[tex]\noindent\rule{12cm}{0.4pt}[/tex]
1. Given:
[tex]\sin\bigg(\dfrac{A}{3}\bigg)=\dfrac{3}{5}[/tex]
[tex]\noindent\rule{12cm}{0.4pt}[/tex]
2. We know from trigonometric identity:
[tex]\cos^2\bigg(\dfrac{A}{3}\bigg)=1-\sin^2\bigg(\dfrac{A}{3}\bigg)[/tex]
[tex]\cos^2\bigg(\dfrac{A}{3}\bigg)=1-\bigg(\dfrac{3}{5}\bigg)^2[/tex]
[tex]\cos^2\bigg(\dfrac{A}{3}\bigg)=1-\dfrac{9}{25}[/tex]
[tex]\cos^2\bigg(\dfrac{A}{3}\bigg)=\dfrac{16}{25}[/tex]
[tex]\cos\bigg(\dfrac{A}{3}\bigg)=\dfrac{4}{5}\ \ \ \text{\bigg[Taking the positive value of $\dfrac{A}{3} $ since it is in the 1st}[/tex]
[tex]\text{quadrant]}[/tex]
[tex]\noindent\rule{12cm}{0.4pt}[/tex]
3. We need to find sinA using the sub-multiple angle formula:
[tex]\sin A=3\sin \dfrac{A}{3}-4\sin^3\dfrac{A}{3}[/tex]
[tex]\text{Substituting $\sin \dfrac{A}{3}=\dfrac{3}{5}:$}[/tex]
[tex]\sin A=3\bigg(\dfrac{3}{5}\bigg)-4\bigg(\dfrac{3}{5}\bigg)^3[/tex]
[tex]\sin A=3\times\dfrac{3}{5}-4\times\dfrac{27}{125}[/tex]
[tex]\sin A=\dfrac{9}{5}-\dfrac{108}{125}[/tex]
[tex]\sin A=\dfrac{225}{125}-\dfrac{108}{125}[/tex]
[tex]\boxed{\sin A=\dfrac{117}{125}}[/tex]
proved
