 # Proving the closed form of sin &#x2061;<!-- ⁡ --> 48 &#x2218;<!-- ∘ --> </msup> uznosititr 2022-05-18 Answered
Proving the closed form of $\mathrm{sin}{48}^{\circ }$
According to WA
$\mathrm{sin}{48}^{\circ }=\frac{1}{4}\sqrt{7-\sqrt{5}+\sqrt{6\left(5-\sqrt{5}\right)}}$
What would I need to do in order to manually prove that this is true?
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This may be incredibly tedious to do, but you apply the sine sum formula to:
$\mathrm{sin}\left(48\right)=\mathrm{sin}\left(30+18\right)$
Then, $\mathrm{sin}\left(18\right)$ and $\mathrm{cos}\left(18\right)$ can be computed by using the half angle formulas on $\mathrm{sin}\left(36\right)$ and $\mathrm{cos}\left(36\right)$, and noting that $\mathrm{cos}\left(36\right)=\frac{1}{2}\varphi$, where $\varphi =\frac{1+\sqrt{5}}{2}$, i.e. the golden ratio.