jika tan x = m dengan x adalah sudut lancip,tentukan nilai dari sin 3x

[tex]
\text{Karena }\tan{x}=m\text{, maka }\\
\sin{x}=\frac{m}{\sqrt{m^2+1}}\\
\cos{x}=\frac{1}{\sqrt{m^2+1}}\\
\text{Selanjutnya, }\\
\sin{3x}=\sin{(2x+x)}\\
=\sin{2x}\cos{x}+\cos{2x}\sin{x}\\
=2\sin{x}\cos{x}\cdot\cos{x}+(\cos^2{x}-\sin^2{x})\sin{x}\\
=2\cdot\frac{m}{\sqrt{m^2+1}}\cdot\frac{1}{\sqrt{m^2+1}}\cdot\frac{1}{\sqrt{m^2+1}}+\left(\left(\frac{1}{\sqrt{m^2+1}}\right)^2-\left(\frac{m}{\sqrt{m^2+1}}\right)^2\right)\cdot\frac{m}{\sqrt{m^2+1}}\\
=[/tex]