This is my attempt:-
Let us consider a right △ABC such that angle A is 15∘ and C is 75∘.
On the line AB, let us assume a point D such that BCBD=1√3 (Without Loss Of Generality). So ∠BDC becomes 30∘ and ∠BCD becomes 60∘. Then ∠DCA becomes equal to ∠BAC, that is 15∘; so CD=DA.
CD will be 2 times BC (angle BDC=30∘; sin30∘). On adding BD and AD we get AB=BC(2+√3)
BC2+AB2=AC2
∴AC=2BC√2+√3
\sin(15^\circ) = \frac{BC}{AC}
\sin(15^\circ) = \frac{1}{\sqrt{2 + \sqrt{3}}}
Rationalising the denominator 2 times we get:-
\sin(15^\circ) = \frac{(4-2\sqrt{3})(\sqrt{2+\sqrt{3}})}{4}
Further simplifying:-
\sin(15^\circ) = \frac{(2-\sqrt{3})(\sqrt{2+\sqrt{3}})}{2}
\sin(15^\circ) = \frac{(\sqrt{2-\sqrt{3}})(\sqrt{2-\sqrt{3}}) (\sqrt{2+\sqrt{3}})}{2}
\sin(15^\circ) = \frac{\sqrt{2-\sqrt{3}}}{2}
Is my answer correct? And is there any other method or way to get the value of \sin(15^\circ) geometrically?
