I need to do in order to manually prove that this is true? I suspect the use of limits, but I don't know where to start.

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    $\begingroup$ It can be denested, and rewritten as $\dfrac{\sqrt3~\big(\sqrt5-1\big)+\sqrt{2\sqrt5~\big(\sqrt5+1\big)}}8$ $\endgroup$ – Lucian Nov 11 '15 at 19:00
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This may be incredibly tedious to do, but you apply the sine sum formula to:

$$ \sin(48)=\sin(30+18)$$ Then, $\sin(18)$ and $\cos(18)$ can be computed by using the half angle formulas on $\sin(36)$ and $\cos(36)$, and noting that $\cos(36)=\frac{1}{2}\phi$, where $\phi=\frac{1+\sqrt{5}}{2}$, i.e. the golden ratio.

This latter fact is proven here and here

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    $\begingroup$ Slightly easier is to note that $\sin^2(48^\circ) = \frac{1}{2}(1 + \sin(36^\circ - 30^\circ))$ $\endgroup$ – Michael Biro Nov 11 '15 at 18:41
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@ASKASK suggested one method that depends on knowing the radical values for $\sin 18^\circ$ and $\cos 18^\circ.$ He also outlined how to obtain these two values. In the event that anyone is interested, below are three ways of going about this.

Finding $\cos 18^\circ$ and $\sin18^\circ$ (Method 1)

Letting $A = 18^\circ,$ we have $90^\circ - 3A = 2A.$ Therefore, $\sin(90^\circ - 3A) = \sin 2A,$ and since $\sin(90^\circ - 3A) = \cos 3A,$ we get $\cos 3A = \sin 2A.$ Now use the triple angle identity for cosine and the double angle identity for sine to get $4 {\cos}^{3} A - 3\cos A = 2 \sin A \cos A.$ Next, divide both sides by $\cos A = \cos 18^\circ \neq 0$ to get $4{\cos}^{2} A - 3 = 2\sin A.$ Replacing ${\cos}^{2} A$ with $1 - {\sin}^{2} A$ in the last equation gives $4(1 - {\sin}^{2} A ) - 3 = 2\sin A.$ We now have a quadratic equation in $\sin A$ whose positive solution is easily found. After getting the solution to this quadratic equation, you can use the identity ${\cos}^{2} A + {\sin}^{2} A = 1$ for $A = 18^\circ$ to obtain a radical value for $\cos 18^\circ.$

Finding $\cos 18^\circ$ and $\sin18^\circ$ (Method 2)

Substituting $A = 18^\circ$ into the identity $\cos 5A = 16{\cos}^{5} A - 20{\cos}^{3} A + 5\cos A$ leads to the equation $0 = 16x^5 - 20x^3 + 5x,$ whose solutions therefore must include $\cos 18^\circ.$ To solve this equation, divide both sides by $x$ (note that the value of $x$ we're looking for is not zero) to get $16x^4 - 20x^2 + 5 = 0.$ This last equation is quadratic in $x^2,$ and hence straightforward to solve. This gives a radical value for $\cos 18^\circ.$ To obtain a radical value for $\sin 18^\circ,$ use the identity ${\cos}^{2} A + {\sin}^{2} A = 1$ for $A = 18^\circ.$

Finding $\cos 18^\circ$ and $\sin18^\circ$ (Method 3)

Using the double angle identity (twice) for sine, we get $$\sin 36^\circ \; = \; 2\sin 18^\circ \cos 18^\circ$$ $$\sin 72^\circ \; = \; 2\sin 36^\circ \cos 36^\circ$$ Multiply these two equations, then divide both sides of the resulting product by $\sin 36^\circ,$ and then divide the resulting equation by $\sin 72^\circ = \cos 18^\circ:$ $$\sin 36^\circ \sin 72^\circ \; = \; (2\sin 18^\circ \cos 18^\circ)(2\sin 36^\circ \cos 36^\circ)$$ $$\sin 72^\circ \; = \; 4\sin 18^\circ \cos 18^\circ \cos 36^\circ$$ $$1 \; = \; 4\sin 18^\circ \cos 36^\circ \;\;\;\;\;\;\;\; (1)$$ Next, use the identity $\sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$ with $A = 54^\circ$ and $B = 18^\circ,$ then use equation (1) (the last displayed equation) to replace $2\cos 36^\circ \sin 18^\circ$ with $\frac{1}{2},$ then use $\sin 54^\circ = \cos 36^\circ:$ $$\sin 54^\circ - \sin 18^\circ \; = \; 2\cos 36^\circ \sin 18^\circ$$ $$\sin 54^\circ - \sin 18^\circ \; = \; \frac{1}{2}$$ $$\cos 36^\circ - \sin 18^\circ \; = \; \frac{1}{2} \;\;\;\;\;\;\;\; (2)$$ Now square both sides of equation (2) and add the resulting equation to equation (1), then expand and factor the left side, then take the positive square root of both sides: $$(\cos 36^\circ - \sin 18^\circ)^2 \; + \; 4\sin 18^\circ \cos 36^\circ \; = \; \frac{1}{4} + 1$$ $$(\cos 36^\circ + \sin 18^\circ)^2 \; = \; \frac{5}{4}$$ $$\cos 36^\circ + \sin 18^\circ \; = \; \frac{1}{2}\sqrt{5} \;\;\;\;\;\;\;\; (3)$$ Adding and subtracting equations (2) and (3) gives $$\cos 36^\circ \; = \; \frac{1}{4}(\sqrt{5} + 1)$$ $$\sin 18^\circ \; = \; \frac{1}{4}(\sqrt{5} - 1)$$ Now use ${\cos}^{2} A + {\sin}^{2} A = 1$ for $A = 18^\circ$ and for $A = 36^\circ$ to get $$\cos 18^\circ \; = \; \frac{1}{4}\sqrt{10 + 2\sqrt{5}}$$ $$\sin 36^\circ \; = \; \frac{1}{4}\sqrt{10 - 2\sqrt{5}}$$

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